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Hints & Solutions
Level - 1
(RECTANGULAR CARTESIAN
CO-ORDNATES)
1.
We have,

We have,
4.
x2
a2

4 



2
2
1 
(3, – 4)   3  (4) , tan   3  





1  4  
 5,  tan  3  
 


and
(–3, 4) 


2
2
1  3  
 ( 3)  (4) , tan   4  





y2
b2
1
b2x2 + a2y2 = a2b2
b2r2cos2 + a2r2sin2 = a2b2
r 2 cos 2 
a
2
cos 2 
a
2


r 2 sin 2 
b2
sin 2 
b
2
1
1
 2
r
We have,
2x2 + 3xy + 2y2 = 1

2r2cos2 + 3r2sincos + 2r2sin2 =
5.

1  4  
 5,   tan   
 3 

1
2.
We have,
r2 = a2 cos 2

r2 = a2 (cos2q – sin2)


3.

r2(2cos2 + 3sincos + 2sin2 = 1

3


r 2  2  sin 2   1
2


 2 y2 
2 x
a
r =  2  2 
r 
r
2
r4 = a2(x2 – y2)
(x2 + y2)2 = a2(x2 – y2)
We have,
r = 2a cos 2
x
 

r = 2a  
r


r2 = 2ax
(x2 + y2 = 2ax)
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1.16 Theory and Exercise Book
SECTION - A : DISTANCE FORMULA
1.
The circumcentre of the triangle with vertices
(0, 0), (3, 0) and (0, 4) is
(A) (1, 1)
(B) (2, 3/2)
(C) (3/2, 2)
(D) none of these
2.
The mid points of the sides of a triangle are (5, 0),
(5, 12) and (0, 12), then orthocentre of this triangle is
(A) (0, 0)
(B) (0, 24)
 13
3.

(D)  , 8 
3 
(C) (10, 0)


SECTION - D : LOCUS
9.
If A(cos, sin), B(sin, – cos), C(1, 2) are the
vertices of a ABC, then as  varies, the locus of
its centroid is
2
2
(A) x + y – 2x – 4y+3=0
2 2
(B) x +y – 2x–4y+1 = 0
2
2
(C) 3(x + y ) – 2x – 4y+1=0
(D) none of these
10.
A stick of length 10 units rests against the floor and
a wall of a room. If the stick begins to slide on the
floor then the locus of its middle point is
2
2
2
2
(A) x + y = 2.5
(B) x + y = 25
2
2
(C) x + y = 100
(D) none
11.
If P(1, 0) ; Q(–1, 0) & R(2, 0) are three give points,
then the locus of the points S satisfying the relation,
2
2
2
SQ + SR = 2 SP is
(A) A straight line parallel to x-axis
(B) A circle passing through the origin
(C) A circle with the centre at the origin
(D) A straight line parallel to y-axis
8
The points  0,  , (1,3) and (82,30) are vertices of
3

(A) an obtuse angled triangle
(B) an acute angled triangle
(C) a right angled triangle
(D) none of these
SECTION - B : SECTION FORMULA
4.
The ratio in which the line joining the points (3, –4)
and (–5, 6) is divided by x-axis
(A) 2 : 3
(B) 6 : 4
(C) 3 : 2
(D) none of these
5.
If a vertex of a triangle is (1, 1) and the mid points
of two sides through this vertex are
(–1, 2) and (3, 2), then the centroid of the triangle is


7
(B)  3 , 3 


 7
(C)  1, 3 


1 7
(D)  3 , 3 


SECTION - C
AREA OF TRIANGLE & CONDITION
OF COLLINEARITY
6.
8.
1
ab
2
–1
The points with the co-ordinates (2a, 3a), (3b, 2b)
& (c, c) are collinear
(A) for no value of a, b, c (B) for all values of a, b, c
(C) If a,
c
2
, b are in H.P.(D) if a, c, b are in H.P..
5
5
3
to the x-axis is
5
(A) 5y – 3x + 15 = 0
(C) 3y – 5x + 15 = 0
(B) 5y – 3x = 15
(D) none of these
13.
The equation of a straight line which passes through
the point (–3, 5) such that the portion of it between
the axes is divided by the point in the ratio
5 : 3 (reckoning from x-axis) will be
(A) x + y – 2 = 0
(B) 2x + y + 1 = 0
(C) x + 2y – 7 = 0
(D) x – y + 8 = 0
14.
The equation of perpendicular bisector of the line
segment joining the points (1, 2) and (–2, 0) is
(A) 5x + 2y = 1
(B) 4x + 6y = 1
(C) 6x + 4y = 1
(D) none of these
15.
The number of possible straight lines, passing through
(2, 3) and forming a triangle with coordinate axes,
whose area is 12 sq. units, is
(A) one
(B) two
(C) three
(D) four
16.
A line is perpendicular to 3x + y = 3 and passes
through a point (2, 2). Its y intercept is
(A) 2/3
(B) 1/3
(C) 1
(D) 4/3
(D) ab
The point A divides the join of the points (–5, 1) and
(3, 5) in the ratio k : 1 and coordinates of points B
and C are (1, 5) and (7, –2) respectively. If the
area of ABC be 2 units, then k equals
(A) 7, 9
(B) 6, 7
(C) 7, 31/9
(D) 9, 31/9
The equation of the line cutting an intercept of 3
on negative y-axis and inclined at an angle
tan
Area of a triangle whose vertices are (a cos ,
b sin), (–a sin , b cos ) and (–a cos , –b sin )
is
(A) ab sin  cos 
(B) a cos  sin 
(C)
7.
12.
 -1 7 
(A)  1, 
3

SECTION - E
VARIOUS FORMS OF STRAIGHT LINE
394-Rajeev Gandhi Nagar, Kota
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Straight Line 3.16
17.
The equation of the line passing through the point
(c, d) and parallel to the line ax + by + c = 0 is
(A) a(x + c) + b(y + d) = 0
(B) a(x + c) – b(y + d) = 0
(C) a(x – c) + b(y – d) = 0
(D) none of these
25.
The foot of the perpendicular drawn from the point
(7, 8) to the line 2x + 3y – 4 = 0 is  23 2 
(A)  13 , 13 


 23
18.
A straight line through the point A(3, 4) is such that
its intercept between the axes is bisected at A. Its
equation is (A) 3x – 4y + 7 = 0
(B) 4x + 3y = 24
(C) 3x + 4y = 25
(D) x + y = 7
SECTION - F
ANGLE BETWEEN TWO LINES
19.
The angle between the lines y – x + 5 = 0 and
3 x  y  7  0 is
(A) 15º
(C) 45º
If the line passing through the points (4, 3) and (2, )
is perpendicular to the line y = 2x + 3, then  is equal to
(A) 4
(B) –4
(C) 1
(D) –1
21.
The equation of two equal sides of an isosceles
triangle are 7x – y + 3 = 0 and x + y–3 = 0 and its
third side is passes through the point
(1, –10). The equation of the third side is
(A) x – 3y – 31 = 0 but not 3x + y + 7 = 0
(B) neither 3x + y + 7 = 0 nor x – 3y – 31 = 0
(C) 3x = y + 7 = 0 or x – 3y – 31 = 0
(D) 3x + y + 7 = 0 but not x – 3y – 31 = 0
Triangle formed by lines x + y = 0, 3x + y = 4 and
x + 3y = 4 is (A) equilateral
(B) right angled
(C) isoscales
(D) None of these
SECTION - G
DISTANCEOF APOINT FROM A LINE
23.
Coordinates of a point which is at 3 distance from
point (1, –3) of line 2x + 3y + 7 = 0 is

9
6 
,3
(A)  1 

13
13 


9

13
(C) 1 
24.
, 3
(A)

9
6 
, 3
1 

13
13 

a
a 2  b2
(B)
2
a b
2

2 23 
(D)   13 , 13 


The coordinates of the point Q symmetric to
the point P(–5, 13) with respect to the line
2x – 3y – 3 = 0 are (A) (11, –11)
(B) (5, –13)
(C) (7, –9)
(D) (6, –3)
28.
The line 3x + 2y = 6 will divide the quadrilateral
formed y the lines x + y = 5, y – 2x = 8, 3y + 2x = 0
& 4y – x = 0 in
(A) two quadrilaterals
(B) one pentagon and one triangle
(C) two triangles
(D) none of these
29.
If the point (a, 2) lies between the lines x–y–1=0
and 2(x – y) – 5 = 0, then the set of values of a is
(A) (–, 3)  (9/2, ) (B) (3, 9/2)
(C) (–, 3)
(D) (9/2, )
30.
The area of triangle formed by the lines x + y – 3 = 0,
x – 3y + 9 = 0 and 3x – 2y + 1 = 0
(A)
16
sq. units
7
(C) 4 sq. units
(B)
10
sq. units
7
(D) 9 sq. units
31.
The co-ordinates of foot of the perpendicular drawn
on line 3x – 4y – 5 = 0 from the point (0, 5) is
(A) (1, 3)
(B) (2, 3)
(C) (3, 2)
(D) (3, 1)
32.
The co-ordinates of the point of reflection of the
origin (0, 0) in the line 4x – 2y – 5 = 0 is
(A) (1, –2)
(B) (2, –1)
a 2  b2
ab
(C)

The position of the point (8, –9) with respect to the
lines 2x + 3y – 4 = 0 and 6x + 9y + 8 = 0 is
(A) point lies on the same side of the lines
(B) point lies on one of the lines
(C) point lies on the different sides of the line
(D) none of these
The length of perpendicular from the origin on the
line x/a + y/b = 1 is b

27.

9
6 
,  3
(B) 1 

13
13 

6 
 (D)
13 
23 
SECTION - H
POSITION OF A POINT W.R.TO A LINE
(B) 60º
(D) 75º
20.
22.
2
(C)   13 , 13 


26.

(B) 13, 13 
(D) None of these
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4 2
(C)  5 , 5 


(D) (2, 5)
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1.18 Theory and Exercise Book
SECTION - I
CONDITION OF CONCURRENCY
33.
39.
Keeping coordinate axes parallel, the origin is shifted
to a point (1, –2), then transformed equation of x2 +
y2 = 2 is (A) x2 + y2 + 2x – 4y + 3 = 0
(B) x2 + y2 + 2x + 4y + 3 = 0
(C) x2 + y2 – 2x – 4y + 3 = 0
(D) x2 + y2 – 2x + 4y + 3 = 0
40.
To remove xy term from the second degree equation 5x2 + 8xy + 5y2 + 3x + 2y + 5 = 0, the coordinates axes are rotated through an angle , then 
equals (A) /2
(B) /4
(C) 3/8
(D) /8
41.
The point (4, 1) undergoes two successive transformations (i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the
positive direction of x axis
The final position of the point is given by the
coordinates (A) (4, 3)
(B) (3, 4)
(C) (7/2, 7/2)
(D) (1, 4)
2
If the lines
xsin A + ysinA + 1 = 0
2
xsin B + ysinB + 1 = 0
2
xsin C + ysinC + 1 = 0
are concurrent where A, B, C are angles of triangle
then ABC must be
(A) equilateral
(B) isosceles
(C) right angle
(D) no such triangle exist
SECTION - J
FAMILY OF STRAIGHT LINE
34.
The line (p + 2q)x + (p – 3q)y = p – q for different
values of p and q passes through a fixed point whose
co-ordinates are
3 5
2 2
(A)  , 
2 2
(B)  , 
5 5
3 3
 2 3
(C)  , 
5 5
35.
36.
(D)  , 
5 5
Given the family of lines, a(3x+4y+6) + b(x+y+2)=0.
The line of the family situated at the greatest
distance from the point P(2, 3) has equation
(A) 4x + 3y + 8 = 0
(B) 5x + 3y + 10 = 0
(C) 15x + 8y + 30 = 0 (D) None of these
The base BC of a triangle ABC is bisected at the
point (p, q) and the equation to the side AB & AC
are px + qy = 1 & qx + py = 1. The equation of the
median through A is
(A) (p–2q)x+(q–2p)y+1=0
(B) (p + q) x+y – 2=0
2
2
(C) (2pq – 1)(px + qy – 1)=(p + q – 1)(qx + py – 1)
(D) None of these
42.
43.
SECTION - K : SHIFTING OF ORIGIN
37.
38.
Without changing the direction of coordinates axes,
to which point origin should be transferred so that
the equationx2 + y2 –4x + 6y –7 = 0 is changed to
an equation which contains no term of first degree(A) (3,2)
(B) (2, –3)
(C) (–2, 3)
(D) None of these
Reflecting the point (2, –1) about y -axis,coordinate
axes are rotated at 45º angle in negative direction
without shifting the origin. The new coordinates of
the points are 
(A)  


(C)  

1
2
3
2
,
,
3 

2
1 

2
 1
(B) 
 2
,
3 

2
SECTION - L : ANGLE BISECTOR
The equation of the bisector of the angle
between the lines 3x – 4y + 7 = 0 and
12x – 5y – 8 = 0 is (A) 99x –77y + 51 = 0, 21x + 27y – 131 = 0
(B) 99x –77y + 51 = 0, 21x + 27y + 131 = 0
(C) 99x –77y + 131 = 0, 21x + 27y – 51 = 0
(D) None of these
The equation of the bisector of the acute angle
between the lines 3x – 4y + 7 = 0 and 12x + 5y – 2
= 0 is (A) 11x – 3y – 9 = 0
(B) 11x – 3y + 9 = 0
(C) 21x + 77y – 101 = 0
(D) None of these
SECTION - M
PAIR OF STRAIGHT LINES
44.
The image of the pair of lines represented by
2
2
ax + 2h xy + by = 0 by the line mirror y = 0 is
2
2
(A) ax – 2hxy + by = 0
2
2
(B) bx – 2h xy + ay = 0
2
2
(C) bx + 2h xy + ay = 0
2
2
(D) ax – 2h xy – by = 0
45.
Area of the triangle formed by the line x+y=3 and
the angle bisector of the pairs of st. lines
2 2
x –y +2y = 1 is
(A) 2 sq. unit
(B) 4 sq. unit
(C) 6 sq. unit
(D) 8 sq. unit
(D) None of these
394-Rajeev Gandhi Nagar, Kota
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Straight Line 3.18
DISTANCE FORMULA
1.
2.
The circumcentre of the triangle formed by the lines,
xy + 2x + 2y + 4 = 0 and x + y + 2 = 0 is
(A) (–1, –1)
(B) (–2, –2)
(C) (0, 0)
(D) (–1, –2)
7.
The area enclosed by 2 | x | + 3| y |  6 is
(A) 3 sq. units
(B) 4 sq. units
(C) 12 sq. units
(D) 24 sq. units
8.
The points (- a, -b),(0, 0), (a, b) and (a2, ab)are(A) collinear
(B) concyclic
(C) vertices of a rectangle
(D) vertices of a parallelogram
9.
A rod PQ of length 2a slides with its ends on the
axes then locus of circumcentre of OPQ is(A) x2 + y2 = 2a2
(B) x2 + y2 = 4a2
2
2
2
(C) x + y = 3a
(D) x2 + y2 = a2
10.
A(1,0) and B (–1, 0) are two points and Q is a point
which satisfies the relation AQ – BQ=± 1. The locus
of Q is (A) 12x2 – 4y2 = 3
(B) 12x2 – 4y2 + 3 = 0
(C) 12x2 + 4y2 = 3
(D) 12x2 + 4y2 + 3 =0
11.
Locus of centroid of the triangle whose
vertices are (a cos t, a sin t), (b sint,–b cost) and
(1, 0), where t is a parameter, is (A) (3x + 1)2 + (3y)2 = a2 – b2
(B) (3x – 1)2 + (3y)2 = a2 – b2
(C) (3x – 1)2 + (3y)2 = a2 + b2
(D) (3x + 1)2 + (3y)2 = a2 + b2
12.
Let A (2, – 3) and B(–2, 1) be vertices of a triangle
ABC. If the centroid of this triangle moves on the
line 2x + 3y = 1, then the locus of the vertex C is
the line
(A) 2x + 3y = 9
(B) 2x – 3y = 7
(C) 3x + 2y = 5
(D) 3x – 2y = 3
If the equation of the locus of a point
equidistant from the points (a1,b1) and (a2, b2) is
(a1- a2 )x + (b1 - b2 )y + c = 0, then the value of c is(A)
LOCUS
a12  b12  a22  b22
(B) a12 - a22 + b12 - b22
(C)
1 2
(a + a22 + b12 + b22 )
2 1
(D)
1 2
(a + b22 - a12 - b12 )
2 2
SECTION FORMULA
3.
4.
The centroid of a triangle is (2, 3) and two of its
vertices are (5, 6) and (- 1, 4). The third vertex of
the triangle is(A) (2, 1)
(B) (2, - 1)
(C) (1, 2)
(D) (1, - 2)
The co-ordinates of the vertices P, Q, R & S of
square PQRS inscribed in the triangle ABC with
vertices A(0, 0), B(3, 0) & C(2, 1) given that two of
its vertices P, Q are on the side AB are respectively
1  3  3 1 1 1
(A)  , 0  ,  , 0  ,  ,  &  , 
 4  8  8 8  4 8
1
 3
 3 1
1 1
(B)  , 0  ,  , 0  ,  ,  &  , 
 2   4   4 4  2 4
3

 3 1
 

 1


VARIOUS FORMS OF STRAIGHT LINE
(C) (1, 0)  , 0  ,  ,  & 1, 
2
2 2
2
3
 9
 9 3
13.
Points A & B are in the first quadrant ; point 'O' is
the origin. If the slope of OA is 1, slope of OB is 7
and OA = OB, then the slope of AB is
(A) –1/5
(B) –1/4
(C) 1/3
(D) –1/2
14.
A(x1, y1), B(x2, y2) and C(x3, y3) are three noncollinear points in cartesian plane. Number of
parallelograms that can be drawn with these three
points as vertices are
(A) one
(B) two
(C) three (D) four
15.
If line y – x + 2 = 0 is shifted parallel to itself towards
the positive direction of the x-axis by a perpendicular
distance of 3 2 units, then the equation of the new
line is
(A) y = x – 4
(B) y = x + 1
(C) y = x – (2 + 3 2 ) (D) y = x – 8
 3 3
(D)  , 0  ,  , 0  ,  ,  &  , 
 2   4   4 4  2 4
5.
Three vertices of triangle ABC are A(–1, 11),
B(–9, –8) and C(15, –2). The equation of angle
bisector of angle A is
(A) 4x – y=7
(B) 4x + y=7
(C) x + 4y=7
(D) x–4y=7
AREA OF TRIANGLE & CONDITION
OF COLLINEARITY
6.
Area of the rhombus bounded by the four lines,
ax ± by ± c = 0 is
(A)
c2
2ab
(B)
2c 2
| ab |
(C)
4c 2
ab
(D)
ab
4c 2
www.motioniitjee.com
26
1.20 Theory and Exercise Book
16.
If the axes are rotated through an angle of 30º in
the anti-clockwise direction, the coordinates of point
22.
Find the equation to the sides of an isosceles rightangled triangled, the equation of whose hypotenuse
is 3x + 4y = 4 and the opposite vertex is the point
(2, 2).
(A) 7y – x – 12 = 0 and 7x + y = 16.
(B) 7y + x – 12 = 0 and 7x + y = 16.
(C) 7y + x – 12 = 0 and 7x – y = 16.
(D) 7y – x + 12 = 0 and 7x – y = 16.
23.
Let the algebraic sum of the perpendicular distances
from the point (3, 0), (0, 3) & (2, 2) to a variable
straight line be zero, then the line passes through a
fixed point whose co-ordinates are
(A) (3, 2)
(B) (2, 3)
(4, – 2 3 ) with respect to new axes are
17.
18.
(A) (2, 3 )
(B) ( 3 , –5)
(C) (2, 3)
(D) ( 3 , 2)
A ray of light passing through the point A(1, 2) is
reflected at a point B on the x-axis and then passes
through (5, 3). Then the equation of AB is
(A) 5x + 4y = 13
(B) 5x – 4y = –3
(C) 4x + 5y = 14
(D) 4x – 5y = –6
DISTANCEOF APOINT FROM A LINE
A square of side a lies above the x-axis and has
one vertex at the origin. The side passing through
the origin makes an angle  (0<<

) with the posi4
3 3
(C)  5 , 5 


24.
tive direction of x-axis. The equation of its diagonal
not passing through the origin is (A) y (cos + sin) + x (cos – sin) = a
(B) y (cos – sin) – x (sin – cos) = a
(C) y (cos + sin) + x (sin – cos) = a
(D) y (cos + sin) + x (sin + cos) = a
19.
20.
y
x
x
y
+
= –1 and
+
= –1
3
2
-2
1
(B)
y
x
x
y
–
= –1 and
+
= –1
3
2
-2
1
(C)
y
x
x
y
+
= 1 and
+
=1
3
2
2
1
(D)
y
x
x
y
–
= 1 and
+
=1
3
2
-2
1
The line parallel to the x-axis and passing through
the intersection of the lines ax + 2by + 3b = 0 and
bx – 2ay – 3a = 0, where (a, b)  (0, 0) is (A) below the x-axis at a distance of 3/2 from it
(B) below the x-axis at a distance of 2/3 from it
(C) above the x-axis at a distance of 3/2 from it
(D) above the x-axis at a distance of 2/3 from it
POSITION OF A POINT W.R.TO A LINE
If one diagonal of a square is along the line x = 2y
and one of its vertex is (3, 0), then its sides through
this vertex are given by the equations
(A) y – 3x + 9 = 0, x – 3y – 3 = 0
(B) y – 3x + 9 = 0, x – 3y – 3 = 0
(C) y + 3x – 9 = 0, x + 3y – 3 = 0
(D) y – 3x + 9 = 0, x + 3y – 3 = 0
26.
A triangle is formed by the lines 2x – 3y – 6 = 0 ;
3x – y + 3 = 0 and 3x + 4y – 12 = 0. If the points
P(, 0) and Q(0, ) always lie on or inside the ABC,
then
(A)  [–1, 2] &  [–2, 3]
(B)  [–1, 3] &  [–2, 4]
(C)  [–2, 4] &  [–3, 4]
(D)  [–1, 3] &  [–2, 3]
27.
If (a, a2) falls inside the angle made by the lines
y=
If origin and (3, 2) are contained in the same angle
of the lines 2x + y – a = 0, x – 3y + a = 0, then 'a'
must lie in the interval
(A) (–, 0)  (8, )
(B) (–, 0)  (3, )
(C) (0, 3)
(D) (3, 8)
x
, x > 0 and y = 3x, x > 0, then a belongs to
2
1
1

(C)  3,  


(B)  ,3 
2 
(A) (3, )
2
1


(D)  0, 
2


CONDITION OF CONCURRENCY
ANGLE BETWEEN TWO LINES
21.
Three lines x + 2y + 3 = 0, x + 2y – 7 = 0 and
2x – y – 4 = 0 form 3 sides of two squares. Find the
equation of remaining sides of these squares.
(A) 2x – y + 6 = 0, 2x – y – 14 = 0
(B) 2x + y – 6 = 0, 2x + y – 14 = 0
(C) 2x + y + 6 = 0, 2x – y + 14 = 0
(D) 2x – y – 6 = 0, 2x + y + 14 = 0
25.
The equation of the straight line passing through
the point (4, 3) and making intercepts on the
coordinate axes whose sum is – 1 is (A)
5 5
(D)  , 
3 3
28.
Lines, L1 : x + 3 y  2 , and L2 : ax + by = 1, meet
at P and enclose an angle of 45º between them.
Line L3 : y = 3 x , also passes through P then
2
2
2
2
(A) a + b = 1
(B) a + b = 2
2
2
2
2
(C) a + b = 3
(D) a + b = 4
394-Rajeev Gandhi Nagar, Kota
27
Straight Line 3.20
29.
If the lines ax + y + 1 = 0, x + by + 1 = 0 &
x + y + c = 0 where a, b & c are distinct real numbers
different from 1 are concurrent, then the value of
36.
MIXED PROBLEM
a fixed point that point is (A) (–1, 2)
(B) (–1, –2)
1

(D) 1,  
2

38.
The line x + y = p meets the axis of x and y at A and
B respectively. A triangle APQ is inscribed in the
triangle OAB, O being the origin, with right angle
at Q, P and Q lie respectively on OB and AB. If the
th
area of the triangle APQ is 3/8 of the area of the
SHIFTING OF ORIGION
31.
Keeping the origin constant axes are rotated at an
angle 30º in anticlockwise direction then new
coordinate of (2, 1) with respect to old axes is
 2 3 3 
,

2 
 2
(B) 
 2 3 1 2  3 
,

2 
 2
(D) none of these
(A) 
(C) 
32.
AQ
triangle OAB, then BQ is equal to
(A) 2
(C) 1/3
 2 3  1 2  3 
,

2
 2

39.
The line PQ whose equation is x – y = 2 cuts the
x-axis at P and Q is (4, 2). The line PQ is rotated
about P through 45º in the anticlockwise direction.
The equation of the line PQ in the new position is
(A) y   2
(C) x = 2


1
3

3, - 3 
2
2



1
3

3,  3 
2
2

1
The equation 2x + 4xy – py + 4x + qy + 1 = 0
will represent two mutually perpendicular straight
lines, if
(A) p = 1 and q = 2 or 6 (B) p = –2 and q = –2 or 8
(C) p = 2 and q = 0 or 8 (D) p = 2 and q = 0 or 6
34.
The line x + 3y – 2 = 0 bisects the angle between a
pair of straight lines of which one has equation
x – 7y + 5 = 0. The equation of the other line is
(A) 3x + 3y – 1 = 0
(B) x – 3y + 2 = 0
(C) 5x + 5y – 3 = 0
(D) none
35.
The pair of straight lines x – 4xy + y = 0 together
3 1


3, 
3
(C)  4 
6
2 3 

2
2
Let A  (3, 2) and B  (5, 1). ABP is an equilateral
triangle is constructed on the side of AB remote
from the origin then the orthocentre of triangle ABP
is
(B)  4 
PAIR OF STRAIGHT LINES
33.
(B) 2/3
(D) 3
(A)  4 -
(B) y = 2
(D) x = –2
2
(D) none
The co-ordinates of a point P on the line 2x–y+5=0
such that |PA – PB| is maximum where A is (4, –2)
and B is (2, –4) will be
(A) (11, 27)
(B) (–11, –17)
(C) (–11, 17)
(D) (0, 5)
x
y
1
+ + = 0 always passes through
a
b
c
(C) (1, –2)
5
37.
If non-zero numbers a,b,c are in H.P., then the
straight line
(B)
5
(C) 2 5
(B) 3
(D) 1
FAMILY OF STRAIGHT LINE
30.
1
(A)
1
1
1
+
+
equals
1 a
1 b
1 c
(A) 4
(C) 2
Distance between two lines represented by the line
2
2
pair, x – 4xy + 4y + x – 2y – 6 = 0 is


(D)  4 
40.
2
with the line x + y + 4 6 = 0 form a triangle which
is
(A) right angled but not isosceles
(B) right isosceles
(C) scalene
(D) equilateral
www.motioniitjee.com
1
3 1 
3, 
3
6
2 3 
The point (4, 1) undergoes the following three
transformations successively
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the
positive direction of x-axis
(iii) Rotation through an angle /4 about the origin
in the counter clockwise direction.
The final position of the points is given by the
coordinates
1 
 7
,
2
 2
(A) 

(C)  

1
2
,
7 

2
1 
 7
,

2
 2
(B) 
(D) none of these
28
1.22 Theory and Exercise Book
6.
SECTION FORMULA
1.
If one vertex of an equilateral triangle of side 'a'
lies at the origin and the other lies on the line
x – 3 y = 0 then the co-ordiantes of the third
vertex are
 3a
(B)  2 ,  2 


(C) (0, –a)
(D)   2 , 2 



2.
DISTANCE OF A POINT FROM A LINE
a
(A) (0, a)
The equation of straight line which is equidistant
from the points
A(2, –2), B(b, 1), C(–3, 4) can be
(A) 2x + 6y – 5 = 0
(B) 12x+10y–43=0
(C) 6x–8y–11=0
(D) 6x–8y+11=0
7.
Let there are three points A (0, 4/3) B(–1, 0) and
C(1, 0) in x – y plane. The distance from a variable
point P to the line BC is the geometic mean of the
distances from this point to lines AB and AC then
locus of P can be
(A) A pair of straight lines
(B) Circle
(C) Ellipse
(D) Hyperbola
8.
If
3 a a
If one diagonal of a square is the portion of the line
x y
  1 intercepted by the axes, then the extremia b
ties of the other diagonal of the square are
 ab ab 
,
(A) 

2 
 2
 ab a b
,
(B) 

2 
 2
 ab ba 
,
(C) 

2 
 2
 ab ba 
,
(D) 

2 
 2
x
y
x
y
+
= 1 and +
= 1 and the lengths of the
a
b
b
a
perpendiuclars drawn from the origin to these lines
are equal in lengths then
AREA OF TRIANGLE & CONDITION
OF COLLINEARITY
3.
(A)
The area of a triangle is 5. Two of its vertices are
(2, 1) & (3, –2). The third vertex lies on y = x + 3.
Find the third vertex.
7
13 
(B)  , 
 2 2
 7 13 
(C)  , 
2 2 
3 3
(D)  , 
2 2
(C)
9.




(C) 3  3 , 4  3

(D) 3 

3
(B) 4  3 , 3  3
3, 4 
–
1
b2
1
b2
=
=
1
c2
1
c2
+
–
1
d2
1
d2
1
1
1
1
+ = +
a
b
c
d
Straight lines 2x + y = 5 and x – 2y = 3 intersect at
the point A. Points B and C are chosen on these
two lines such that AB = AC. Then the equation of
a line BC passing through the point (2, 3) is
(A) 3x – y – 3 = 0
(B) x + 3y – 11 = 0
(C) 3x + y – 9 = 0
(D) x – 3y + 7 = 0
POSITION OF A POINT W.R.T. A LINE
If the point P(0,  ) lies inside or on the
FAMILY OF STRAIGHT LINE
10.
5.
1
a2
+
triangle formed by the lines y = x + 1,
y = –3x + 4 and y = 7x + 17 then the range of 
is [m, M]. Then (m+M)
(A) lies in interval (4,10)(B) is a prime number
(C) is an odd number (D) is a perfect square
A and B are two fixed points whose co-ordinates
are (3, 2) and (5, 4) respectively. The co-ordinates
of a point P if ABP is an equilateral triangle, is/are
(A) 4  3, 3  3
1
a2
(D) none
VARIOUS FORMS OF STRAIGHT LINE
4.
(B)
 -3 3 
(A)  2 , - 2 


x
y
+
= 1 is a line through the intersection of
c
d
If 25a2 + 16b2 – 40ab – c2 = 0, then the family of
straight line 2ax + by + c = 0 is concurrent at
 -5

 -5

(A)  , 4 
2 
(C)  ,-4 
2

5

5

(B)  ,-4 
2 
(D)  ,4 
2 
394-Rajeev Gandhi Nagar, Kota
29
Straight Line 3.22
PAIR OF STRAIGHT LINE
11.
If the equation ax2 – 6xy + y2 + bx + cy+d=0 represents pair of lines whose slopes are m and m2, then
value of a is/are
(A) a = – 8
(B) a = 8
(C) a = 27
(D) a = – 27
12.
The lines joining the origin to the point of intersection
of 3x2 + xy – 4x + 1 = 0 and 2x + y – 1 = 0 are at
right angles for
(A)  = – 4
(B)  = 4
(C)  = 7
(D) no value of 
14.
In the xy plane, the line ‘1 ’ passes through the
point (1, 1) and the line ‘2’ passes through the point
(–1, 1). If the difference of the slopes of the lines is
2. Find the locus of the point of intersection of the
lines 1 and 2.
(A) y = –x2
(B) y = 2 + x2
(C) y = 2 – x2
(D) y = x2
15.
Two consecutive sides of a parallelogram are
4x + 5y = 0 & 7x + 2y = 0. If the equation to one
diagonal is 11x + 7y = 9, then
(A) Equation of other diagonal is x – y = 0
(B) End points of other diagonal are (0, 0) and (1, 1)
(C) Other two sides are 4x + 5y – 9 = 0 & 7x + 2y
+9=0
(D) None of these
16.
The points (1, 3) & (5, 1) are two opposite vertices
of a rectangle. The other two vertices lie on the
line y = 2x + c.
(A) Vertices are (2, 0) & (4, 4)
(B) Value of c is 4
(C) Vertices are (0, 2) & (4, 4)
(D) Value of c is –4
MIXED PROBLEM
13.
The straight lines x + y = 0, 3x + y – 4 = 0 and
x + 3y – 4 = 0 form a triangle which is
(A) isosceles
(B) right angled
(C) obtuse angled
(D) equilateral
www.motioniitjee.com
30
1.24 Theory and Exercise Book
1.
Two vertices of a triangle are (4, –3) & (–2, 5). If
the orthocentre of the triangle is at (1, 2), find the
coordinates of the third vertex.
2.
Line
x y
 = 1 intersects the x and y axes at M and
6 8
N respectively. If the coordinates of the point P lying
inside the triangle OMN (where ‘O’ is origin) are
(a, b) such that the areas of the triangle POM, PON
and PMN are equal. Find
(a) The coordinates of the point P and
(b) The radius of the circle escribed opposite to the
angle N.
3.
4.
The point A divides the join of P(–5, 1) & Q(3, 5) in
the ratio K : 1. Find the two values of K for which
the area of triangle ABC, where B is (1, 5) and C is
(7,–2), is equal to 2 units in magnitude.
9.
The points (–6, 1), (6, 10), (9, 6) and (–3, –3) are
the vertices of a rectangle. If the area of the portion
of this rectangle that lies above the x-axis is a/b,
find the value of (a + b), given a and b are coprime.
10.
A point P is such that its perpendicular distance from
the line y – 2x + 1 = 0 is equal to its distance from
the origin. Find the equation of the locus of the point
P. Prove that the line y = 2x meets the locus in two
points Q and R, such that the origin is the mid point
of QR.
11.
A line through the point P(2, – 3) meets the lines
x – 2y + 7 = 0 and x +3y – 3 = 0 at the points A and
B respectively. If P divides AB externally in the ratio
3 : 2 then find the equation of the line AB.
12.
If the straight line drawn through the point P( 3, 2)
and inclined at an angle
Determine the ratio in which the point P(3, 5) divides
the join of A(1, 3) and B(7, 9). Find the harmonic
conjugate of P w.r.t. A & B.
the line 3 x – 4y + 8 = 0 at Q. Find the length PQ.
13.
5.
6.
7.
A triangle has side lengths 18, 24 and 30. Find the
area of the triangle whose vertices are the incentre,
circumcentre and centroid of the triangle.
locus of the mid point of AB is the curve 2xy (a + b)
= ab(x + y)
14.
( 3 + 1) y – 6 = 0, BE lies along the line (1 –
8.
If (x1 – x2)2 + (y1 – y2)2 = a2
(x2 – x3)2 + (y2 – y3)2 = b2
and (x3 – x1)2 + (y3 – y1)2 = c2
x1
then  x2
x3
y1 1
y2 1
15.
= (a + b + c)(b + c – a)(c+ a – b) (a + b – c). Find the
value of .
The line 3x + 2y = 24 meets the y-axis at A and the
x-axis at B. The perpendicular bisector of AB meets
the line through (0, –1) parallel to x-axis at C. Find
the area of the triangle ABC.
Two equal sides of an isosceles triangle are given by
the equations 7x – y + 3 = 0 and x + y – 3 = 0 and its
third side passes through the point (1, – 10).
Determine the equation of the third side.
16.
The interior angle bisector of angle A for the triangle
ABC whose coordinates of the vertices are
A(–8, 5); B(–15, –19) and C(1, – 7) has the equation
ax+2y+c=0. Find ‘a’ and ‘c’.
17.
Find the equations of the sides of a triangle having
(4, –1) as a vertex, if the lines x – 1 = 0 and
x – y – 1 = 0 are the equations of two internal
bisectors of it angles
394-Rajeev Gandhi Nagar, Kota
2
y3 1
x y
x y
 = 1 &  = 1, meets
a b
b a
the coordinate axes in A & B. Show that the
AD, BE and CF.AD lies along the line (1 – 3 )x +
2 3 )x + ( 3 + 2)y – 8 = 0. If the length of the
hypotenuse is 60, find the area of the triangle ABC.
A variable line, drawn through the point of intersection
of the straight lines
A straight line L is perpendicular to the line 5x – y = 1.
The area of the triangle formed by the line L & the
coordinate axes is 5. Find the equation of the line.
The triangle ABC, right angled at C, has median

with the x-axis, meets
6
31
Straight Line 3.24
18.
19.
20.
21.
(a)
(b)
(c)
22.
Show that all the chords of the curve
3x2 – y2 – 2x + 4y = 0 which subtend a right angle
at the origin are concurrent. Does this result also
hold for the curve, 3x2 + 3y2 – 2x + 4y = 0? If yes,
what is the point of concurrency and if not, give
reasons.
The equations of the perpendicular bisectors of
the sides AB and AC of a triangle ABC are
x – y + 5 = 0 and x + 2y = 0, respectively. If the point
A is (1, –2) find the equation of the line BC.
Triangle ABC lies in the Cartesian plane and has an
area of 70 sq. units. The coordinates of B and C are
(12, 19) and (23, 20) respectively and the
coordinates of A are (p, q). The line containing the
median to the side BC has slope –5. Find the largest
possible value of (p + q).
Consider a triangle ABC with sides AB and AC
having the equations L1 = 0 and L2 = 0. Let the
centroid, orthocentre and circumcentre of the 
ABC are G, H and S respectively. L = 0 denotes
the equation of side BC.
If L1 : 2x – y = 0 and L2 : x + y = 3 and G (2, 3) then
find the slope of the line L = 0.
25.
COMPREHENSION
26.
(a)
(b)
(c)
If L1 : 2x + y = 0 and L2 : x – y + 2 = 0 and H(2, 3)
then find the y-intercept of L = 0.
If L1 : x+y – 1=0 and L2 : 2x – y + 4=0 and S(2, 1)
then find the x-intercept of the line L = 0.
27.
The equations of perpendiculars of the sides AB
and AC of triangle ABC are x–y–4=0 and
2x–y–5=0 respectively. If the vertex A is (–2, 3)
and point of intersection of perpendiculars bisectors
23.
24.
Two sides of a rhombus ABCD are parallel to the
lines y = x + 2 and y = 7x + 3. If the diagonals of the
rhombus intersect at the point (1, 2) and the vertex
A is on the y-axis, find the possible coordinates of
A. x – y – 1 = 0 are the equations of two internal
bisectors of its angles.
P is the point (–1, 2), a variable line through P cuts
the x and y axes at A and B respectively Q is the
point on AB such that PA, PQ, PB are H.P. Show
that the locus of Q is the line y = 2x.
Consider two points A  (1, 2) and B  (3, –1).
Let M be a point on the straight line L  x + y = 0.
If M be a point on the line L = 0 such that
AM + BM is minimum, then the reflection of M in
the line x = y is
(A) (1, –1)
(B) (–1, 1)
(C) (2, –2)
(D) (–2, 2)
If m be a point on the line L = 0 such that
|AM – BM| is maximum, then the distance of M
from N  (1, 1) is
(A) 5 2
(B) 7
(C) 3 5
(D) 10
If M be a point on the line L = 0 such that
|AM – BM| is minimum, then the area of AMB
equals
(A)
13
4
(B)
13
2
(C)
13
6
(D)
13
8
A straight line passing through O(0, 0) cuts the lines
x = , y =  and x + y = 8 at A, B and C respectively
such that OA. OB. OC = 48 2
f(, )  0
Where f(x, y) =
3 5
is  ,  , find the equation of medians to the sides
2 2
AB and AC respectively.
Find the equation of the two straight lines which
together with those given by the equation
6x2 – xy – y2 + x + 12y – 35 = 0 will make a
parallelogram whose diagonals intersect in the origin.
y 3

x 2
and
+ (3x – 2y) 6 +
ex  2 y  2e  6
(a)
Find the point of intersection of lines x =  and y=
(A) (3, 2)
(B) (2, 3)
(C) (–2, –3)
(D) (–3, –2)
(b)
Find the value of (OA + OB + OC).
(c)
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(A) 7 2
(B) 8 2
(C) 9 2
(D) None of these
Find the equation of line OA.
(A) y – x = 0
(B) y + x = 0
(C) y + 2 x = 0
(D)
2y + x = 0
32
1.26 Theory and Exercise Book
MATRIX MATCH TYPE
28.
Let ABC be a triangle such that the coordinates of
A are (–3, 1). Equation of the median through B is
2x + y – 3 = 0 and equation of the angular bisector
of C is 7x – 4y – 1 = 0. Then match the entries of
column-I with their corresponding correct entries
of column-II.
Column–I
Consider the 3 linear equations
ax + by + c = 0
bx + cy + a = 0
cx + ay + b = 0
where a, b, c  R.
Column-I
Column-I
(A) If a+b+c = 0 and
(P) entire xy plane
a2+b2+c2  ab+bc+ca then
(B) If a+b+c = 0 and
(Q) the lines are
a2+b2+c2 = ab+bc+ca then
concurrent
(C) If a+b+c  0 and
(R) line are concident
2
2
2
a +b +c  ab+bc+ca then
(D) If a+b+c  0 and
(S) lines are neither
2
2
2
a +b +c = ab+bc+ca then
coincident nor
concurrent
29.
Column–II
(A) Equation of the line AB is
(P) 2x + y – 3 = 0
(B) Equation of the line BC is
(Q) 2x – 3y + 9 = 0
(C) Equation of the line CA is
(R) 4x + 7y + 5 = 0
(S) 18x – y – 49 = 0
394-Rajeev Gandhi Nagar, Kota
33
Straight Line 3.26
1.
Let A (h, k), B(1,1) and C(2,1) be the vertices of a
right angled triangle with AC as its hypotenuse. If
the area of the triangle is 1, then the set of values
which 'k' can take is given by
[AIEEE-2007]
(A) {1, 3}
(B) {0, 2}
(C) {–1, 3}
(D) {–3, –2}
2.
The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept-4. Then a
possible value of k is [AIEEE-2008]
(A) 2
(B) –2
(C) –4
(D) 1
3.
The line L given by
6.
A line is drawn through the point (1, 2) to meet the
coordinate axes at P and Q such that it forms a
triangle OPQ, where O is the origin. If the area of
the triangle OPQ is least, then the slope of the line
PQ is :
[AIEEE-2012]
(A) –2
(B) – 1/2
(C) –1/4
(D) –4
7.
A ray of light along x+ 3y = 3 gets reflected upon
reaching x-axis, the equation of the reflected ray is :
[JEE-MAIN 2013]
x y
  1 passes through the
5 b
(A) y =
point (13, 32). The line K is parallel to L and has the
x y
equation   1 . The thedistance between L
c 3
and K is (A)
(C)
4.
23
15
17
15
[AIEEE-2010]
8.
23
17
The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect
the line L 3 : y + 2 = 0 at P and Q
respectively. The bisector of the acute angle between L1 and L2 intersects L3 at R. [AIEEE-2011]
Statement-1 : The ratio PR : RQ equal 2 2 : 5
Statement - 2 : In any triangle, bisector of an
angle divides the triangle into two similar triangles.
(A) Statement (1) is true and statement (2) is true
and statement (2) is correct explanation for
Statement (1)
(B) Statement (1) is true and statement (2) is true
and statement (2) is NOT a correct explanation
for Statement (1)
(C) Statement (1) is true but (2) is false
(D) Statement (1) is false but (2) is true
5.
(C) y = x + 3
(B) 17
(D)
If the line 2x + y = k passes through the point which
divides the line segment joining the points
(1, 1) and (2, 4) in the ratio 3 : 2, then k equal:
[AIEEE-2012]
(A) 6
(B) 11/5
(C) 29/5
(D) 5
3x –
3
(B)
3y = x – 1
(D)
3y = x –
3
The x-coordinate of the incentre of the triangle that
has the coordinates of mid points of its sides as (0,
1) (1, 1) and (1, 0) is :
[JEE-MAIN 2013]
(A) 1  2
(B) 1  2
(C) 2  2
(D) 2  2
9.
Let a, b, c and d be non-zero numbers. If the point
of intersection of the lines
4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the
fourth quadrant and is equidistant from the two axes
then :
[JEE-MAIN 2014]
(A) 2bc – 3ad = 0
(B) 2bc + 3ad = 0
(C) 3bc – 2ad = 0
(D) 3bc + 2ad = 0
10.
Let PS be the median of the triangle with vertices
P(2, 2), Q(6, –1) and R(7, 3). The equation of the
line passing through (1, –1) and parallel to PS is :
[JEE-MAIN 2014]
(A) 4x – 7y – 11 = 0
(B) 2x + 9y + 7 = 0
(C) 4x + 7y + 3 = 0
(D) 2x – 9y – 11 = 0
11.
The number of points, having both co-ordinates as
integers, that lie in the interior of the triangle with
vertices (0, 0), (0, 41) and (41, 0), is
[JEE-MAIN 2015]
(A) 820
(B) 780
(C) 901
(D) 861
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34
1.28 Theory and Exercise Book
12.
Locus of the image of the point (2, 3) in the line
(2x – 3y + 4) + k (x – 2y + 3) = 0, k  R, is a :
[JEE-MAIN 2015]
(A) circle of radius
13.
2
14.
Let k be an interger such that eh triangle with
vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq.
units. Then the orthocentre of this triangle is at the
point
[JEE-MAIN 2017]
(B) circle of radius 3
(C) straight line parallel to x-axis
(D) straight line parallel to y-axis
(A)  2,– 2 
Two sides of a rhombus are along the lines,
(C)  1,– 4 

1



3



3



1


(B) 1, 4 
(D)  2, 2 
x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals
intersect at (–1, –2), then which one of the following is a vertex of this rhombus?[JEE-MAIN 2016]
(A) (–3, –8)
 10
7
(C)  - 3 , - 3 


1
8
(B)  3 , - 3 


(D) (–3, –9)
394-Rajeev Gandhi Nagar, Kota
35
Straight Line 3.28
1. (a)
Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the
3.
The locus of the orthocentre of the triangle formed
by the lines
[JEE 2009, 3]
(1 + p) x – py + p(1 + p) = 0,
(1 + q) x –qy + q(1 + q) = 0
and y = 0, where p  q, is
(A) a hyperbola
(B) a parabola
(C) an ellipse
(D) a straight line
4.
A straight line L through the point (3, –2) is inclined
triangle OPQ. The point R inside the triangle OPQ
is such that the triangles OPR, PQR, OQR are of
equal area. The coordinates of R are
[JEE 2007]
(b)
(A) (4/3, 3)
(B) (3, 2/3)
(C) (3, 4/3)
(D) (4/3, 2/3)
Lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the
at an angle 60º to the line
line L3 : y + 2 = 0 at P and Q, respectively. The
intersects the x-axis, then the equation of L is
bisector of the acute angle between L1 and L2
(A) y  3x  2  3 3  0
intersects L3 at R.
(B) y  3x  2  3 3  0
Statement-1: The ratio PR : RQ equals 2 2 : 5
because
Statement-2: In any triangle, bisector of an angle
divides the triangle into two similar triangles.
(A) Statement-1 is true, statement-2 is true;
statement-2 is a correct explanation for statement-1
(B) Statement-1 is true, statement-2 is true;
statement-2 is NOT a correct explanation for
statement-1
(C) Statement-1 is true, statement-2 is false
(D) Statement-1 is false, statement-2 is true
5.
Consider the lines given by
[JEE 2008]
L1 = x + 3y – 5 = 0
L2 = 3x – ky – 1 = 0
L3 = 5x + 2y – 12 = 0
Match the statements/Expression in Column-I with
the statements/Expressions in Column-II and
indicate your answer by darkening the appropriate
bubbles in the 4 × 4 matrix given in OMR.
Column–I
Column–II
(A) L1, L2, L3 are concurrent, if
(P) k = – 9
(B) One of L1, L2, L3 is parallel to
at least one of the other two, if
(Q) k = –
6
5

6
(C) L1, L2, L3 form a triangle, if
(R) k =
(D) L1, L2, L3 do not form a triangle, if
(S) k = 5
3y  x  3  2 3  0
(D)
3y  x  3  2 3  0
[JEE 2011]
For a > b > c > 0, the distance between (1, 1) and
the point of intersection of the lines ax + by +c=0
and bx + ay + c = 0 is less than 2 2 . Then
(A) a + b – c > 0
(B) a – b + c < 0
(C) a – b + c > 0
(D) a + b – c < 0
[JEE 2013]
6.
2.
(C)
3 x + y = 1. If L also
www.motioniitjee.com
For a point P in the plane, let d1(P) and d2(P) be the
distances of the point P from the lines
x – y = 0 and x + y = 0 respectively. The area of the
region R consisting of all points P lying in the first quadrant of the plane and satisfying 2  d1(P) + d2(P)  4 is
[JEE 2014]
36
1.30 Theory and Exercise Book
EXERCISE - I
JEE Main
1.
8.
15.
22.
29.
36.
43.
C
D
C
C
B
C
B
2.
9.
16.
23.
30.
37.
44.
A
C
D
B
B
B
A
3.
10.
17.
24.
31.
38.
45.
D
B
C
c
D
A
A
4.
11.
18.
25.
32.
39.
A
D
B
A
B
A
5.
12.
19.
26.
33.
40.
C
A
A
A
B
B
6.
13.
20.
27.
34.
41.
D
D
A
A
D
B
7.
14.
21.
28.
35.
42.
C
C
D
A
A
A
B
B
A
A
B
D
4.
11.
18.
25.
32.
39.
D
C
A
D
C
D
5.
12.
19.
26.
33.
40.
B
A
D
D
C
C
6.
13.
20.
27.
34.
B
D
A
B
C
7.
14.
21.
28.
35.
C
C
A
B
D
B,C
ABC
A,B
4.
10.
16.
A,B
A,B
A,D
5.
11.
A,B
B,D
6.
12.
A,B,D
A, B, C
EXERCISE - II
JEE Advance
Single correct Option - type Questions
1.
A
2.
D
3.
8.
A
9.
D
10.
15.
D
16.
B
17.
22.
A
23.
D
24.
29.
D
30.
C
31.
36.
B
37.
B
38.
Multiple correct Option - type Questions
1.
7.
13.
A,B,C,D
B, D
A,C
2.
8.
14.
A,C
A,C
C,D
3.
9.
15.
EXERCISE - III
Subjective - type Questions
1. (33, 26)
 8
2. (a)  2,  ; (b) 4
 3
5. 3 units
6. x + 5y + 5 2 = 0 or x + 5y–5 2 =0
8. 4
9. 533
12. 6 units
14. 91 sq. units
3. K = 7 or 31/9
4. 1 : 2; Q(–5, – 3)
7. 400 sq. units
10. x2 + 4y2 + 4xy + 4x – 2y – 1 = 0
11. 2x + y – 1 = 0
15. x – 3y – 31 = 0 or 3x + y + 7 = 0
17. 2x – y + 3 = 0, 2x + y – 7 = 0, x – 2y – 6 = 0
16. a = 11, c = 78
 1 2
18. (1, –2), yes  , 
3 3
19. 14x + 23 y = 40
20. 47
21. (a) 5 ; (b) 2 ; (c)
23. (0, 0) or (0, 5/2)
25. 6x2 – xy – y2 – x – 12y – 35 = 0
3
2
22. x + 4y = 4; 5x + 2y = 8
Comprehension - based Questions
26. (a)
63
3
50
(8 5  5 10 )
; (b)
; (c)
10
10
7
27. (a)  = 2 and  = 3; (b) 9 2 ; (c) x – y = 0
394-Rajeev Gandhi Nagar, Kota
37
Straight Line 3.30
Matrix Match - type Questions
28. (A)–R ; (B)–S ; (C)–Q
29. (A)–Q ; (B)–P ; (C)–S ; (D)–R
EXERCISE - IV
Previous Year’s Question
JEE Main
1. C
2. C
3. D
4. C
5. A
6. A
9. C
10. B
11. B
12. A
13. B
14. D
7. D
8. D
4. B
5. A
JEE Advanced
1. (a) C ; (b) C
2. (A)–S ; (B)–P,Q ; (C)–R ; (D)–P,Q,S
6. 6
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3. D
38
STRAIGHT LINE
THEORY AND EXERCISE BOOKLET
CONTENTS
S.NO.

TOPIC
PAGE NO.
THEORY WITH SOLVED EXAMPLES ............................................................... 3 – 20
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com
39
STRAIGHT LINE
Page # 2
JEE Syllabus :
Cartesian coordinates, distance between two points, section formulae, shift of origin, equation of a
straight line in various forms, angle between two lines, distance of a point from a line. Lines through the
point of intersection of two given lines, equation of the bisector of the angle between two lines,
concurrency of lines, centroid, orthocentre, incentre and circumcentre of a triangle.
394 - Rajeev Gandhi Nagar
Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052,
0744-2439053
www.motioniitjee.com,
email-info@motioniitjee.com
IVRS No. 0744-2439051,
0744-2439052,
0744-2439053 www.motioniitjee.com,
email-info@motioniitjee.com
40
STRAIGHT LINE
A.
Page # 3
COORDINATE GEOMETRY
Coordinate Geometry is the unification of algebra and geometry in which algebra is used in the study
of geometrical relations and geometrical figures are represented by means of equations. The most
popular coordinate system is the rectangular Cartesian system. Coordinates of a point are the real
variables associated in an order to describe its location in space. Here we consider the space to be
two-dimensional. Through a point O, referred to as the origin, we take two mutually perpendicular
lines XOX' and YOY' and call them x and y axes respectively. The position of a point is completely
determined with reference to these axes by means of an ordered pair of real numbers (x, y) called the
coordinates of P where | x | and | y | are the distances of the point P from the y-axis and the x-axis
respectively, x is called the x-coordinate or the abscissa of P and y is called the y-coordinate or the
ordinate of the point P.
(1) Distance between two points :
y
(a) Let A and B be two given points,
whose coordinates are given by
A(x1, y1) and B(x2 , y2) respectively.
Then AB =
B(x2, y2)
( x1  x 2 )2  ( y1  y 2 )2
(b) Distance of (x1, y1) from origin :
A(x1, y1)
x12

y12
x
O
Note :- If two vertex A(x 1, y 1), B(x 2, y 2) are given then third vertex of equilateral triangle
 x1  x 2  3 ( y 2  y1) y1  y 2  3 ( x 2  x1) 
,

C is 
2
2


(2) Section formula :
Coordinates of the point P dividing the join of two points
A(x1, y1) and B(x2, y2) internally in the given ratio 1 : 2
A(x1,y1)
  2 x 1  1x 2  2 y 1  1y 2 

,
are P    
 2  1  .
2
1

Coordinates of the point P dividing the join of two points
A(x1, y1) and B(x2, y2) externally in the ratio of 1 : 2 are
  x  1x 2  2 y1  1y 2
P  2 1
,
 2  1
  2  1

 . In both

2
1
P(x,y)
B(x2,y2)
1
2
A(x1,y1)
B(x2,y2)
P(x,y)
the cases, 1/2 is positive.
Notes :
(i) If the ratio, in which a given line segment is divided, is to be determined, then sometimes, for
convenience (instead of taking the ratio 1 : 2), we take the ratio k : 1. If the value of k turns
out to be positive, it is an internal division otherwise it is an external division.
(ii) The coordinates of the mid-point of the line-segment joining A(x1, y1) and B(x2, y2) are
 x1  x 2 y1  y 2  .
,


2 
 2
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com
41
STRAIGHT LINE
Page # 4
(3) Special points in a triangle with co-ordinates :
P(x1, y1)
(a) Centroid (G) :
Definition : The point of concurrence of the medians
of a triangle is called the centroid of the triangle.
(i) G divides median into 2 : 1.
(ii) G always lies inside the triangle.

 x1  x 2  x 3 y1  y 2  y 3 
,
 or 
(iii) Co-ordinates of G is 

3
3



N
G
(x2, y2) Q
M
R (x3, y3)
L
 x1 ,  y1 
3
3


(b) Incentre (I) :
Definition : The point of concurrency of the internal
bisectors of the angles of a triangle is called the
incentre of the triangle.
(i) I always lies inside the triangle.
(ii) Internal angle bisector divides the
base in the ratio of adjacent sides.
P
N
r
r
r
Q
M
I
R
L
 ax1  bx 2  cx 3 ay1  by 2  cy 3 
,

(iii) Co-ordinates of I is 
abc
abc


where a, b, c are the lengths of the sides of the 
(c) Ex-centres (I1, I2, I3) :
Definition : The centre of the escribed circle which is opposite to vertices.
To get I1 (or I2 or I3) replace a by –a (b by –b or c by –c) in formula of coordinate of I
(d) Circumcentre (C) :
Definition : The point of concurrency of the perpendicular bisectors
of the sides of a triangle is called circumcentre of the triangle.
(i) For acute angle   lies inside
(ii) For obtuse angle   lies outside
(iii) For right angle   Mid point of hypotenuse
(iv) Co-ordinates of circumcentre is
P
Q
C
R
 x1 sin 2A  x 2 sin 2B  x 3 sin 2C y1 sin 2 A  y 2 sin 2B  y 3 sin 2C 
,


sin 2 A  sin 2B  sin 2C
sin 2 A  sin 2B  sin 2C


(e) Orthocentre (O) :
P
Definition : The point of concurrency of the altitudes of a triM
angle is called the orthocentre of the triangle.
N
(i) For acute angle  lies inside
O
(ii) For obtuse angle  lies outside
(iii) For right angle  vertex at ar
Q
L
(iv) Co-ordinates of orthocentre is
R
 x1 tan A  x 2 tan B  x 3 tan C y1 tan A  y 2 tan B  y 3 tan C 
,


tan A  tan B  tan C
tan A  tan B  tan C


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STRAIGHT LINE
Page # 5
Notes :
(i) In any triangle O, G, C are collinear.
(ii) In any triangle G divides the line joining O & C in ratio 2 : 1.
(iii) In an equilateral triangle O, G, C, I are coincident.
(iv)In an isosceles triangle O, G, C, I are collinear.
(f) Harmonic Conjugate : If P is a point that divides AB internally in the ratio m1 : m2 and Q is
another point which divides AB externally in the same ratio m1 : m2, then the point P and Q are said
to be Harmonic conjugate to each other with respect to A and B.
A
P
B
Q
1
1
2
+
=
AQ
AP
AB
Note :- Internal and External angle bisector of an angle divides the base harmonically.
i.e. AP , AB and AQ forms a HP 
Ex.1
If midpoints of the sides of a triangle are (0, 4), (6, 4) and (6, 0), then find the vertices of triangle,
centroid and circumcentre of triangle.
Sol.
A(x1, y1)
Let points A(x1, y1), B(x2, y2) and C(x3, y3) be vertices of ABC.
x1 + x3 = 0, y1 + y3 = 8
x2 + x3 = 12, y2 + y3 = 8
(6, 0)
(0, 4)
x1 + x2 = 12, y1 + y2 = 0
Solving we get A(0, 0), B (12, 0) and C(0, 8).
B(x2, y2) (6, 4) C(x3, y3)
Hence ABC is right angled triangle A = /2.

Circumcentre is midpoint of hypotenuse which is (6, 4) itself and
 x1  x 2  x 3 y1  y 2  y 3   8 
,
   4,  .
centroid  
3
3

  3
Ex.2
Prove that the incentre of the triangle whose vertices are given by A(x1, y1), B(x2, y2), C(x3, y3) is
 ax1  bx2  cx 3 ay1  by 2  cy 3 
,

 where a, b, and c are the sides opposite to the angles A, B and C
abc
abc


respectively.
Sol.
By geometry, we know that
BD AB

(since AD bisects A).
DC AC
BD AB
c

=
.
DC AC
b
If the length of the sides AB, BC and AC are c, a and b respectively, then
 bx 2  cx 3 by 2  cy 3 
,

 Coordinates of D are 
bc 
 bc
BD
c
ac
Since
= , BD =
DC
b
bc
 ac 


ID
BD
a
ac

B bisects B, Hence



IA BA
c
c b
Let the coordinates of I be ( x, y ) .
Then x 
A(x1, y1)
F
F
l
B(x2, y2)
ax 1  bx 2  cx 3
ay  by 2  cy 3
(using section formula).
,y  1
abc
abc
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C(x3, y3)
43
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Page # 6
B.
AREA OF A TRIANGLE
Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle
ABC. Then the area of triangle ABC, is
1
[ x 1( y 2
2
x1
1
x2
= 2
x3
– y 3 )  x 2 ( y 3 – y 1 )  x 3 ( y 1 – y 2 )]
...(1)
y1 1
y2 1
...(2)
y3 1
While using formula (1) or (2). order of the points (x1, y1), (x2, y2) and (x3, y3) has not been taken
into account. If we plot the points A(x1, y1), B (x2, y2) and C(x3, y3), then the area of the triangle as
obtained by using formula (1) or (2) will be positive or negative as the points A, B, C are in anticlockwise or clockwise directions.
A
A
A
+
–
B
C
C
B
B
C
D
So, while finding the area of triangle ABC, we use the formula :
x1
1
1
x 1( y 2 – y 3 )  x 2 ( y 3 – y 1 )  x 3 ( y 1 – y 2 ) = Modulus of 2 x 2
Area of ABC =
2
x3
y1 1
y2 1
y3 1
Notes :
(i) If three points (x1, y1), (x2, y2) and (x3, y3) are collinear, then
x1
y1 1
x2
y2 1  0
x3
y3 1
x
(ii) Equation of straight line passing through (x1, y1) and (x2, y2) is given by
y
1
x1
y1 1  0
x2
y2 1
(iii) In case of polygon with (x1, y1), (x2, y2) ............ (xn, yn) the area is given by
1
| ( x1y 2 – y1x 2 )  ( x 2 y 3 – y 2 x 3 )  .........  ( x n –1y n – y n –1x n )  ( x n y 1 – y n x 1 ) |
2
Ex.3
The vertices of quadrilateral in order are (–1, 4), (5, 6), (2, 9) and (x, x2). The area of the quadrilateral
is
Sol.
15
sq. units, then find the point (x, x2)
2
Area of quadrilateral =
1
15
1
x x2
1 –1 4 5 6 2 9
 – 26  33  2x 2 – 9 x  4 x  x 2  3 x 2 – 5 x  7 =



2
2 5 6 2 9 x x
2
2
2
–1 4
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
Page # 7
3x2 – 5x + 7 = ± 15
 3x2 – 5x – 8 = 0, 3x2 – 5x + 22 = 0  x =
8
, x = –1
3
 8 64 
 or (–1, 1). But (–1, 1) will not form a quadrilateral as per given order of the
Hence point is  ,
3 9 
 8 64 

points. Hence the required point is  ,
3 9 
(1) Locus : When a point moves in a plane under certain geometrical conditions, the point traces out
a path. This path of a moving point is called its locus.
Note : All those points which satisfy the given geometrical condition will definitely lie on the locus.
But converse is not true always.
(2) Equation of Locus : The equation to a locus is the relation which exists between the coordinates
of any point on the path, and which holds for no other point except those lying on the path.
(3) Procedure for finding the equation of the locus of a point
(i)
If we are finding the equation of the locus of a point P, assign coordinates (h, k) to P.
(ii)
Express the given conditions as equations in terms of the known quantities to facilitate
calculations. We sometimes include some unknown quantities known as parameters.
(iii) Eliminate the parameters. so that the eliminate contains only h, k and known quantities.
(iv) Replace h by x, and k by y, in the eliminate. The resulting would be the equation of the locus
of P.
(v)
If x and y coordinates of the moving point are obtained in terms of a third variable t (called
the parameter), eliminate t to obtain the relation in x and y and simplify this relation. This
will give the required equation of locus.
Ex.4
Find the focus of the middle points of the segment of a line passing through the point of intersection
of the lines ax + by + c = 0 and lx + my + n = 0 and intercepted between the axes
Sol.
Any line (say L = 0) passing through the point of intersection of ax + by + c = 0 and lx + my + n = 0 is
(ax + by + c) +  (lx + my + n) = 0, where  is any real number.
c  n 
 c  n 

, 0  and  0, –

Point of intersection of L = 0 with axes are  –
b  m 
 a  l 

Let the mid point be (h, k). Then h  –
Eliminating , we get
1  c  n 
1  c  n 

 and k  – 2  b  m 


2  a  l 
2ah  c
2kb  c
=
. The required locus is : 2(am – lb) = (lc – an) x + (nb – mc)y..
2hl  c
2km  c
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STRAIGHT LINE
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(4) Inclination of a line : Its a measure of the smallest
y
non-negative angle which the line makes with +ve direction of the x-axis [angle being measured in

x
anti-clockwise direction]. 0  < 
(5) Slope of the line : If the inclination of line is  and  

then its slope is defined as tan  and
2
denoted by ‘m’
(i)
If  = 0, then m = 0 i.e. line parallel to x-axis.
(ii)
If  = 90º, then m does not exist i.e. line parallel to y-axis
(iii) Slope of line joining two points A(x1, y1) & B(x2, y2) is m  tan  
y 2  y1
x 2  x1
(iv) If a line equally inclined with co-ordinate axes then slope is ± 1.
(6) Intercepts : The point where a line cuts the x-axis (or y-axis) is called its x-intercept (or
y-intercept).
(i)
Intercepts may be +ve, –ve or zero.
(ii)
A line making an intercept of –a with y-axis means the line passing through (0,–a)
(iii) A line makes equal non-zero intercept with both co-ordinate axes then slope is –1.
(iv) A line makes non-zero intercept with both co-ordinate axes equal in magnitude then slope is ±1.
C.
STANDARD EQUATIONS OF STRAIGHT LINES
(1) General Form : Any first degree equation of the form Ax + By + C = 0,
y
where A, B, C are constant always represents general equation of a straight
line (at least one out of A and B is non zero.)
y = mx + c
(2) Slope - Intercept Form :
(0,c)
y = mx + c
x
where m = tan
where m = slope of the line = tan 
c = y intercept
y
(3) Intercept Form :
(0,b)
x y
 1
a b
x y
 1
a b
x
x intercept = a
(a,0)
y intercept = b
y
(4) Normal Form :
x cos  + y sin  = p, where  is the angle which the perpendicular to the
line makes with the axis of x and  is the length of the
p
perpendicular from the origin to the line. p is always positive

x
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STRAIGHT LINE
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(5) Slope Point Form :
Equation : y – y1 = m (x – x1), is the equation of line passing through the point (x1, y1) and having the
slope 'm'
(6) Two points Form :
y 2 – y1
Equation : y – y1 = x – x ( x – x 1 ) , is the equation of line passing through two points (x1, y1) and (x2, y2)
2
1
(7) Parametric Form :
To find the equation of a straight line which passes through
a given point A(h, k) and makes a given angle  with the
positive direction of the x-axis. P(x, y) is any point on the
line LAL'. Let AP = r, x – h = r cos , y – k = r sin 
P
y
A
L
r

(h,k)
L’
x–h y–k

 r is the equation of the straight line LAL'.
cos  sin 
x
Any point on the line will be of the form (h + r cos , k + r sin ), where |r| gives the distance of the
point P from the fixed point (h, k)
Note : If point P is taken relatively upward to A then r is positive otherwise negative. If line is
parallel to x-axis then for he point right to A, r is positive and for left to A, r is negative.
D.
REDUCTION OF GENERAL EQUATION TO DIFFERENT STANDARD FORMS
(1) Slope - Intercept Form :
To reduce the equation Ax + By + C = 0 to the form y = mx + c
Given equation is Ax+ By + C = 0  y =
Note : Slope of the line Ax + By + c = 0 is –
C
–A
,c= –
(B  0)
B
B
 coefficient of x 
A
C
.i.e –  coefficient of y  . y intercept the line  –
B
B


(2) Intercept Form :
To reduce the equation Ax + By + C = 0 the form
x y
  1 . This reduction is possible only when C  0
a b
Given equation is Ax + By = – C

x
y
x y
–C
C

 1 which is the form   1 . where a =
,b=–
–C –C
a b
A
B
A
B
Note : Intercept on the x-axis = –
C
C
, intercept on the y-axis = – . Thus intercept of a straight line
A
B
on the x-axis can be obtained by putting y = 0 in the equation of the line and then finding the value
of x. similarly, intercept on the y-axis can be obtained by putting x = 0 and solving for y.
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(3) Normal form :
To reduce the equation Ax + By + C = 0 to the form x cos  + y sin  = p
Given equation is Ax + By + C = 0 or, Ax + By = – C
A
Case I : When – C > 0, then normal form is
2
B
2
x+
2
–C
2
A B
A B
A
B
–C
where cos =
2
2 , sin  =
2
2 ; p =
A B
A B
A 2  B2
Case II : When –C < 0, the write the equation as –Ax – By = C
–A
2
–B
x
2
2
A 2  B2
C
y
A B
A  B2
C
–A
–B
,p 
, sin  
where cos  =
2
2
2
2
2
A B
A  B2
A B
Note : In the normal form x cos  + y sin  = p, p is always taken as positive.
A B
2
y=
2
Ex.5
Reduce the line 2x – 3y + 5 = 0, in slope intercept, intercept and normal forms.
Sol.
Slope - Intercept Form : y =
Intercept Form :
sin  
Ex.6
3
13
2x 5
2
5
 , tan = m = , c =
3 3
3
3
2x
3y
5
5
5
y
x



 1, a  – , b  . Normal Form : –
13
13
13
2
3
5
5

  
–   
 2 3
, cos  
–2
13
,p
5
13
Find the equations of the lines which pass through the point (3, 4) and the sum of their respective
intercepts on the axes is 14.
Sol.
Let the equation of the line be
x y
 1
a b
This passes through (3, 4) therefore
It is given that a + b = 14
(a – 7) (a – 6) = 0
3 4
 1
a b
...(ii)
 b = 14 – a
Putting b = 14 – a in (ii), we get

...(i)
3
4

 1  a2 – 13a + 42 = 0
a 14 – a
 a = 7, 6

two such lines are there.
For a = 7, b = 14 – 7 = 7 and for a = 6, a = 14 – 6 = 8
Putting the values of a and b in (i), we get the equations of lines
x y
x y
  1 and   1 or x + y = 7 and 4x + 3y = 24.
6 8
7 7
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Ex.7
Page # 11
A rod of steel is fixed at A (4, 0) and a toy is placed on it at B (0, 4). Now rod is rotated about A
through an angle of 15° in clockwise direction, then find the new position of a toy.
Sol.
Let new position of a toy be C.
y
4–0
 –1
Slope of AB =
0–4
C
B(0,4)
  = 135°
15°
Rod is rotated through 15° in clockwise
x
new = 135° – 15° = 120°
O
A(4, 0)
AB = 4 2  h = 4, k = 0
Hence C = (h + r cos , k = r sin ) = (4 + 4 2 cos 120°, 0 + 4 2 sin 120°) = ( 4 – 2 2 , 2 6 )
Ex.8
If the straight line through the point P(3, 4) makes an angle /6 with the x-axis and meets the line
12x + 5y + 10 = 0 at Q, find the length of PQ.
Sol.
The equation of a line passing through P(3, 4) and making an angle = /6 with the x-axis is
x–3 y–4

= r, where r represents the distance of any point on this line from the given point


cos
sin
6
6




P(3, 4). The co-ordinates of any point Q on this line are  r cos  3, r sin  4 
6
6



132
–132
3 
r

If Q lies on 12x +15y + 10 = 0, then 1213  r
 length PQ=
 5 4    10  0  r 


12 3  5
2 
2
12 3  5


Ex.9
A canal is 4
1
kms from a place and the shortest route from this place to the cenal is exactly north2
east. A village is 3 kms north and 4 kms east from the place. Does it lie on canal?
Sol.
Let AB be the canal and O be the given place.
Let L be the foot of perpendicular from O to AB.
Y
North
Given, OL = 9/2. And AOL = 45°
xcos45° + ysin45° =
9
2
2
L
45°
9
or x + y =
B
O
...(1)
A
East
X
9
Let S be the given village, then S = (4, 3). Putting x = 4 and y = 3 in equation (1), we get 4 + 3 =
2
,
which is not true. Thus the co-ordinates of S doesn't satisfy equation (1) and hence the given village
does not lie on the canal.
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(4) Position of a point w.r. to a line L : Ax + By + c = 0
(i)
If the points P(x 1, y 1) & Q(x 2, y 2) lies on the same side of the line Ax + By + C = 0 then
the expressions Ax1 + By1 + C & Ax2 + By2 + C have same sign otherwise if P and Q lies
on opposite side then Ax 1 + By 1 + C and Ax 2 + By 2 + C will have opposite sign.
(ii)
If only one point is given then position of that point is checked w.r. to origin.
Ex.10 Find the range of  in the interval (0, ) such that the points (3, 5) and (sin , cos ) lie on the same
side of the line x + y – 1 = 0
Sol.
Here 3 + 5 – 1 = 7 > 0. Hence sin + cos – 1 > 0  sin(
E.
ANGLE BETWEEN TWO STRAIGHT LINES
1

 ) 
4
2

3
 

 
0<<
4 4
4
2
y = m1x + c1
m – m2
If  is the acute angle between two lines, then tan   1
1  m1m 2
where m1 and m2 are the slopes of the two lines and are finite.
180 – 

y = m2x + c2
Notes :
(i)
If the two lines are perpendicular to each other then m1m2 = –1
(ii) Any line perpendicular to ax + by + c = 0 is of the form bx – ay + k = 0
(iii) If the two lines are parallel or coincident, then m1= m2
(iv) Any line parallel to ax + by + c = 0 is of the form ax + by + k = 0
(v) If any of the two lines is perpendicular to x-axis, then the slope of that line is infinite.
m
1– 2
m1 – m 2
m1
1


Let m1 = , Then tan  
1
1  m1m 2
m2
 m2
m1
or  = |90° –  |, where tan  = m2
i.e. angle  is the complimentary to the angle which the oblique line makes with the x-axis.
(vi) If lines are equally inclined to the coordinate axis then m1 + m2 = 0
Ex.11 Find the equation to the straight line which is perpendicular bisector of the line segment AB, where A,
B are (a,b) and (a', b') respectively.
Sol.
Equation of AB is y – b =
b'– b
( x – a)
a'– a
i.e. y (a' – a) – x (b' – b) = a'b – ab'.
Equation to the line perpendicular to AB is of the form (b' – b)y + (a' – a)x + k = 0
....(1)
 b  b' 
 a  a' 
  (a'– a )
k  0
Since the midpoint of AB lies on (1), (b'– b )
 2 
 2 
Hence the required equation of the straight line is 2(b'–b)y  2(a'– a)x  (b' 2 –b 2  a' 2 – a 2 )
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STRAIGHT LINE
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(1) Equation of straight Lines passing through a given point and equally inclined to a given line :
Let the straight passing through the point (x1, y1) and make equal angles with the given straight line
y = mx + c. If m is the slope of the required line and  is the angle which this line makes with the
m1 – m
given line then tan    1  m m
1
(x1, y1)
(2) The above expression for tan, given two values of m, say mA and mB.
The required equations of the lines through the point (x1, y1) and

making equal angles  with the given line are

y – y1 = mA(x – x1), y – y1 = mB (x – x1)
Ex.12 Find the equation to the sides of an isosceles right-angled triangled, the equation of whose hypotenuse
is 3x + 4y = 4 and the opposite vertex is the point (2, 2).
Sol.
The problem can be restarted as :
Find the equation to the straight lines passing through the given point (2, 2) and making equal angles
of 45° with the given straight line 3x + 4y – 4 = 0. Slope of the line 3x + 4y – 4 = 0 is m1 = –
m – m1
m 3/4
 tan 45   1  m m , i.e., 1  
1
3
1– m
4
3
4
(2, 2)
45°
1
mA = , and mB = – 7
7
45°
3x + 4y = 4
Hence the required equations of the two lines are
y – 2 = mA(x – 2) and y – 2 = mB (x – 2)  7y – x – 12 = 0 and 7x + y = 16.
F.
DISTANCE BETWEEN POINT & LINE AND TWO PARALLEL LINES
(1) Length of the Perpendicular from a Point on a Line :
ax 1  by 1  c
The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is
a2  b2
c
The length of the perpendicular from origin on ax + by + c = 0 is
2
a  b2
(2) The distance between two parallel lines :
| c1 – c 2 |
The distance between two parallel lines : ax + by + c1 = 0 and ax + by + c2 = 0 is
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a2 – b2
51
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Page # 14
(3) Area of parallelogram with given sides :
y = m1x + C2
D
y = m2x + d2
( C 1  C 2 ) ( d1  d 2 )
Area 
m1  m 2
C
y = m2x + d1
p1
p2
A
B
y = m1x + C1
(4) Condition of parallelogram as shown becomes a rhombus :
p1  p2 
c1  c 2
2
a b
2
d1  d2

a2  b2
(5) Reflection (Image) of a point P() about a line (ax + by + c = 0)
P( , ) ( Given)
x
y
2 (a   b   c )


a
b
a2  b2
A
 0 B
Q(h,k) reflection)
(6) Foot of perpendicular from a point (, ) to a given line  ax + by + c = 0
x
y
(a   b   c )


a
b
a2  b2
y
Y
P(x1 y)
(7) Shifting of the origin :
y–b
x, y  old co-ordinates axes
a
(a,b)
X'
X, Y  New co-ordinate axes
y
X=0x–a=0x=a
x'
Y=0y–b=0y=b
x a
X
b
x
x
(0,0) O
Slope and area of closed figure remains
y'
unchanged under the translation of co-ordinate axes.
Y'
Ex.13 Three lines x + 2y + 3 = 0, x + 2y – 7 = 0 and 2x – y – 4 = 0 form 3 sides of two squares. Find the
equation of remaining sides of these squares.
|73|
Sol.
Distance between the two parallel lines is
5
2 5.
x + 2y + 3 = 0
The equations of sides A and C are of the form
2x – y + k = 0. Since distance between sides A and
B = distance between sides
| k – (–4) |
B and C
5
2 5 
2x–y–4 = 0
A
k4
5
 2 5
 k = 6, –14
Hence the fourth sides of the two squares are
(i) 2x – y + 6 = 0,
B
C
x + 2y – 7 = 0
(ii) 2x – y – 14 = 0
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Ex.14 Find the foot of the perpendicular drawn from the point (2, 3) to the line 3x – 4y + 5 = 0. Also, find the
image of (2, 3) in the given line.
Sol.
Let AB  3x – 4y + 5 = 0, P  ( 2, 3) and PM  AB .
P(2, 3)
3
Slope of AB 
4
 sin  
4
3
, cos  
 tan  (say)
5
3
Now, r  p 
G.
4
 tan  (say)
 slooe of PM 
3
32  43  5
9  16

A
 sin  
4
3
, cos  
5
5
M
B
Q
6  12  5 1

5
5
1
1


 53 71 
M   2  cos , 3  sin     , 
5
5


 25 25 
Which is the foot of the perpendicular.

Let Q be the image of P
2
2

  56 67 
,
 Q   2  cos , 3  sin    

5
5

  25 25 
BISECTORS OF THE ANGLES BETWEEN TWO GIVEN LINES
Angular bisector is the locus of a point which moves in such a way so that its distance from two
intersecting lines remains same.
The equation of the two bisectors of the angles between the lines a 1x + b 1y + c 1 = 0 and
a 2 x + b 2y + c2 = 0 are
a1x  b1y  c 1
a12
 b12

a2 x  b2 y  c 2
a 22  b 22
.
If the two given lines are not perpendicular i.e. a1 a2 + b1 b2  0, then one of these equation is the
equation of the bisector of the acute angle and the other that of the obtuse angle.
Note : Whether both lines are perpendicular or not but the angular bisectors of these lines will always
be mutually perpendicular.
(1) The bisectors of the acute and the obtuse angles :
Take one of the lines angle let its slope be m1 and take one of the bisectors and let its slope be m2.
If  be the acute angle between them, then find tna 
m1  m 2
.
1  m1m 2
If tan   1 then the bisector taken is the bisector of the obtuse angle and the other one will be the
bisector of the acute angle. If 0  tan   1 then the bisector taken is the bisector of the acute
angle and the other one will be the bisector of the obtuse angle.
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STRAIGHT LINE
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If two lines are a1x  b1y  c 1  0
a1x  b1y  c 1
a12  b12

and a 2 x  b 2 y  c 2  0 , then
C
a2 x  b2y  c 2
N
will represent the equation of the bi-
a 22  b 22
A
sector of the acute or obtuse angle between the lines according as

P(x, y)
M
B
c 1c 2 (a1a 2  b1b 2 ) is negative or positive.
(2) The equation of the bisector of the angle containing the origin
Write the equations of the two lines so that the constants c1 and c2 become positive. Then the
equation
a1x  b1y  c 1
a12  b12

a2x  b2 y  c 2
a 22  b 22
is the equation of the bisector containing the origin.
Notes :
(i)
If a1a 2  b1b 2  0 , then the origin will lie in the acute angle and if a1a 2  b1b 2  0 , then origin
(ii)
The note (i) is helpful in finding the equation of bisector of the obtuse angle or acute angle
will lie in the obtuse angle.
directly.
(3) The equation of the bisector of the angle which contains a given point
The equation of the bisector of the angle between the two lines containing the point (,) is
a1x  b1y  c 1
a12  b12
a xb yc
2
2
  2

2
2
a2  b2


a x b y c
a1x  b1y  c 1

 2
2
2

–
 or

2
2
2
2
a1  b1
a2  b2




 according as a1   b1   c 1 and

a 2   b 2   c 2 are of the same sings or of opposite signs.
Ex.15 For the straight line 4 x  3y  6  0 and 5 x  12y  9  0 , find the equation of the
(i) bisector of the obtuse angle between them.
(ii) bisector of the actus angle between them.
(iii) bisector of the angle which contains (1, 2).
Sol.
Equations of bisectors of the angles between the given lines are
4x  3y  6
5 x  12y  9
 9 x  7 y  41  0 and 7 x  9 y  3  0 .
4 3
5 2  12 2
If  is the acute angle between the line 4 x  3y  6  0 and the bisector 9 x  7 y  41  0 , then
2
2

4 9

3
7  11  1
tan  
3
4 9
1 

3
7



Hence
(i) The bisector of the obtuse angle is 9 x  7 y  41  0
(ii) The bisector of the acute angle is 7 x  9 y  3  0
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54
STRAIGHT LINE
Page # 17
–4 x – 3 y  6
5 x  12y  9
 7x  9y  3  0
( 4)  ( 3)
5 2  122
(i) For the point (1, 2), 4x + 3y – 6 = 4 × 1 + 3 × 2 – 6 > 0  5x + 12y + 9 = 5 × 1 + 12 × 2 + 9 > 0
Hence equation of the bisector of the angle containing the point (1, 2) is
4 x  3 y  6 5 x  12 y  9

 9 x  7 y  41  0
5
13
Alternative : 5 lines. Similarly bisector of obtuse angle is 9x – 7y – 41 = 0.
(iii) The bisector of the angle containing the origin
2
2

(4) The equation of reflected ray :
Let L 1  a1x  b1y  c 1  0 be the incident ray in the line mirror L 2  a 2 x  b 2 y  c 2  0
Let L3 be the reflected ray from the line L2. Clearly L2 will be one of the bisectors of the angles
between L1 and L3. Since L3 passes through A, so L3  L1 + L2 = 0.
Let (h, k) be a point on L2. Then,
| a1h  b1k  c 1 | | a1h  b1k  c 1   (a 2h  b 2k  c 2 ) |

a12  b12
( a 1   a 2 ) 2  (b 1   b 2 ) 2
Since (h, k) lies on L2, a2h + b2k + c2 = 0


A
a12  a 22 2  2a1a 2  b12  b 22 2  2b1b 2   a12  b12
  0 or  
(h, k)
L1
2(a1a 2  b1b 2 )
L2
L3
a 22  b 22
But  = 0 gives L3 = L1. Hence L 3  L1 
2(a1a 2  b1b 2 )
a 22  b 22
L2  0 .
Note : Some times the reflected ray L3 is also called the mirror image of L1 in L2.
H.
FAMILY OF LINES
The general equation of the family of lines through the point of intersection of two given lines is
L + L' = 0, where L = 0 and L' = 0 are the two given lines, and  is a parameter. Conversely, any line
of the form L1 + , L2 = 0 passes through a fixed point which is the point of intersection of the lines
L1 = 0 and L2 = 0.
The family of lines perpendicular to a given line ax + by + c = 0 is given by bx – ay + k = 0, where k is
a parameter.
The family of lines parallel to a given line ax + by + c = 0 is given by ax + by + k = 0, where k is a
parameter.
Ex.16 Show that all the chords of the curve 3x2 – y2 – 2x + 4y = 0, which subtend a right angle at the origin
pass through a fixed point. Find that the point.
Sol.
Let the equation of chord be lx + my = 1. So equation of pair of straight line joining origin to the
points of intersection of chord and curve.
3x2 – y2 – 2x (lx + my) + 4y(l + my) = 0, which subtends right angle at origin.
 (3 – 2l + 4m – 1) = 0  1 = 2m + 1. Hence chord becomes (2m + 1)x + my = 1
x – 1 + m(2x + y) = 0
L1
L2
Which will pass through point of intersection of L1 = 0 and L2 = 0.
 x = 1, y = – 2. Hence fixed point is (1, – 2).
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55
STRAIGHT LINE
Page # 18
(1) One Parameter Family of Straight Lines
If a linear expression L1 contains an unknown coefficient, then the line L1 = 0 can not be a fixed
line. Rather it represents a family of straight lines known as one parameter family of straight lines.
e.g. family of lines parallel to the x-axis i.e. y = c and family of straight lines passing through the
origin i.e. y = mx.
Each member of the family passes a fixed point. We have two methods to find the fixed point.
Method (i) :
Let the family of straight lines of the form ax + by + c = 0 where a, b, c are variable parameters
satisfying the condition al + bm + cn = 0, where I , m, n, are given and n  0. Rewriting the
 1
m
condition as a    b  + c = 0 and comparing with the given family of straight lines, we find
n
 
n
1 m
that each member of it passes through the fixed point  , 
n n 
Ex.17 If the algebraic sum of perpendiculars from n given points on a variable straight line is zero then prove
that the variable straight line passes through a fixed point.
Sol.
Let n given points be (x1, y1) where i = 1, 2 .......... n and the variable line is ax + by + c = 0, Given
n
that
 ax  by  c 
1
1

 = 0.
2
2
a

b
i 1 

 
 ax1 + ay1 + cn = 0 a
x
i
b
y
i
+ c = 0.
n

xi
y i 

Hence the variable straight line always passes through the fixed point  n . n  .


Method (ii) :
n
 
If a family of straight lines can be written as L1 + L2 = 0 where L1, L2 are two fixed lines and  is
a parameter, then each member of it will pass through a fixed point given by point of intersection
of L1 = 0 and L2 = 0.
Note : If L1= 0 an L2 = 0 are parallel lines, they will meet at infinity.
Ex.18 Prove that each member of the family of straight lines
(3 sin + 4 cos )x + (2 sin  – 7cos) + (sin +2cos) = 0 ( is a parameter)
passes through a fixed point.
Sol.
The given family of straight lines can be rewritten as (3x + 2y + 1) sin +(4x – 7y + 2) cos  = 0
or, (4x – 7y + 2) + tan  (3x + 2y + 1) = 0 which is of the form L1 +  L2 = 0
Hence each member of it will pass through a fixed point which is the intersection of 4x – 7y + 2 = 0 and
  11 2 
, 
3x + 2y + 1 = 0 i.e. 
 29 29 
(2) Concurrency of Straight Lines : The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2 + c2 = 0,
a3x + b3x + b3y + c3 = 0 to be concurrent is
(i)
a1
b1
c1
a2
b2
c2  0
a3
b3
c3
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STRAIGHT LINE
(ii)
Page # 19
There exist 3 constants l, m, n (not all zero at the same time) such that L1 + mL2 + nL3 = 0,
where L1 = 0, L2 = 0 and L3 = 0 ar the three given straight lines.
(iii) the three lines are concurrent if any one of the lines passes through the pint of intersection
of the other two lines.
I.
PAIR OF STRAIGHT LINES
The combined equation of pair of straight lines L1 = a1x + b1y + c1 = 0 and L2 = a2x + b2y + c2 = 0 is
(a1x + b1y + c1) (a2x + b2y + c2) = 0 i.e. L1L2 = 0. Opening the brackets and comparing the terms
with the terms of general equation of 2nd degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we can get all
the following results for a pair of straight lines.
The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of
a h g
straight lines if =
h b
f
g
c
f
= 0 and h2  ab.
abc + 2fgh – al2 – bg2 – ch2 = 0 and h2  ab.
The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines
through the origin if h2  ab.
If the lines through the origin whose joint equation is ax 2 + 2hxy + by 2 = 0, are y = m 1x and
y = m2x, then y2 – (m1 + m2) xy + m1m2x2 = 0 and y2 +
m1 + m 2 = –
2h
,
b
m1m2 =
2h
a
xy  x 2  0 are identical, so that
b
b
a
b
(m1  m 2 )2  4m1m 2
If  be the angle between two lines, through the origin, then tan = ±
1  m1m 2
=±
2 h 2  ab
ab
The lines are perpendicular if a + b = 0 and coincident if h2 = ab.
In the more general case, the lines represented by ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will be
perpendicular if a + b = 0, parallel if the terms of second degree make a perfect square i.e. ax2 + 2hxy + by2
gets converted into (l1x + 2hxy + by2 gets converted into (l1x ± m1y)2, coincident if the whole
equation makes a perfect square i.e. ax2 + 2hxy + by2 + 2fy + c can be written as (lx + my + n)2.
Note : Point of intersection of the two lines represented by ax2 + 2hxy + by2 + 2gx + 2fy +c = 0 is
obtained by solving the equations
f
f
f
= ax + hy + g = 0 and y = hx + by + f = 0 where
x
x
denotes the derivative of f with respect to y, keeping x constant. The fact can be used in splitting
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 into equations of two straight lines. With the above method, the
point of intersection can be found. Now only the slopes need to be determined.
If should be noted that the line ax + hy + g = 0 and hx + by + f = 0 are not the lines represented by
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. These are the lines concurrent with the lines represented by
given equation.
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57
STRAIGHT LINE
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(Homogenization) Joint equation of pair of lines joining the origin and the points of intersection of a
curve and a line :
I f the line lx + my + n = 0, ((n  0) i.e. the line does not pass through
origin) cuts the curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at two points
A
A and B, then the joint equation of straight lines passing through A and B
and the origin is given by homogenizing the equation of the curve by the
B
equation of the line. i.e.
2
 x  my   x  my 
  c
 =0
ax2 + 2hxy + by2 + (2gx + 2gy) 
 n   n 
is the equation of the line OA and OB.
O
Ex.19 If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0(a, b and c being distinct and different from 1)
are concurrent, then prove that
1
1
1


= 1.
1a 1b ac
a 1 1
Sol.
Since the given lines are concurrent
1 b 1
= 0.
1 1 c
a 1 a 1 a
Operating C2  C2 – C1 and C3  C3 – C1 , we get 1 b  1
1

0
0
=0
c 1
a 1 1
a
1
1
1 b 1


a (b – 1) (c – 1) – (b – 1) (1 – a) –(c – 1) (1 – 0) = 0 
= 0 
= 0.
1 a a  b 1 c
1 1 c
Ex.20 The chord
2
2 of the curve y + 1 = 4x subtends a right angle at origin then find the
6 y = 8 x +
value of .
Sol.
3 y  2x = 1 is the given chord. Homogenizing the equation of the curve, we get,
y2 – 4x( 3 y – 2x) + ( 3 y – 2x)2 = 0  (42 + 8)x2 + (+ 3)y2 – 4 3 xy – 4
Now, angle at origin is 90º

42 + 8 + + 3 = 0  42 + 9 + 3 = 0
3 xy = 0

coefficient of x2 + coefficient of y2 = 0

 9  81  48  9  33
.

8
8
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SINGLE CORRECT (OBJECTIVE QUESTIONS)
EXERCISE – I
 8
1. The points  0,  , (1,3) and (82,30) are vertices of
 3
(A) an obtuse angled triangle (B) an acute angled triangle
(C) a right angled triangle (D) none of these
2. The ratio in which the line joining the points (3, –4)
and (–5, 6) is divided by x-axis
(A) 2 : 3 (B) 6 : 4 (C) 3 : 2 (D) none of these
3. The circumcentre of the triangle with vertices
(0, 0), (3, 0) and (0, 4) is
(A) (1, 1)
(B) (2, 3/2)
(C) (3/2, 2)
(D) none of these
4. The mid points of the sides of a triangle are (5, 0),
(5, 12) and (0, 12), then orthocentre of this triangle is
(A) (0, 0) (B) (0, 24)
 13 
(D)  , 8 
 3

(C) (10, 0)
5. Area of a triangle whose vertices are (a cos , b sin),
(–a sin , b cos ) and (–a cos , –b sin ) is
(A) ab sin  cos 
(B) a cos  sin 
(C)
1
ab
2
6. The point A divides the join of the points (–5, 1)
and (3, 5) in the ratio k : 1 and coordinates of points
B and C are (1, 5) and (7, –2) respectively. If the
area of ABC be 2 units, then k equals
(A) 7, 9
(B) 6, 7
(C) 7, 31/9
(D) 9, 31/9
– cos), C(1, 2) are the
a varies, the locus of its
2
2
(B) x +y – 2x–4y+1 = 0
(D) none of these
8. The points with the co-ordinates (2a, 3a), (3b, 2b)
& (c, c) are collinear
(A) for no value of a, b, c
(B) for all values of a, b, c
(C) If a,
c
, b are in H.P..
5
10. The equation of the line cutting an intercept of 3
on negative y-axis and inclined at an angle tan
to the x-axis is
(A) 5y – 3x + 15 = 0
(C) 3y – 5x + 15 = 0
–1
3
5
(B) 5y – 3x = 15
(D) none of these
11. The equation of a straight line which passes
through the point (–3, 5) such that the portion of it
between the axes is divided by the point in the ratio
5 : 3 (reckoning from x-axis) will be
(A) x + y – 2 = 0
(B) 2x + y + 1 = 0
(C) x + 2y – 7 = 0
(D) x – y + 8 = 0
12. The co-ordinates of the vertices P, Q, R & S of
square PQRS inscribed in the triangle ABC with vertices
A(0, 0), B(3, 0) & C(2, 1) given that two of its vertices
P, Q are on the side AB are respectively
 1  3  3 1  1 1
(A)  , 0 ,  , 0 ,  ,  &  , 
4  8  8 8 4 8
(D) ab
7. If A(cos, sin), B(sin,
vertices of a ABC, then as
centroid is
2
2
(A) x + y – 2x – 4y+3=0
2
2
(C) 3(x + y ) – 2x – 4y+1=0
9. A stick of length 10 units rests against the floor
and a wall of a room. If the stick begins to slide on
the floor then the locus of its middle point is
2
2
2
2
(A) x + y = 2.5
(B) x + y = 25
2
2
(C) x + y = 100
(D) none
(D) if a,
2
c, b are in H.P..
5
 1  3  3 1  1 1
(B)  , 0 ,  , 0 ,  ,  &  , 
2  4  4 4 2 4
3  3 1  1
(C) (1, 0)  , 0 ,  ,  & 1, 
2  2 2  2
3  9  9 3 3 3
(D)  , 0 ,  , 0 ,  ,  &  , 
2  4  4 4 2 4
13. The equation of perpendicular bisector of the line
segment joining the points (1, 2) and (–2, 0) is
(A) 5x + 2y = 1
(B) 4x + 6y = 1
(C) 6x + 4y = 1
(D) none of these
14. The number of possible straight lines, passing
through (2, 3) and forming a triangle with coordinate
axes, whose area is 12 sq. units, is
(A) one
(B) two
(C) three
(D) four
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15. Points A & B are in the first quadrant ; point 'O' is
the origin. If the slope of OA is 1, slope of OB is 7 and
OA = OB, then the slope of AB is
(A) –1/5 (B) –1/4
(C) 1/3
(D) –1/2
23. If the point (a, 2) lies between the lines x–y–1=0
and 2(x – y) – 5 = 0, then the set of values of a is
(A) (–, 3)  (9/2, )
(B) (3, 9/2)
(C) (–, 3)
(D) (9/2, )
16. Coordinates of a point which is at 3 distance from
point (1, –3) of line 2x + 3y + 7 = 0 is
24. A(x1, y1), B(x2, y2) and C(x3, y3) are three noncollinear points in cartesian plane. Number of
parallelograms that can be drawn with these three
points as vertices are
(A) one
(B) two
(C) three
(D) four

9
6 

,3
(A) 1 
13
13 


9
6 

,3
(B) 1 
13
13 


9
6 

,3
(C) 1 
13
13



9
6 

,3
(D) 1 
13
13


17. The angle between the lines y – x + 5 = 0 and
3 x  y  7  0 is
(A) 15º
(B) 60º
(C) 45º
25. If P(1, 0) ; Q(–1, 0) & R(2, 0) are three give
points, then the locus of the points S satisfying the
2
2
2
relation, SQ + SR = 2 SP is
(A) A straight line parallel to x-axis
(B) A circle passing through the origin
(C) A circle with the centre at the origin
(D) A straight line parallel to y-axis
(D) 75º
26. The area of triangle formed by the lines x + y – 3 = 0,
x – 3y + 9 = 0 and 3x – 2y + 1 = 0
18. A line is perpendicular to 3x + y = 3 and passes
through a point (2, 2). Its y intercept is
(A) 2/3
(B) 1/3
(C) 1
(D) 4/3
(A)
16
sq. units
7
(C) 4 sq. units
19. The equation of the line passing through the point
(c, d) and parallel to the line ax + by + c = 0 is
(A) a(x + c) + b(y + d) = 0 (B) a(x + c) – b(y + d) = 0
(C) a(x – c) + b(y – d) = 0 (D) none of these
20. The position of the point (8, –9) with respect to
the lines 2x + 3y – 4 = 0 and 6x + 9y + 8 = 0 is
(A) point lies on the same side of the lines
(B) point lies on one of the lines
(C) point lies on the different sides of the line
(D) none of these
21. If origin and (3, 2) are contained in the same
angle of the lines 2x + y – a = 0, x – 3y + a = 0, then
'a' must lie in the interval
(A) (–, 0)  (8, )
(B) (–, 0)  (3, )
(C) (0, 3)
(D) (3, 8)
22. The line 3x + 2y = 6 will divide the quadrilateral
formed y the lines x + y = 5, y – 2x = 8, 3y + 2x = 0 &
4y – x = 0 in
(A) two quadrilaterals
(B) one pentagon and one triangle
(C) two triangles
(D) none of these
(B)
10
sq. units
7
(D) 9 sq. units
27. The co-ordinates of foot of the perpendicular drawn
on line 3x – 4y – 5 = 0 from the point (0, 5) is
(A) (1, 3) (B) (2, 3)
(C) (3, 2)
(D) (3, 1)
28. Distance of the point (2, 5) from the line 3x + y + 4 = 0
measured parallel to the line 3x – 4y + 8 = 0 is
(A) 15/2 (B) 9/2
(C) 5
(D) none
29. Three vertices of triangle ABC are A(–1, 11),
B(–9, –8) and C(15, –2). The equation of angle bisector
of angle A is
(A) 4x – y=7 (B) 4x + y=7 (C) x + 4y=7 (D) x–4y=7
30. If line y – x + 2 = 0 is shifted parallel to itself
towards the positive direction of the x-axis by a
perpendicular distance of 3 2 units, then the equation
of the new line is
(A) y = x – 4
(B) y = x + 1
(C) y = x – (2 + 3 2 )
(D) y = x – 8
31. The co-ordinates of the point of reflection of the
origin (0, 0) in the line 4x – 2y – 5 = 0 is
(A) (1, –2) (B) (2, –1)
4 2
(C)  , 
5 5
(D) (2, 5)
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32. If the axes are rotated through an angle of 30º in
the anti-clockwise direction, the coordinates of point
(4, – 2 3 ) with respect to new axes are
(A) (2, 3 )
(B) ( 3 , –5) (C) (2, 3) (D) ( 3 , 2)
33. Keeping the origin constant axes are rotated at
an angle 30º in clockwise direction then new coordinate
of (2, 1) with respect to old axes is
2 3 3 

,
(A) 

2
2


 2 3 1  2  3 

,
(B) 

2
2


39. Equation of the pair of straight lines through origin
and perpendicular to the pair of straight lines
2
2
5x – 7xy – 3y = 0 is
2
2
2
2
(A) 3x – 7xy – 5y = 0 (B) 3x + 7xy + 5y = 0
2
2
2
2
(C) 3x – 7xy + 5y = 0 (D) 3x + 7xy – 5y = 0
40. One of the diameter of the circle circumscribing
the rectangle ABCD is 4y = x + 7. If A and B are the
points (–3, 4) and (5, 4) respectively then the area of
rectangle is equal to
(A) 30
(B) 8
(C) 25
(D) 32
2
41. If the lines
 2 3 1 2  3 


,
(C) 
2
2 

(D) none of these
34. If one diagonal of a square is along the line x = 2y
and one of its vertex is (3, 0), then its sides through
this vertex are given by the equations
(A) y – 3x + 9 = 0, x – 3y – 3 = 0
(B) y – 3x + 9 = 0, x – 3y – 3 = 0
(C) y + 3x – 9 = 0, x + 3y – 3 = 0
(D) y – 3x + 9 = 0, x + 3y – 3 = 0
35. The line (p + 2q)x + (p – 3q)y = p – q for different
values of p and q passes through a fixed point whose
co-ordinates are
3 5
2 2
(A)  ,  (B)  , 
2 2
5 5
3 3
(C)  , 
5 5
2 3
(D)  , 
5 5
36. Given the family of lines, a(3x+4y+6) + b(x+y+2)=0.
The line of the family situated at the greatest distance
from the point P(2, 3) has equation
(A) 4x + 3y + 8 = 0
(B) 5x + 3y + 10 = 0
(C) 15x + 8y + 30 = 0
(D) none
37. The base BC of a triangle ABC is bisected at the
point (p, q) and the equation to the side AB & AC are
px + qy = 1 & qx + py = 1. The equation of the median
through A is
(A) (p–2q)x+(q–2p)y+1=0
(B) (p + q) x+y – 2=0
2
2
(C) (2pq – 1)(px + gy – 1)=(p + q – 1)(qx + py – 1)
(D) none
2
2
38. The equation 2x + 4xy – py + 4x + qy + 1 = 0
will represent two mutually perpendicular straight lines, if
(A) p = 1 and q = 2 or 6 (B) p = –2 and q = –2 or 8
(C) p = 2 and q = 0 or 8 (D) p = 2 and q = 0 or 6
xsin A + ysinA + 1 = 0
2
xsin B + ysinB + 1 = 0
2
xsin C + ysinC + 1 = 0
are concurrent where A, B, C are angles of triangle
then ABC must be
(A) equilateral
(B) isosceles
(C) right angle
(D) no such triangle exist
42. The co-ordinates of a point P on the line 2x–y+5=0
such that |PA – PB| is maximum where A is (4, –2)
and B is (2, –4) will be
(A) (11, 27) (B) (–11, –17)
(C) (–11, 17) (D) (0, 5)
43. The line x + y = p meets the axis of x and y at A
and B respectively. A triangle APQ is inscribed in the
triangle OAB, O being the origin, with right angle at Q,
P and Q lie respectively on OB and AB. If the area of
th
the triangle APQ is 3/8 of the area of the triangle
OAB, then
AQ
is equal to
BQ
(A) 2
(B) 2/3
44. Lines, L1 : x +
(C) 1/3
(D) 3
3 y  2 , and L2 : ax + by = 1, meet
at P and enclose an angle of 45º between them. Line
L3 : y =
2
3 x , also passes through P then
2
(A) a + b = 1
2
2
(C) a + b = 3
2
2
(B) a + b = 2
2
2
(D) a + b = 4
45. A triangle is formed by the lines 2x – 3y – 6 = 0 ;
3x – y + 3 = 0 and 3x + 4y – 12 = 0. If the points P(, 0)
and Q(0, ) always lie on or inside the ABC, then
(A)  [–1, 2] &  [–2, 3] (B)  [–1, 3] &  [–2, 4]
(C)  [–2, 4] &  [–3, 4] (D)  [–1, 3] &  [–2, 3]
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46. The line x + 3y – 2 = 0 bisects the angle between
a pair of straight lines of which one has equation
x – 7y + 5 = 0. The equation of the other line is
(A) 3x + 3y – 1 = 0
(B) x – 3y + 2 = 0
(C) 5x + 5y – 3 = 0
(D) none
47. A ray of light passing through the point A(1, 2) is
reflected at a point B on the x-axis and then passes
through (5, 3). Then the equation of AB is
(A) 5x + 4y = 13
(B) 5x – 4y = –3
(C) 4x + 5y = 14
(D) 4x – 5y = –6
48. Let the algebraic sum of the perpendicular
distances from the point (3, 0), (0, 3) & (2, 2) to a
variable straight line be zero, then the line passes
through a fixed point whose co-ordinates are
(A) (3, 2) (B) (2, 3)
3 3
(C)  , 
5 5
5 5
(D)  , 
3 3
49. The image of the pair of lines represented by
2
2
ax + 2h xy + by = 0 by the line mirror y = 0 is
2
2
2
2
(A) ax – 2hxy + by = 0 (B) bx – 2h xy + ay = 0
2
2
2
2
(C) bx + 2h xy + ay = 0 (D) ax – 2h xy – by = 0
2
2
50. The pair of straight lines x – 4xy + y = 0 together
with the line x + y + 4 6 = 0 form a triangle which is
(A) right angled but not isosceles (B) right isosceles
(C) scalene
(D) equilateral
51. Let A  (3, 2) and B  (5, 1). ABP is an equilateral
triangle is constructed on the side of AB remote from
the origin then the orthocentre of triangle ABP is
1
3


3,  3 
(A)  4 
2
2


1
3


3,  3 
(B)  4 
2
2


1
3 1


3, 
3
(C)  4 
6
2 3


1
3 1


3, 
3
(D)  4 
6
2 3


54. The circumcentre of the triangle formed by the
lines, xy + 2x + 2y + 4 = 0 and x + y + 2 = 0 is
(A) (–1, –1)
(B) (–2, –2) (C) (0, 0) (D) (–1, –2)
55. Area of the rhombus bounded by the four lines,
ax ± by ± c = 0 is
(A)
c2
2ab
(B)
2c 2
| ab |
(C)
4c 2
ab
ab
(D)
4c 2
56. If the lines ax + y + 1 = 0, x + by + 1 = 0 &
x + y + c = 0 where a, b & c are distinct real numbers
different from 1 are concurrent, then the value of
1
1
1
+
+
equals
1 a
1 b
1 c
(A) 4
(B) 3
(C) 2
(D) 1
57. The area enclosed by 2 | x | + 3| y |  6 is
(A) 3 sq. units
(B) 4 sq. units
(C) 12 sq. units
(D) 24 sq. units
58. The point (4, 1) undergoes the following three
transformations successively
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the
positive direction of x-axis
(iii) Rotation through an angle /4 about the origin in
the counter clockwise direction.
The final position of the points is given by the
coordinates
 7
1 

,
(A) 
2
 2
 7
1 

,
(B) 
 2 2
 1 7 

,
(C)  
2
2

(D) none of these
52. The line PQ whose equation is x – y = 2 cuts the
x-axis at P and Q is (4, 2). The line PQ is rotated
about P through 45º in the anticlockwise direction.
The equation of the line PQ in the new position is
(A) y   2
(B) y = 2 (C) x = 2
(D) x = –2
53. Distance between two lines represented by the
2
2
line pair, x – 4xy + 4y + x – 2y – 6 = 0 is
1
(A)
(B) 5
(C) 2 5
(D) none
5
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MULTIPLE CORRECT (OBJECTIVE QUESTIONS)
EXERCISE – II
1. If one vertex of an equilateral triangle of side 'a'
lies at the origin and the other lies on the line
3 y = 0 then the co-ordiantes of the third
x –
6. The straight lines x + y = 0, 3x + y – 4 = 0 and
x + 3y – 4 = 0 form a triangle which is
(A) isosceles
(B) right angled
(C) obtuse angled
(D) equilateral
vertex are
 3 a a

3 a a 
,   (C) (0, –a) (D)  
,
(B) 


2
2 2 
 2

(A) (0, a)
2. If one diagonal of a square is the portion of the line
x y
  1 intercepted by the axes, then the extremities
a b
of the other diagonal of the square are
ab ab
,

(A) 
2 
 2
ab ab
,

(B) 
2 
 2
ab ba
,

(C) 
2 
 2
ab ba
,

(D) 
2 
 2
x
y
+
= 1 is a line through the intersection of
c
d
3. If
x
y
x
y
+
= 1 and
+
= 1 and the lengths of the
a
b
b
a
perpendiuclars drawn from the origin to these lines
are equal in lengths then
1
(A)
(C)
a
2
1
+
b
2
1
=
c
2
1
+
1
1
1
1
+
=
+
a
b
c
d
2
d
1
(B)
a
1
2
–
b
2
1
=
c
2
1
–
d2
(D) none
4. A and B are two fixed points whose co-ordinates
are (3, 2) and (5, 4) respectively. The co-ordinates
of a point P if ABP is an equilateral triangle, is/are

(C) 3 

3
(A) 4  3 , 3  3
3, 4 

(D) 3 

3
(B) 4  3 , 3  3
3, 4 
5. Straight lines 2x + y = 5 and x – 2y = 3 intersect at
the point A. Points B and C are chosen on these two
lines such that AB = AC. Then the equation of a line
BC passing through the point (2, 3) is
(A) 3x – y – 3 = 0
(B) x + 3y – 11 = 0
(C) 3x + y – 9 = 0
(D) x – 3y + 7 = 0
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Page # 26
EXERCISE – III
SUBJECTIVE QUESTIONS
x y

= 1 intersects the x and y axes at M and
6 8
N respectively. If the coordinates of the point P lying
inside the triangle OMN (where ‘O’ is origin) are (a, b)
such that the areas of the triangle POM, PON and
PMN are equal. Find
(a) the coordinates of the point P and
1. Line
(b) the radius of the circle escribed opposite to the
angle N.
2. Two vertices of a triangle are (4, –3) & (–2, 5). If
the orthocentre of the triangle is at (1, 2), find the
coordinates of the third vertex.
3. The point A divides the join of P(–5, 1) & Q(3, 5) in
the ratio K : 1. Find the two values of K for which the
area of triangle ABC, where B is (1, 5) and C is (7,–2),
is equal to 2 units in magnitude.
4. Determine the ratio in which the point P(3, 5) divides
the join of A(1, 3) and B(7, 9). Find the harmonic
conjugate of P w.r.t. A & B.
5. A line is such that its segment between the straight
lines 5x – y – 4 = 0 and 3x + 4y – 4 = 0 is bisected at
the point (1, 5). Obtain the equation.
6. A line through the point P(2, – 3) meets the lines
x – 2y + 7 = 0 and x +3y – 3 = 0 at the points A and B
respectively. If P divides AB externally in the ratio 3 : 2
then find the equation of the line AB.
7. The area of a triangle is 5. Two of its vertices are
(2, 1) & (3, –2). The third vertex lies on y = x + 3. Find
the third vertex.
8. A variable line, drawn through the point of intersection
x y
x y
of the straight lines  = 1 & 
= 1, meets the
a b
b a
coordinate axes in A & B. Show that the locus of
the mid point of AB is the curve 2xy (a + b) = ab(x + y)
9. In the xy plane, the line ‘1’ passes through the
point (1, 1) and the line ‘2’ passes through the point
(–1, 1). If the difference of the slopes of the lines is 2.
Find the locus of the point of intersection of the lines
1 and 2.
10. Two consecutive sides of a parallelogram are
4x + 5y = 0 & 7x + 2y = 0. If the equation to one
diagonal is 11x + 7y = 9, find the equation to the
other diagonal.
11. The line 3x + 2y = 24 meets the y-axis at A and
the x-axis at B. The perpendicular bisector of AB meets
the line through (0, –1) parallel to x-axis at C. Find
the area of the triangle ABC.
12. If the straight line drawn through the point P( 3, 2)

and inclined at an angle
with the x-axis, meets the
6
line
3 x – 4y + 8 = 0 at Q. Find the length PQ.
13. Find the area of the triangle formed by the
straight lines whose equations are x + 2y – 5 = 0,
2x + y – 7 = 0 and x – y + 1 = 0 without determining
the coordinates of the vertices of the triangle. Also
compute the tangent of the interior angles of the
triangle and hence comment upon the nature of
triangle.
14. A triangle has side lengths 18, 24 and 30. Find the
area of the triangle whose vertices are the incentre,
circumcentre and centroid of the triangle.
15. The points (1, 3) & (5, 1) are two opposite vertices
of a rectangle. The other two vertices lie on the line
y = 2x + c. Find c and the remaining vertices.
16. A straight line L is perpendicular to the line 5x – y = 1.
The area of the triangle formed by the line L & the
coordinate axes is 5. Find the equation of the line.
17. The triangle ABC, right angled at C, has median
AD, BE and CF.AD lies along the line y = x + 3, BE lies
along the line y = 2x + 4. If the length of the
hypotenuse is 60, find the area of the triangle ABC.
18. Two equal sides of an isosceles triangle are given by
the equations 7x – y + 3 = 0 and x + y – 3 = 0 and its
third side passes through the point (1, – 10). Determine
the equation of the third side.
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19. The equations of the perpendicular bisectors
of the sides AB and AC of a triangle ABC are
x – y + 5 = 0 and x + 2y = 0, respectively. If the point
A is (1, –2) find the equation of the line BC.
20.
If
and
(x1 – x2)2 + (y1 – y2)2 = a2
(x2 – x3)2 + (y2 – y3)2 = b2
(x3 – x1)2 + (y3 – y1)2 = c2
x1 y1 1
then  x 2 y 2 1
x3 y3 1
2
= (a + b + c)(b + c – a)(c+ a – b)
(a + b – c). Find the value of .
21. Given vertices A(1, 1), B(4, –2) and C(5, 5) of a
triangle, find the equation of the perpendicular dropped
from C to the interior bisector of the angle A.
22. Triangle ABC lies in the Cartesian plane and has an
area of 70 sq. units. The coordinates of B and C are
(12, 19) and (23, 20) respectively and the coordinates
of A are (p, q). The line containing the median to the
side BC has slope –5. Find the largest possible value
of (p + q).
23. Determine the range of values of   [0, 2] for
which the point (cos , sin ) lies inside the triangle
formed by the lines x + y = 2; x – y = 1 and
6x + 2y – 10 = 0.
24. The points (–6, 1), (6, 10), (9, 6) and (–3, –3)
are the vertices of a rectangle. If the area of the
portion of this rectangle that lies above the x-axis is
a/b, find the value of (a + b), given a and b are
coprime.
25. Let ABC be a triangle such that the coordinates
of A are (–3, 1). Equation of the median through B is
2x + y – 3 = 0 and equation of the angular bisector of
C is 7x – 4y – 1 = 0. Then match the entries of
column-I with their corresponding correct entries of
column-II.
Column–I
Column–II
(A) Equation of the line AB is
(P) 2x + y – 3 = 0
(B) Equation of the line BC is
(Q) 2x – 3y + 9 = 0
(C) Equation of the line CA is
(R)4x + 7y + 5 = 0
(S)18x – y – 49 = 0
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EXERCISE – IV
ADVANCED SUBJECTIVE QUESTIONS
1. Consider a triangle ABC with sides AB and AC having
the equations L1 = 0 and L2 = 0. Let the centroid,
orthocentre and circumcentre of the  ABC are G, H
and S respectively. L = 0 denotes the equation of side
BC.
(a) If L1 : 2x – y = 0 and L2 : x + y = 3 and G (2, 3)
then find the slope of the line L = 0.
(b) If L1 : 2x + y = 0 and L2 : x – y + 2 = 0 and H(2, 3)
then find the y-intercept of L = 0.
(c) If L1 : x+y – 1=0 and L2 : 2x – y + 4=0 and S(2, 1)
then find the x-intercept of the line L = 0.
2. The equations of perpendiculars of the sides AB
and AC of triangle ABC are x–y–4=0 and 2x–y–5=0
respectively. If the vertex A is (–2, 3) and point of
3 5
intersection of perpendiculars bisectors is  ,  , find
2 2
the equation of medians to the sides AB and AC
respectively.
3. The interior angle bisector of angle A for the triangle
ABC whose coordinates of the vertices are A(–8, 5);
B(–15, –19) and C(1, – 7) has the equation ax+2y+c=0.
Find ‘a’ and ‘c’.
4. Find the equation of the straight lines passing through
(–2, –7) and having an intercept of length 3 between
the straight lines 4x + 3y = 12, 4x + 3y = 3.
8. Find the equations of the sides of a triangle having
(4, –1) as a vertex, if the lines x – 1 = 0 and
x – y – 1 = 0 are the equations of two internal bisectors
of its angles.
9. P is the point (–1, 2), a variable line through P cuts
the x and y axes at A and B respectively Q is the
point on AB such that PA, PQ, PB are H.P. Show that
the locus of Q is the line y = 2x.
10. The equations of the altitudes AD, BE, CF of a
triangle ABC are x + y = 0, x – 4y = 0 and 2x – y = 0
respectively. The coordinates of A are (t, –t). Find
coordinates of B and C. Prove that if t varies the
locus of the centroid of the triangle ABC is x + 5y = 0.
11. The distance of a point(x1, y1) from each of two
straight lines which passes through the origin of coordinates is ; find the combined equation of these
straight lines.
12. Consider a  ABC whose sides AB, BC and CA are
represented by the straight lines 2x + y = 0, x + py=q
and x – y = 3 respectively. The point P is (2, 3).
(a) If P is the centroid, then find the value of (p + q).
(b) If P is the orthocentre, then find the value of (p + q).
(c) If P is the circumcentre, then find the value of (p + q).
5. Two sides of a rhombus ABCD are parallel to the
lines y = x + 2 and y = 7x + 3. If the diagonals of the
rhombus intersect at the point (1, 2) and the vertex
A is on the y-axis, find the possible coordinates of A.
6. A triangle is formed by the lines whose equations
are AB : x + y – 5 = 0, BC : x + 7y – 7 =0 and
CA : 7x + y + 14 = 0. Find the bisector of the interior
angle at B and the exterior angle at C. Determine the
nature of the interior angle at A and find the equation
of the bisector.
7. A point P is such that its perpendicular distance
from the line y – 2x + 1 = 0 is equal to its distance
from the origin. Find the equation of the locus of the
point P. Prove that the line y = 2x meets the locus in
two points Q and R, such that the origin is the mid
point of QR.
13. Consider a line pair 2x2+3xy–2y2–10x+15y–28=0
and another line L passing through origin with gradient
3. The line pair and line L form a triangle whose vertices
are A, B and C.
(a) Find the sum of the cotangents of the interior
angles of the triangle ABC.
(b) Find the are a of triangle ABC
(c) Find the radius of the circle touching all the 3
sides of the triangle.
14. Show that all the chords of the curve
3x2 – y2 – 2x + 4y = 0 which subtend a right angle at
the origin are concurrent. Does this result also hold
for the curve, 3x2 + 3y2 – 2x + 4y = 0? If yes, what is
the point of concurrency and if not, give reasons.
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15. A straight line is drawn from the point (1, 0) to
the curve x2 + y2 + 6x – 10y + 1 = 0, such that the
intercept made on it by the curve subtends a right
angle at the origin. Find the equations of the line.
16. The two line pairs y2 – 4y + 3 = 0 and
x2 + 4xy + 4y2 – 5x – 10y + 4 = 0 enclose a 4 sided
convex polygon find
(i) area of the polygon
(ii) length of its diagonals.
17. Find the equation of the two straight lines which
together with those given by the equation
6x2 – xy – y2 + x + 12y – 35 = 0 will make a
parallelogram whose diagonals intersect in the origin.
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Page # 29
67
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Page # 30
JEE PROBLEMS
EXERCISE – V
1. (a) The incentre of the triangle with vertices
(A)
(B) x +
3y=0
3 ), (0, 0) and (2, 0) is [JEE 2000(Scr.), 1 + 1]
 3
2 3 
2 1 
 1 



 (C)  ,

(A) 1, 2  (B)  3 ,
(D)  1,




3
2
3
3






3
x+y=0
2
(C)
3x+y=0
(D) x +
3
y=0
2
(1,
(b) Let PS be the median of the triangle with vertices,
P(2, 2), Q(6, –1) and R(7, 3). The equation of the line
passing through (1, –1) and parallel to PS is
(A) 2x – 9y – 7 = 0
(B) 2x – 9y – 11 = 0
(C) 2x + 9y – 11 = 0
(D) 2x + 9y + 7 = 0
(c) For points P(x1, y1) and Q(x2, y2) of the coordinate plane, a new distance d(P, Q) is defined by
d(P, Q)=|x1–x2| + |y1 – y2|. Let O(0, 0) and A(3, 2).
Prove that the set of points in the first quadrant which
are equidistant (with respect to the new distance)
from O and A consists of the union of a line segment
of finite length and an infinite ray. Sketch this set in a
labelled diagram.
[JEE 2000(Mains), 10]
(b) A straight line through the origin O meets the
parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points
P and Q respectively. Then the point O divides the
segment PQ in the ratio
(A) 1 : 2 (B) 3 : 4
(C) 2 : 1
(D) 4 : 3
(c) The area bounded by the curves y = | x | – 1 and
y = – | x | + 1 is
[JEE 2002(Scr.)]
(A) 1
(B) 2
(C) 2 2
(D) 4
(d) A straight line L through the origin meets the line
x + y = 1 and x + y = 3 at P and Q respectively.
Through P and Q two straight lines L1 and L2 are
drawn, parallel to 2x – y = 5 and 3x + y = 5 respectively.
2. Find the position of point (4, 1) after it undergoes
the following transformations successively.
(i) Reflection about the line, y = x – 1
Lines L1 and L2 intersect at R. Show that the locus of
(ii) Translation by one unit along x–axis in the positive
direction.
(e) A straight line L with negative slope passes through
R, as L varies, is a straight line. [JEE 2002 (Mains)]
the point (8, 2) and cuts the positive coordinates
axes at points P and Q. Find the absolute minimum
(iii) Rotation through an angle /4 about the origin in
the anti-clockwise direction. [REE 2000(Mains), 3]
3. (a) Area of the parallelogram formed by the lines
y = mx, y = mx + 1, y = nx and y = nx + 1 equals
|mn|
2
1
1
(A)
(B) | m  n |
(C) | m  n | (D) | m  n |
2
(m  n)
(b) The number of integer values of m, for which the
x co-ordinate of the point of intersection of the lines
3x + 4y = 9 and y = mx + 1 is also an integer, is
[JEE 2001(Scr.)]
(A) 2
(B) 0
(C) 4
(D) 1
4. (a) Let P(–1, 0), Q(0, 0) and R(3, 3 3 ) be three
points. Then the equation of the bisector of the angle
PQR is
value of OP + OQ, as L varies, where O is the origin.
[JEE 2002 (Mains), 5]
5. The area bounded by the angle bisectors of the
lines x2 – y2 + 2y = 1 and the line x + y = 3, is
(A) 2
(B) 3
(C) 4
(D) 6
[JEE 2004 (Scr.)]
6. The area of the triangle formed by the intersection
of a line parallel to x-axis and passing through P(h,k)
with the lines y = x and x + y = 2 is 4h2. Find the
locus of the point P.
[JEE 2005(Mains), 2]
7. (a) Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of
the triangle OPQ. The point R inside the triangle OPQ
is such that the triangles OPR, PQR, OQR are of equal
area. The coordinates of R are
[JEE 2007, 3 + 3]
(A) (4/3, 3) (B) (3, 2/3) (C) (3, 4/3)
(D) (4/3, 2/3)
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(b) Lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect
the line L3 : y + 2 = 0 at P and Q, respectively. The
bisector of the acute angle between L 1 and L 2
intersects L3 at R.
Statement-1: The ratio PR : RQ equals 2 2 : 5
because
Statement-2: In any triangle, bisector of an angle
divides the triangle into two similar triangles.
(A) Statement-1 is true, statement-2 is true;
statement-2 is a correct explanation for statement-1
(B) Statement-1 is true, statement-2 is true;
statement-2 is NOT a correct explanation for
statement-1
(C) Statement-1 is true, statement-2 is false
(D) Statement-1 is false, statement-2 is true
8. Consider the lines given by
[JEE 2008, 6]
L1 = x + 3y – 5 = 0
L2 = 3x – ky – 1 = 0
L3 = 5x + 2y – 12 = 0
Match the statements/Expression in Column-I with the
statements/Expressions in Column-II and indicate your
answer by darkening the appropriate bubbles in the
4 × 4 matrix given in OMR.
Column–I
Column–II
(A) L1, L2, L3 are concurrent, if
(P) k = – 9
(B) One of L1, L2, L3 is parallel to
at least one of the other two, if
(Q) k = –
6
5

6
(C) L1, L2, L3 form a triangle, if
(R) k =
(D) L1, L2, L3 do not form a triangle, if
(S) k = 5
9. The locus of the orthocentre of the triangle formed
by the lines
[JEE 2009, 3]
(1 + p) x – py + p(1 + p) = 0,
(1 + q) x –qy + q(1 + q) = 0
and y = 0, where p  q, is
(A) a hyperbola
(B) a parabola
(C) an ellipse
(D) a straight line
10. A straight line L through the point (3, –2) is
inclined at an angle 60º to the line
3 x + y = 1. If L
also intersects the x-axis, then the equation of L is
(A) y  3x  2  3 3  0
(B) y  3x  2  3 3  0
(C)
(D)
3y  x  3  2 3  0
3y  x  3  2 3  0
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SINGLE CORRECT (OBJECTIVE QUESTIONS)
Answer Ex–I
1. D
2. A
3. C
4. A
5. D
6. C
7. C
8. D
9. B
10. A
11. D
12. D
13. C
14. C
15. D
16. B
17. A
18. D
19. C
20. A
21. A
22. A
23. B
24. C
25. D
26. B
27. D
28. C
29. B
30. D
31. B
32. B
33. B
34. D
35. D
36. A
37. C
38. C
39. A
40. D
41. B
42. B
43. D
44. B
45. D
46. C
47. A
48. D
49. A
50. D
51. D
52. C
53. B
54. A
55. B
56. D
57. C
58. C
MULTIPLE CORRECT (OBJECTIVE QUESTIONS)
Answer Ex–II
1. A,B,C,D
2. A,C
3. A,C
4. A,B
5. A,B
6. A,C
SUBJECTIVE QUESTIONS
Answer Ex–III
 8
1. (a)  2,  ; (b) 4 2. (33, 26)
 3
3. K = 7 or 31/9
4. 1 : 2; Q(–5, – 3)
9. y = x2 and y = 2 – x2
5. 83x – 35y + 92 = 0
6. 2x + y – 1 = 0
 7 13 
 3 3
7.  ,  or   , 
2 2 
 2 2
10. x – y = 0
11. 91 sq. units
12. 6 units
14. 3 units
15. c = – 4; B(2, 0); D(4, 4)
17. 400 sq. units
18. x – 3y – 31 = 0 or 3x + y + 7 = 0
21. x – 5 = 0
22. 47 23. 0 <  <
5
– tan–1 3
6
13.
3

3
sq. units,  3,3,  , isosceles
4

2
16. x + 5y + 5 2 = 0 or x + 5y–5 2 =0
19. 14x + 23 y = 40
24. 533
20. 4
25. (A)–R ; (B)–S ; (C)–Q
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ADVANCED SUBJECTIVE QUESTIONS
Answer Ex–IV
1. (a) 5 ; (b) 2 ; (c)
3
2
5. (0, 0) or (0, 5/2)
2. x + 4y = 4; 5x + 2y = 8
13. (a)
8. 2x – y + 3 = 0, 2x + y – 7 = 0, x – 2y – 6 = 0
11. ( y12   2 )x 2  2x1y1xy  ( x12   2 )y 2  0
 1 2
63
3
50
(8 5  5 10 ) 14. (1, –2), yes  , 
; (b)
; (c)
10
10
3 3
7
16. (i) area = 6sq. units, (ii) diagonals are
5&
15. x + y = 1; x + 9y = 1
JEE PROBLEMS
2. (4, 1) (2, 3)  (3, 3)  (0, 3 2 )
4. (a) C; (b) B; (c) B; (d) x – 3y + 5 = 0 ; (e) 18
7. (a) C ; (b) C
12. (a) 74 ; (b) 50 ; (c) 47
17. 6x2 – xy – y2 – x – 12y – 35 = 0
53
Answer Ex–V
1. (a) D ; (b) D
4. 7x + 24y + 182 = 0 or x = – 2
6. 3x + 6y – 16 = 0; 8x + 8y + 7 = 0; 12x + 6y – 11 = 0
7. x2 + 4y2 + 4xy + 4x – 2y – 1 = 0
t 
10. B   2t ,  t  , C  , t 
3
6
2 


3. a = 11, c = 78
3. (a) D ; (b) A
5. A
8. (A)–S ; (B)–P,Q ; (C)–R ; (D)–P,Q,S
6. y = 2x + 1, y = – 2x + 1
9. D
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10. B
71
Solution Slot -3
(English Medium)
CONTENTS
S.NO.
TOPIC
.....................
PAGE NO.

Straight Line ................................................................................................................ 3 – 37
Exercise - I .................................................................................................................... 3 – 13
Exercise - II .................................................................................................................. 14 – 15
Exercise - III ................................................................................................................. 16 – 22
Exercise - IV ................................................................................................................ 23 – 30
Exercise - V ................................................................................................................. 31 – 35
Answer Key .................................................................................................................. 36 – 37

Circle ........................................................................................................................... 38 – 72
Exercise - I ................................................................................................................... 38 – 47
Exercise - II .................................................................................................................. 48 – 50
Exercise - III ................................................................................................................. 51 – 58
Exercise - IV ................................................................................................................ 59 – 63
Exercise - V ................................................................................................................. 64 – 70
Answer Key .................................................................................................................. 71 – 72

Binomial Theorem ..................................................................................................... 73 –95
Exercise - I ................................................................................................................... 73 – 77
Exercise - II .................................................................................................................. 78 – 79
Exercise - III ................................................................................................................. 80 – 88
Exercise - IV ................................................................................................................ 89 – 92
Exercise - V ................................................................................................................. 93 – 94
Answer Key .................................................................................................................. 95 – 95

Matrices & Determinants .......................................................................................... 96 – 134
Exercise - I .................................................................................................................. 96 – 108
Exercise - II ................................................................................................................. 109 – 111
Exercise - III ................................................................................................................ 112 – 117
Exercise - IV .............................................................................................................. 118 – 126
Exercise - V ............................................................................................................... 127 – 132
Answer Key ................................................................................................................ 133 – 134
394 - Rajeev Gandhi Nagar Kota,Ph. No. 0744-2209671, 93141-87482,
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72
Page #S.NO.
2
TOPIC
.....................
NO.
Solution Slot – 3 PAGE
(Mathematics)

Complex Number .....................................................................................................
Exercise - I .................................................................................................................
Exercise - II ................................................................................................................
Exercise - III ...............................................................................................................
Exercise - IV ..............................................................................................................
Exercise - V ...............................................................................................................
Answer Key ................................................................................................................
135 – 162
135 – 142
143 – 143
144 – 149
150 – 153
154 – 159
160 – 162

Parabola ...................................................................................................................
Exercise - I .................................................................................................................
Exercise - II ................................................................................................................
Exercise - III ...............................................................................................................
Exercise - IV ..............................................................................................................
Exercise - V ...............................................................................................................
Answer Key ................................................................................................................
163 – 178
163 – 165
166 – 167
168 – 171
172 – 173
174 – 177
178 – 178

Ellipse .......................................................................................................................
Exercise - I .................................................................................................................
Exercise - II ................................................................................................................
Exercise - III ...............................................................................................................
Exercise - IV ..............................................................................................................
Exercise - V ...............................................................................................................
Answer Key ................................................................................................................
179 – 196
179 – 181
182 – 184
185 – 187
188 – 190
191 – 195
196 – 196

Hyperbola .................................................................................................................
Exercise - I .................................................................................................................
Exercise - II ................................................................................................................
Exercise - III ...............................................................................................................
Exercise - IV ..............................................................................................................
Exercise - V ...............................................................................................................
Answer Key ................................................................................................................
197 – 213
197 – 199
200 – 202
203 – 206
207 – 209
210 – 212
213 – 213

Vector ........................................................................................................................
Exercise - I .................................................................................................................
Exercise - II ................................................................................................................
Exercise - III ...............................................................................................................
Exercise - IV ..............................................................................................................
Exercise - V ...............................................................................................................
Answer Key ................................................................................................................
214 – 241
214 – 221
222 – 223
224 – 229
230 – 234
235 – 239
240 – 241

Three Dimensional Geometry (3-D) ........................................................................
Exercise - I .................................................................................................................
Exercise - II ................................................................................................................
Exercise - III ...............................................................................................................
Exercise - IV ..............................................................................................................
Exercise - V ...............................................................................................................
Answer Key ................................................................................................................
242 – 264
242 – 248
249 – 250
251 – 253
254 – 257
258 – 262
263 – 264
394 - Rajeev Gandhi
Kota,
Ph.
No. Kota,
0744-2209671,
93141-87482,
93527-21564
394Nagar
- Rajeev
Gandhi
Nagar
Ph. No. 0744-2209671,
93141-87482,
93527-21564
www.
motioniitjee.com
,
email-info@motioniitjee.com
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73
Solution Slot – 3 (Mathematics)
Page # 3
STRAIGHT LINE
HINTS & SOLUTIONS
EXERCISE – I
Sol.1 D
since the points are collinear option D is
correct
Sol.2 A
6  4
5  3
= x,
=0
3
 1
(3, 4)  : 1
(–5, 6)  = 2
3
(x, 0)
2
3
10  9
1
3
=
=–
2
23
5
1
3
x1  10 ,
y1  0  A(10,0) on x  axis

x2  0 , y2  24  B(a24) on y  axis
x3  0 , y3  0 
C(40) is origin
ABC is right angled  orthocentre is (0, 0)
Sol.5 D
a cos  b sin  1
1
a sin  b cos  1
= 2
a cos  b sin  1
 5,
x=
R R R
3
11  

Sol.3 C
 right angled
(0, 4)
0
a sin 
=
0
2
b cos  1
a cos  b sin  1
1
. 2 (ab sin2 + ab cos2) = ab
2
Sol.6 C
 3k  5 5k  1 
,


 k 1 k 1 
(0, 0)
 circum centre
(3, 0)
k
3 
= mid point of hypotaneous =  ,2 
2 
Q (3, 5)
1
A
P
P(–5, 11)
Sol.4 A
x1  x3  10 , y1  y3  0

 x2  x3  0 , y2  y3  24
x  x  10 , y  y  24
2
2
2
 1
B (x2,y2)
C(7, –2)
B(1, 5)
3k  5 5k  1
1
k 1
k 1
1
5
1
1
= |2|
2
7
2 1
  6k  10 35k  7 


1. (–2 – 3) –1. 
k 1 
 k 1
(0,12)
(5, 12)
(x3, y3) C
(5, 0) x(x1, y1)
x1 = x2 = 10, y1 – y2 = –24
x1 = 10, y1 = 0
x2 = 0, y2 = 24
x3 = 0, y3 = 0
 15k  25 5k  1 

 =±4
+ 
k 1 
 k 1
6k – 10 + 35 k + 7 + 15k – 25 – 5k – 1
= ± 4 + 37 (k + 1)
51 k – 29 = 41 k + 41 or 51 k – 29
= 33k + 33
10 k = 70 or 18 k = 62
k=7
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k=
31
9
74
Solution Slot – 3 (Mathematics)
Page # 4
Sol.11 D
Sol.7 C
Let centroid is (h, k)
3a  0
0  5b
,5=
53
53
a = –3, b = 8
(–3, 5) 3
x
y
+
=1
5
8
8
–x+y=8
x–y+8=0
(a, 0)
–3=
cos   sin   1
sin   cos   2
&k=
3
3
cos + sin = 3h – 1 & sin – cos =3k –2
squaring & adding
2 = (3h – 1)2 + (3k – 2)2 Locus of (h, k)
(3x –1)2 + (3k – 2)2 = 2
3(x2 + y2) – 2x – 4y + 1 = 0
then h =
Sol.8 D
(2a, 3a), (3b, 2b) & (c, c) are collinear
(0, b)
Sol.12 D
Let side of square is a units
equation of OC is 2y = x
S(2a, a)  R(3a, a)
C(2,1)
2a 3a 1

3b 2b 1
c
c
(2a, a)
=0
a
 (3bc – 2bc) – (2ca – 3ca)
+ (4ab – 9ab) = 0
 bc + ca + 5ab = 0
2a P
B(3, 0)
3 
9 
9 3
3 3
P  ,0  , Q  ,0  , R  ,  & S  , 
4
4
2
4






2 4
Sol.9 B
P is a mid point AB
Sol.13 C
Perpendicular bisector of slopoe of line BC
20
2
mBC =
=
12
3
A
B
C(–2, 0)
3
(1, 2)
mAP =
P
2
(0, 2k)
P(h, k)
(2h, 0)
1  2 2  0 
 1 
,
    ,1 
A= 
2
2


 2 
AB = 10 units
(2h)2 + (2k)2 = 102
h2 + k2 = 25
Locus of (h, k)
x2 + y2 =2 5
3  x  1 

  4y – 4 = – 6x – 3
2
2 
 6x + 4y = 1
locus of P
y–1=
Sol.14 C
Equation y – 3 = m (x – 2)
cut the axis at
Sol.10 A
3
tan  =
5
a
a
Q
0 1
Slope mBC =
= –1
32
B = 45º in QBR
QB = a
OB = OP + PQ = QB
3
3 = 2a + a + a  a =
4
2C
, b are in H.P..
5
 = tan–1
R(3a, a)
45º
O
2
2
5
1
1
1
1
2


.
=
+
  C  =
+
2
c
a
b
a
b
 5 
 a,
a
S
1
3
C = –3
5
y
  tan1
O
3
5
(2, 3)
x
3
x– 3
5
3x – 5y–15=0
O
y=
(0, –3)
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Solution Slot – 3 (Mathematics)
Page # 5
Sol.16 B
2x + 3y + 7 = 0
2m  3
y = 0 & x =
m
tan  =
x = 0 & y = – (2m – 3)
1 (2m  3)
.
{(2m  3)}
2
m
Area  = 12 =
(2m – 3)2 = ± 24m
4m2 – 12m + 9 = 24m
or 4m2 – 12m + 9 = – 24m
4m2 – 3y m + 9 = 0
D>0
or 4m2 + 12m + 9 = 0
(2m + 3)2 = 0
two distinct root of m
no. of values of m is 3.
2
3
2
 sin =
, cos =
13
13
3
y3
x 1
=
=±3
2
3
13
13
3

9
6 
1 

,3 
13
13 

(1, –3)
3

9
36

,
or 1 
13
13 

Sol.17 A
y–x+5=0,
3x–y+7=0
m1 = 1
m2 = 3
1 = 45º
2 = 60º
 = 60º – 45º =1 5º
Sol.15 D
OA line y = x, m1 = tan1 = 1
OB line y = 7m, m2 = tan2 = 7
3 1
Aliter tan  =
y
42 3
=2– 3
3 1
 = 15º
y =7x
B
Sol.18 D
 to 3x + y = 3, passes (2, 2)
y =x
r
A
r
=
1 3
m=+
1
& (2, 2)
3
2
1
O
y–2=+
x
1
(x – 2)
3
 –x + 3y = 4 
A, B lies in Ist quadrant
OA = OB = r (let)
x
y
x
y
OA line cos  = sin  = r 
=
=r
1
1
1
1
2
2
 r
r 

,
A 
2
 2
OB line
x
1
5 2
 r
7r 

,
= r  B . 
4 2 5 2 
y
7
=
5 2
7r

Sol.19 C
required line should be
ax + by +  = 0 satsify (c, d)
ac + bd +  = 0  = – (ac + bc)
ax + by – (ac + bc) = 0
a (x – c) + b (y – d) = 0
Sol.20 A
L1 : 2x + 3y – 4 = 0
L2 : 6x + 96 + 8 = 0 ,
P(8, – 9)
L1(P) = 2.8 – 3.9 – 4 = 16 – 27 – 4 = – 15<0
L2(O) = 48 – 81 + 8 + 8 = – 25 < 0
point (8, –9) lies same side of both lines.
Sol.21 C
L1 ; 2x + y – a = 0
0 (0, 0), P (3, 2)
r
2
1
2 7r  5r
=
=
=–
r
r  5r
4
2

5 2
2
5 2
Slope mAB =
1
x
4
y
+
=1b=
4
4
3
3
O
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3
8
76
Solution Slot – 3 (Mathematics)
Page # 6
L2 : x – 3y + a = 0
 L1 (0) L1 (P) > 0 & L2(0) L2 (P) > 0
–a (8 – a) > 0 & a(–3 + a) > 0
a (a – 8) > 0 & a (a – 3) > 0
a  (–, 0)  (8, ) & a  (–, 0)  (3, )
 a  (–, 0)  (8, )
Sol.24 C
A(x1, y1), B(x2,m y2), C(x3, y3)
B
Sol.22 A
L1 : x + y = 5, L2 : y – 2x = 8
L3 : 3y + 2x = 0, L4 : 4y – x = 0
L5 : (3x + 2y) = 6
L1
A (4
(–3, 2) C
L3
)
,1
x
L5
0)
vertices of quadrilateral
0(0, 0), A (4, 1), B (–1, 6), C(–3, 2)
L5(0) = – 6 < 0
L5 (A) = 12 + 2 – 6 = 8 > 0
L5(B) = – 3 + 12 – 6 = 3 > 0
L5(C) = –9 + 4 – 6 = – 11 < 0
O & C points are same side
& A & B points are other same side w.r.t to L5
So L5 divides the quadrilateral in two
quadrialteral
Aliter :
If abscissa of A is less then abscissa of B
A lies left of B
otherwise A lies right of B
Sol.25 D
P(1, 0), Q(–1, 0), R(2, 0), Locus of s
(h, k) if SQ2 + SR2 = 2SP2
(h + 1)2 + k2 + (h – 2)2 + k2
= 2(h – 1)2 + 2k2
h2 + 2h + 1 + h2 – 4h – 4 = 2h2 – 4h + 2
2h + 3 = 0 Locus of s(h, k)
2x + 3 = 0
Parallel to y-axis.
Sol.26 B
L1 : x + y – 3 = 0,
L2 : x – 3y + 9 = 0
L3 : 3x – 2y + 1 = 0
L2
 15 26 
,


 2 7 
(0, 3)
L3
L1
Sol.23 B
P(a, 2) lies between
L1 : x – y – 1 = 0 &
L2: 2(x – y) – 5 = 0
Method-I
L1(P) L2 (P) < 0
(a – 3) (2a – 9) < 0
P(a, 2) lies on y = 2
intersection with given lines
x=3
C
F
L4
O (0
,
D
only three sides can be made parallel to
corresponding sides of triangle passing
through vertex of triangle respectively
So no. of IIgrams is 3.
y
(–1, 6)B
L2
A
E
(1, 2)
15 26
1
7
7
3 1
1 1
=
2 1
2 1
=
9
& x=
2
1 15 (3  2)  0  1 26  3 


 7

2 7
1 15  5  10
=
sq.units

7 
2 7
7
Aliter : by parallelogram
9
a>3 & a<
2
(gemetrically)
=
 9
a   3, 
 2
=
1 (c1  c2 )(d1  d2 )
(m1  m2 )
2
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77
Solution Slot – 3 (Mathematics)
Page # 7
Sol.27 D
Aliter :
3
4
 mPQ = –
4
3
equation of PQ
m =
19
13
= tan1, mAC = tan2 =
8
16
mAB =
P(a, 0)
19 13

8 16 =  136
tan2 =
13
19 13
1
.
8 6
4
x
3
4x + 3y – 15 = 0
3x – 4y – 5 = 0
25x = 75
& 3x – 4y – 5 = 0 x = 3 & y = 1
Q(3, 1)
y–5=–
2 tan 
=
2
1  tan 
Sol.28 C
3
mPQ =
= tan
4
3
4
, cos=
5
5
Let PQ = r
equation PQ
3x
sin =
4=
y+
+
0
Q
L1
136
{ is acute tan  > 0
13
 68 tan2+13tan – 68 = 0  tan=0.9
 =  +1
tan   tan 1
tan = 1  tan  tan 
1
equation is (y – 11) = tan (x +1)
r
P(2, 5)
L2 3x
– 4y
+8
=0
x 2 y 5 4
=
=
4 /5 3 /5 5
 4r

 3r

 2 & y = 
 5  lies on L
x = 
1
3
5




Sol.30 D
OP =
2 , PQ = 3 2
OQ = 4 2
45º
27
0º
O
+
45 2
º
 4r
  3r

 2 + 
 5  +4 =0  154 =–15
3
5
5

 

5
3
Q
 r =–5  |r| = 5 units
L1
Sol.29 B
By geometry
L2
OQ makes angle with (+) x-axis in anti
clockwise  = 270º + 45º
equation L2
y
(–1, 11)A


2
x cos + y sin = 4 2

x cos (270º + 45º) + y sin (270º+45º)=4 2
2
1
x
O
C(15, –2)
D
B (–9, –8)
Angle bisector of A is origin containing
line
AB : 19x – 8y + 107 = 0
Line AC : –13x – 16y + 163 = 0
19x  8y  107
2
2
19  8
13x  16y  163
=
132  162
{192 + 82 = 132 + 162 = 425
 32x + 8y – 56 = 0  4x + y = 7
x sin 45º + y (– cos 45º) = 4 2
x–y=8
Aliter :
O
y–x+2=0
x – y – 2 = 0
3 2
Parallel lines
x–y+=0
2
3 2 =
2
 + 2 = ± 6
 = –8, 4
Line shift to (+) x-axis
So line is x – y – 8 = 0
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
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78
Solution Slot – 3 (Mathematics)
Page # 8
Sol.31 B
Point of reflection of (0, 0)
w.r.t. to 4x – 2y – 5 = 0
5
OA =
2
Sol.33 B
Before rotation
(2, 1) = (4 cos, r sin)
r cos  = 2, r sin  = 1
2
=
2
4 2
y'
y
2 5
y
5
= AB
2
equtaion of line OB
=
x0 y0
=
=±
2
1

5
5
OB =
O
O
B
x'
  30º
5
x0
y0
=
=
4
2
2(4.0  2.0  5)
42  (2)2
x
y
10
=
=
 x = 2, y= –1, B(2, –1)
4  2 20
new position
x = 4 cos cos  – r sin  sin 
= 2.
  1
3
 =
+2. 
 2 
2
3 2
2
2 3 1 3  2


,
(x, y) = 
2
2 

Sol.34 D
x = 2y, A(3, 0)
y = m (x – 3)
Sol.32 B
First position
y
y
x1
Q
1
m1 =
(given line)
2
P
1
2
tan 45º=
m
1
2
m
  30º
O

45º
45º
O
A(3, 0)
2 3
2 7
(4,2 3 )
4
(4, –2 3 ) = (4 cos (–), r sin (–))
r cos  = 4
 1
m
1
=m
2
2
m
1


3m
1
 1   =  m   or
=–
2
2


2
2
1
3
lines are y = 3(x – 3)
m=3
r sin  = + 2 3
1
3
, cos  =
2
2
Last position w.r.t is x
(r cos (– – a, r sin(– – ))
= (r cos ( + ), – r(sin ( + ))
= ((4 cos  cos  – r sin  sin )),
m (–r cos sin  – r sin cos)
& sin º =

3
1  
1
3  

=   4. 2  2 3 , 2 ,   4. 2  2 3. 2  
 


= ((2 3 –
X
A
5
x= 2,y=±1
B (2, –1)
Aliter :
Image of origin w.r. to line

(2, 1)
5

  30º
x
3 ), (–2 – 3)) = ( 3 , –5)
m=–
 3x – y – 9 = 0 & y = –
1
(x – 3)
3
 x + 3y – 3 = 0
Sol.35 D
(p + 2q) x + (p – 3q) y = p – q
px + py – p + 2qx – 3qy + q = 0
p(x + y – 1) + q (2x – 3y + 1) = 0
passing through intersection of
2 3
x + y – 1 = 0 & 2x – 3y + 1 = 0 is  , 
5 5
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79
Solution Slot – 3 (Mathematics)
Page # 9
Sol.36 A
PM is maximum if required
line  intersection of
3x + 4y + 6 = 0
 (–2, 0)
x+y+2=0
A
3O 3
mAP =
=
22 4
P(2, 3)
Sol.40 D
5x2 + 12xy – 6y2 + 4x – 2y + 3
= 0 & (x + ky) = 1
4
Slope m = –
3
y–0=–
Sol.39 A
5x2 – 7xy – 3y2 = 0
we know if given lines ax2 + 2hxy + by2
then
 lines are bx2 – hxy + ay2 = 0
– 3x2 – (– 7xy) + 5y2 = 0
3x2 –7xy – 5y2 = 0
4
(x + 2)  4x + 3y + 8 = 0
3
y
Sol.37 C
L1 : Px + qy = 1
L2 : qx + py = 1
O
x
qy
=
1
A
45º
45º
qx + py = 1
px
+
5x2 + 12xy – 6y + 4x (1) + 3(1)2 = 0
5x2 + 12xy – 6y2 + 4x (x + ky)
+ 2y (x + ky) + 3 (x + ky)2 = 0
These lines are equally inclined to
coordinate axis.
(p, q)
B
C
2h
=0 h=0
b
5 (k + 1) = 0  k = –1
& both line perpendicular also a + b = 0
12 + 3k2 – 2k – 6 = 0
3k2 – 2k + 6 = 0
D<0
K is not real
real k doesn't exist
 m1 + m2 = 0  –
L1 + L2 = 0
(px + qy – 1) +  (qx + py – 1) = 0
=
(p2  q2  1)
(2pq–1) (px + qy – 1)
(2pq  1)
= (p2 + q2 –1) (qx + py – 1)
Sol.38 C
2x2 + 4xy / py2 + 4x + 4x + qy + 1 = 0
q
, y = 2, h = 2
2
abc + 2fgh – af2 – bg2 – ch2 = 0
a = 2, b = –p, c = 1, f = –
–2p + 4q –
q2
+ 4P – 4 = 0
2
q2
–4=0
2
 a + b = 0
Sol.41 B
sin2 A sin A 1
sin2 B sin B 1
sin2 C sin C 1
=0
 (sinA–sinB) (sinB – sin C) (sin C – sin C)=0
 A = B or B = C or C = A
any two angles are equal  is isosceles
2P + 4q –
q2
–4=0
2
q

q  4   = 0
2

q = 0 , q = 8
2–p=0
p=2
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2x –
y+
5=0
4 + 4q –
Sol.42 B
P lies on 2x – y + 5 = 0
|PA – PB| is maximum
we know
b
C
A(4, –2)
b<a+c
c
a
b–a<c
B(2, –4)
If b – a = c
r
then (P – PB) is max. P
PBA colinear
80
Solution Slot – 3 (Mathematics)
Page # 10
Slope mAB = 1 = tan
If PB = r
x2 y 4
r
r
=
=r  x =
+2, y =
–4
1
1
2
2
2
2
Sol.44 B
L1 : x+ 3 y = 2, L2 : ax+by = 1, q = 45º,
L3 = y
1
Satisfy given equation
a
3x
3 2
1
b
3 1
 r

 r

 2  – 
 4  + 5 = 0
2 
2
2




=0
0
 3 (– 3 + 2b) + (–1 + 2a) = 0
r
2
2
r
+4–
2
a +
+4+5=0
3b=2
....(i)
1
m1 =
r
2
3
, m2 = –
a
b
r = – 13 2
= – 13

  13 2

 13 2

 2,
 4   (–11, –17)
P

2
2


1
a
b
a
3b
3 b| = | 3 a – b|
(a + 3 b)2 + 2 3 ab = 3a2 + b2 –2 3 ab
a2 + b2 – 2 3 ab ....(ii)
squaring (i) & adding (ii)
2a2 + ab2 = 4  a2 + b2 = 2
y
Q
k
=
QB
1

3
tan 45º =
 |a +
Sol.43 D
x+y=p
Let Q divides AB in k : 1
1
(0, p)
B
Q
45º
P
pk 
 p
,
, m = 1
Q
PQ
k  1 k  1
k
45º
O
line PQ . y –
x
A(p, 0)
p 

kp
= x 
(If cut y-axis)
k
 1 
k 1

then (x=0 put) y=
(k  1)p
 pk  p 

, p  0,
(k  1)
k 1 

2
2
pk
p 
 p 
 pk


 

k 1
k  1 k  1 k  1
PQ=BQ= 
=
2pk
k 1
Sol.45 D
L1 : 2x – 3y – 6 = 0
L2 : 3x – y + 3 = 0
L3 : 3x + 4y – 12 = 0
P(a, 0), Q(0, )
By geometry origin lies in 
L1 (0) < 0 & L2(0) > 0 L3 (0) < 3
L1 (P)  0 & L2 (P)  0 & L3(P)  0
–3&a+10&a4
a  [–1, 3]
L1 (Q)  0 & L2 (Q)  0 & L3 (Q)  0
–3 – 6  0 & –b + 3  0 & 4 – 12  0
  –2 &   3  3 &  3  [–2, 3]
Sol.46 C
3
3 1 2
3 2
Area APQ =
OAB =
. p =
p
8
8 2
16

2pk
2p
1
3
.
=
p2
(k  1)
2 (k  1)
16
 16k = 3 (k + 1)2  3k2 + 6k + 3 = 16k
L1 : x – 7y + 5 = 0
m1 =
1
7
1
3
L3 : (x + 3y – 2) +  (x – 7y + 5) = 0
L2 : x + 3y – 2 = 0
m2 = –
L1
k=3
1
k=
is reject
3
(  P lies on OB only)
L2
L3
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81
Solution Slot – 3 (Mathematics)
Page # 11
1
1 1
m3 

1
3 = 1
tan  = 7 3 =
tan  =
m3
2
2
1
1
1
3
21
m3 = –
Sol.49 A
ax2 + 2hxy + by2 = 0
m1 + m2 =
y
(1  )
= –1 m3 = –1
(3  7)
1 +  = 3 – 7
 =
a
2h
, m1m2 =
b
b
L2
L1
1
4
2
1
1
(x – 7y + 5) = 0
4
 5x + 5y –3 = 0
x
 x + 3y – 2 +
Sol.47 A
mAB + mPB = 0 y
2
3
+
=0
1a 5a
2
P(5, 3)
Relation of slopes of image lines
A(1, 2)
(m1 + m2) = – (m1 + m2)
13
a=
5


O
B(a, 0)
10
5
2
mAB =
=
=
13
8
4
1
5
equation of AB
y–2=–
x
  2h 
2h
 =
=– 
 b 
b
{m1 = tan (1)
m1m2 = (–m1) (–m2)
= m1m2 =
5
(x – 1) 5x + 4y = 13
4
Sol.48 D
Let a line ax + by + c = 0
P1 + P2 + P3 = 0
A
B
(2, 2)
(0, 3)
P1
a
b
2
y
y
  – (m1 + m2)   + m1m2 = 0
x
x
2
y
  
x
–
2h  y 
a
  +
=0
b x
b
by2 – 2hxy + ax2 = 0
ax2 – 2hxy + by2 = 0
P3
P2
Sol.50 D
x2 – 4xy + y2 = 0, x + y + 4 6 = 0
C(3, 0)
3a  c
a2  b2
3b  c
+
a2  b2
angle bisector of given pair of st. lines
2a  2b  c
+
a2  b2
=0
y
5a + 5b + 3c = 0
5
5
a  + b   + C = 0
3
3
5 5
  ,  satisfy the given line
3 3
x
O
5 5
fix point is  ,  which is centroid of ABC
3 3
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82
Solution Slot – 3 (Mathematics)
Page # 12
1
1
3
±
3
2
xy
xy
x2  y2
x2  y2
=

=
h
2
ab
1 1
x=4±
 x2 – y2 = 0
 (x + y) (x – y) = 0
x + y = 0 is || to third side
altitude  angle bisector  isosceles 


3 3
3 
3 3
3 


G  4  6 , 2  3  , G  4  6 , 2  3 





3 3
3 

OG > OG  4  6 , 2  3 


2
2 h ab
2 4 1
=
ab =
2
Now tan =
3
 = 60º
 angle between two equal sides is 60º
 equiliteral 
Sol.51 D
 3
D  4,  , AB =
 2
4 1 =
2 3
,y=
5
Sol.52 C
P(2, 0), Q (4, 2)
line PQ is x – y = 2
mPQ = + 1
 = 45º
required line is
parallel to y-axis
(according questions)
 x=2
Q
45º (4, 2)
45º
P(2, 0)
P
y
G1
(3,2)
A
(5, 1)B
G'
D
x
O
Sol.53 B
x2 – 4xy + 4y2 + x – 2y – 6 = 0
(x – 2y + C) (x – 2y + d) = 0
(x – 2y)2 + (C + d) x – 2 (c + d) y + cd=0
c + d = 1, cd = – 6
c = 3, d = –2
lines are (x – 2y + 3) = 0, (x – 2y – 2) = 0
P
3  (2)
PD =
5
5 =
4
1
.
3
G.D. =
distance =
15
2
15
=
2
15
2
1
1
=
=2
1
mAB

2
2
= tan 
5
5
y
3
2
2
5
5
=
5
=
5
Sol.54 A
xy + 2x + 2y + 4 = 0 & x + y + 2= 0
(x + c) (y + d) = 0
y
L3
xy + dx + cy + cd=0
d = 2, c = 2
(–2,0)
x
A
xz0 yz0
&
L1
L2
1
, cos  =
equation of pp is
xy
=
1
2
1 2
[Centroid  orthocentre in equilateral]
m PD =
2
5
&
B(–2,–2)
O
(0,–2)C
xyz 0
L3
L1  L2 L2
hypotaneous line L3
mid point of hypotenous is circumcentre
5
=±
2 3
0 2 2  0
,

 = (–1, –1)
2 
 2
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83
Solution Slot – 3 (Mathematics)
Page # 13
Sol.55 B
ax ± by ± C = 0
Sol.57 C
2|x| + 3|y|  6
area ABCD = 4 (OAB)
ax+by–c=0
B
ax – by + c
a
m1 
b
ax–by–c=0
2x – 3y = –6
C
2x + 3y = 6
2
O
3
2x – 3y = 6
2x + 3y = –6
ax+by+c=0
a
m1  
b
m1 = –
a
a
, m2 =
b
b
A
D
1

= 4  .2  3  = 12 sq. units
2

Sol.58 C
c
c
d1 = – , d2 =
b
b
d1 =
c
c
, d2 = –
b
b
(i) Reflection about y = x of (4, 1) is (1, 4)
y
(1, 4) (3, 4)
2
(3, 4)
(c1  c2 )(d1  d2 )
Area of rhombus =
(m1  m2 )
(4, 1)
O
2
=
c 2c

b b
2c2
=
sq. units
a
| ab |
2
b
 5
4

O
(1 + 2, 4 + 0)  (3, 4)
we wish to find


 


 5 cos   ,5 sin    
4
4 



Sol.56 D
concurrent
a 1 1
cos 
= 0 a, b  R, a  1, b ± 1, c  c
x=5
2
5 sin 
–
2
1
=–
2
1 1 c
C2  C2  C1 & C3  C3  C1
 a (b – 1) (c – 1) – (1 –a) (c – 1)
+1 (0 – (1 – a) (b – 1)) = 0

a
1
1
+
+
=0
1a
1b
1c
sin 
y = 5
2
5 cos 
+
2
 1
7 

,
(x, y)  
2
 2
a 

1
1
 +
 1 
+
=1
1

a


1b
1c

x
3
(ii) Now 2 units along (+) x direction
(iii)
1 b 1
4
1
1
1
+
+
=1
1a
1b
1c
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7
=
2
84
Solution Slot – 3 (Mathematics)
Page # 14
HINTS & SOLUTIONS
EXERCISE – II
Sol.1 A,B,C,D
1
y=
x
3
P
B
(0, b)
1
tan  =
3
a b
 , 
2 2
A
,
y B2
a
x
A
B3
3y
O
0
(0, 9)
Q
a
x
a
a
D
1
3
, cos =
2
2
 a  b b  a
,


2 
 2
y
=±a
1
2
Sol.3 A,C
Lengths from origin
cd
2
equation of B1B2 , mB1B2 = –
a
3a
y
4 =±
4 =
1
3

2
2
x
ab
2
=
c d
 3a a 

3a a 



D  4 , 4  , D   4 , 4 




2
a  b2
1
1
1
1
c2d2
a2b2
 2
2 = 2
2  a2 + b2 = c2 + d2
c d
a b
all three lines will be concurrent
3
1
a
1
b
1
c
3a
2
 3a  a 
  3a a 




B1  2 , 2  , B2 (0, a), B3  2 , 2  ,




B4 (0, –a)

Sol.2 A,C
b
a
1
1
b
1
1 = 0
a
1
1
d
1 
1  1 1 1  1 1
 1
 – 
  –1 


= 0
a
d
b
c
bd
ac
Q 
 b 



1
1
1
1
1
1
 – 2 + 2 – 2 –
–
+
=0
bc
bd
ac
a
b
b
a
b
parametric form of PQ
mPQ =
a2  b2
a
b
b
a
± ,y=
±
2
2
2
2
 x=
a 3 a
  a 3  a




 A  2 , 2  , A  2 , 2 




mAB =




a
b
y
2 =
2 =± 1
2
b
a
B4
3
2
a
 2
2
 a b
=± 
2


x
a
=
b
2
a
B1
x
=
a2  b2
a
A'
sin =
y
b
x
a
O
a
2

1 1  1 1 1  1 1  1 1  1

+ 
 –

=0
d a b c a b a b a b
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85
Solution Slot – 3 (Mathematics)
Page # 15
Sol.4 A,B
Mid point M (4, 3)
 2x  x + y ± 2y = 5  3
A/B2 are
P
m = –
x + 3y = 2
2 2
& 3x – y = 8 m = 3
BC line
y – 3 (x – 2)  3x – y = 3
B(5,4)
2
1
3
M
2
A(3,2)
& y –3 = –
P'
Sol.6 A,C
L1 : x + y = 0
L2 : 3x + y – 4 = 0
2
=1
2
mPQ = –1
m=
AB =
2
2
2 q
1
(x –2)  x + 3y = 11
3
m1 = –1
m2 = –3
L3 : x + 3y – 4 = 0
=2 2
m3 = –
1
3
A
PM  6
line pp
x4
y3
=
=±
1
1
2
m1
x+y = 0
m2
3x + y – 4 = 0
6
2
C
x=4±
3 y=3±
x=4±
3,y=3±
3
B
(4 +
3,3–
0
y–4=
x+ 3 m
3
3
3 ) & (4 3 , 3 + 3 )
Sol.5 A,B
L1 : 2x + y = 5
L2 : x – 2y = 3
Slope is decreasing order
m3 > m1 > m2
–
1
>–1>–3
3
m3 > m1 > m2
L1
–
1
> – 1 > –3
3
B
1
 1
2
3 1
m3  m1
3
tan C =
=
1 = 3 × 4 =2
1  m3m1
1
3
A
C
C
(2, 3)
L2
B
Line BC passing throug (2, 3)
(y –3) = m (x – 2)
m is equal to slope of
2x  y  5
2
2 1
m1  m2
2
1
1  3
tan A = 1  m m =
=
=
13
4
2
1 2
A = C & B is obtuse.
A = C & B is obtuse.
Obtuse isosceles triangle.
x  2y  3
=±
1  22
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86
Solution Slot – 3 (Mathematics)
Page # 16
HINTS & SOLUTIONS
EXERCISE – III
Sol.1 (a) Area of POM = PON = PMN
=
=
1
OMN
3
 3k  5 5k  1 
,

Sol.3 Since A 
 k 1 k 1 
y
N(0,8)
1 1
. 6. 8 = 8sq. units
3 2
Now
ar (ABC)=
 5k  1

 3k  5     3k  5 
 5k  1 
5

    2 
  7  k  1   37 
 k  1    k  1 


 k 1

P(a, b)
b
1
.6×b=8
2
b=
1
2
O a
8
3
x
 14k  66
2(k  1)
4=
M(6, 0)

14k  66
 4
2(k  1)
k = 7 or 31/9
1
8
&
8 × a = 8  a = 2  (2,
)
2
3
48
6(0)  8(6)  10(0)
=
 6  8  10
12
=4
Sol.4 Internally
K
1
(7, 9)
(b)  =
A(1, 3) P(3, 5)
B
Q
7k  1
9k  3
= 3,
=5
k 1
k 1
N(0, 8)
4k = 2
1
2
P divide internally AB in 1 : 2
 So Q. divide externally AB in 1 : 2
10
M(6, 0)
O
k =

7  2 9  6
,
 = (–5, –3)
Q. 
1 2 1 2 

Sol.2 Let third vertex is (a, b)
mAH mBH = –1
b2
8
.
= –1
a1
6
Sol.5 L1 : 5x–y–4=0
L2 : 3x+4y–4=0 L1
A
Line AB
y – 5 = m (x–1)
P(1, 5)
mx – y+5–m=0
5x – y – 4 = 0
B
– + +
L2
A(a, b)
m  9
 4m  16 
 For B x = 

for A x = 
m5
 4m  3 
H(1, 2)
(–2, 5)
B
AP = BP
1  m  9  4m  16 
=1

4m  3 
2 m  5
3a – 4b + 5 = 0
& mAB . mHC = –1
(4, –3) C
....(i)
4m2 – 33m – 27 + 4m2 – 46 m + 80
= 8m2 – 34m – 30
m=
bc
5
.
=–1
a2
3
3a – 5b +31 = 0
....(ii)
solve (i) & (ii) (a, b)  (33, 26)
83
35
y–5=
83
(x – 1)
35
 35y–175=83 x–83 83x–35y+92=0
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87
Solution Slot – 3 (Mathematics)
Sol.6
Page # 17
PA
3
=
 PA = 3r, PB = 2r
PB
2
Sol.8 L1 :
x
y
+
=1
b
a
L2 :
x
y
+
=1
b
a
L1 : x – 2y + 7 = 0
L2 : x + 3y – 3 = 0
L1 + L2 = 0
y
A
L1
(0, b)
L2
B
P(2, –3)
(0, a)
B
P(h, k)
A
O
Parametric equatoin of PA
x2
y3
=
cos 
sin 
(a, 0)
(b, 0)
x
x y

x y

   1 +     1 = 0
a b

a b

coordinate A (2 + 3r cos, – 3 + 3r sin) satisfy L1
coordinate B (2 + 2r cos, – 3 + 2r sin) satisfy L2
2 + 3r cos + 6 – 6r sin  + 7 = 0
r cos – 2r sin  + 5 = 0
1  
1  
   x +   y =  + 1
a b
b a
....(i)
& 2 + 24 cos – 9 + 6r sin – 3 = 0
r cos + 3r sin  – 5 = 0
....(ii)
solve (i) & (ii) r cos = –1, r sin = 2
 tan = –2




  1 
 0,   1 
,
0
1  
 1 
A  
 
, B 
a b 
 b a




equation of AB is y + 3 = – 2 (x – 2)
Mid point of AB is P(h, k)
2x + y –1 = 0
2h =
Sol.7 Let vertex A (a, a + 3)
ABC = 5 sq. units
 1
 1
& 2k =
1 
1 


a b
b a
A
y=x+3
a a3 1
1 2
2 3
1
1
2
1
2k
2h
1
b
a =
=
2k
2h
1
1
a
b
1
=±5
(2, 1)
B
C(3, –2)
 (3) a – (a + 3) (–1) + (–4 – 3) = ± 10
(a  2h)(2k  a)

 4a = ± 10 + 4
 a=
 7 13 
 3 3
 or   , 
A  ,
2 2 
 2 2
7 3
,
2
2
2
a
(b  2k)(2h  b)
=
b2
 b2 (2ak – a2 – 4hk + 2ah)
= a2 (2bh – 4hk + 2bk)
 2(a2 –b2) hk = (h + k) ab (a – b)
 2(a + b) hk = (h + k) ab locus of (h, k)
 2xy (a + b) = ab (x + y)
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88
Solution Slot – 3 (Mathematics)
Page # 18
C = 0 not possible
 c=–9
& d=–9
Diagonal BD is
(4x + 5y) (7x + 2y – 9)
– (4x + 5y – 9) (7x + 2y) = 0
–9(4x + 5y) – (–9) (7x + 2y) = 0
3x – 3y = 0
x – y = 0
Sol.9 |m1 – m2| = 2
m1 =
k 1
k 1
, m2 =
h 1
h1
P(h, k)
(–1,1)
(1, 1)
Sol.11 L : 3x + 2y = 24
ABC isoscles, (AC = BC)
M is mid point of AB
M (4, 6)
O
A(0, 12)
M
2
k 1 k 1

 =4
 
h 1 h 1
O
C
2
L
1)2
 (k –
=
–
 (y – 1) = ± (x2 – 1)
 y = x2 or y = 2 – x2
equation of  bisector of AB
Sol.10 B should be (0, 0)
given diagonal AC is
 13

,1  , ABC = 1 (AB) . (CM)
C 
 2

2
D 4x + 5y + c = 0 C
7x + 2y + d= 0
A
x+
11
=
7y
2
(x – 4)  2x – 3y + 10 = 0
3
y–6=
=
1
2
=
7
× 4 13
4
9
4x + 5y = 0
7x + 2y = 0
7
208 . 2
13
13 = 91 sq. units
B (0, 0)
11x + 7y = 9
....(i)
equation of AC
(4x + 5y + C) (7x + 2y + d)
– (4x + 5y) (7x + 2y) = 0
(7C + 4d) x +(2C + 5d) y + cd = 0 ...(ii)
compair (i) & (ii)
7c  4d 2c  5d cd


117  9

Sol.12 Let PQ = r
equation of PQ
x 3
y 2
=
=r


cos
sin
6
6

3r
r 

 Q  3  2 ,2  2 


satisfy given line

cd
 7c  4d


11
9
49c  28d  22c  55d 
2
 9c  C  0
 cd
 C(C  9)  0


8
0
(h2
4y 
1)2
(0, –1)
=4
3x 
 (k
 2 


 h2  1 
–1)2
B(8, 0)
P( 3,2)

6
O
Q


 3  3r ,2  r 
3
2
2  +8=0

 3+
3
r
r – 8 – 2r + 8 = 0 
=3
2
2
 r=6
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89
Solution Slot – 3 (Mathematics)
Page # 19
Sol.15 y = 2x + c
Sol.13 Diagonal AC
L1L3 – L2L4 = 0
C
y
L2
D
1,
D(
C
3)
P
L4
B(5, 1)
L3
L1
A
B
(ax + by + c) (ax + by + c) – (ax+by+ c)
(ax + by + c) = 0
c(ax + by) + c (ax + by)
–c (ax +by) – c (ax + by) + c2 –c2= 0
 (c – c) (a + a) x + (c – c) (b + b) y
+ (c + c) (c–c) = 0
(a + a) x + (b + b) y + (c + c) = 0
Diagonal BD
L1 L4 – L2 L3 = 0
(C–C) (a –a)x + (c – c) + (b – b) y = 0
(a – a) x +(b – b) y = 0
x
O
A
Diagonal bisect each other
mid point of BD is P (3, 2)
y = 2x + C passing through P
 2=6+c
 c = –4
AP=BPO=CP=DP, BP = 22  (1)2 = 5
parametric form of AC
tan = 2, P(3, 2)
x3
5
=
y2
=±
2
5
5
x = 3 ± 1, y = 2 ± 2  A(2, 0), C(4, 4)
Sol.16 Line L can be
x + 5y = C
Sol.14 Sides are 18,m 24, 30
which are right triangle triplet
(182 + 242 = 30º)
 c
B  0, 5 


A
(c, 0)
O
(0, 18)
A
C'(12,9)
18 G
I
(0, 0) B
24 C(24, 0)
Let coordinates of A,B, C are A(0, 18),
B(0, 0), C(24, 0) circumcentre c' (12, 9)
centroid G(8, 6)
 0  0  18.24 24 .18  0  0 
,

Incentre I 
 24  30  18 24  30  18 
Area DOAB = 5 sq. units
1
c
c
=5
2
5
 |c2| = 50  c= ± 50
c=±5 2
L : x + 5y = ± 5 2
Sol.17 LAD  y = x + 3
LBE = y = 2x + 4
AB = 60
CF = 30
B
I  (6, 6)
30
6 6 1
1 12 9 1
area of ICG is =
2 8 6 1
=
1
[18 – 24 + 0] = 3 sq. units
2

F
D

 a
 G
2
30
A
b
2
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E
b
2
C
90
Solution Slot – 3 (Mathematics)
Page # 20
Sol.19 L1  x – y + 5 = 0
L2 = x + 2y = 0
2 1
1
tan  =
 tan  =
12
3
A(1, –2)
& a2 + b2 = 602
b
b
In BCE  tan  = 2  tan  =
2a
a
In ACD tan ( + ) =
b
a
2
L2
L1
B
C
Mirror image of A. w.r. to L1 is B
1 b

tan   tan 
2b
1
2b
=
 + 3 2a =
b
1  tan   tan  2a
3
a
1
3.2a

2a  3b
2b
=
2a2+3ab=12ab – 2b2
6a  b
a
9ab = 2(a2 + b2) 9ab = 2.602
(1  2  5)
x 1
y2
=
=– 2
1  (1)2
1
1
 x = – 7, y = 6
B (–7, 6)
& C is image ofr A w.r.t. to L2
2(4)
x 1
y2
=
= 2
1
2
1  22
2
 60 
1
6a2
 = 202
 ab =
ABC = 
2
9
 3 
x =
 11  2 
11
2
,

y=
,C 
5
5
 5 5 
Line BC is
= 400 sq. units
14
(x + 7)
23
23 y – 138 = –14x – 98  14x + 23y = 40
y–6=–
Sol.18 One angle bisector is
7x – y + 3 = 0
x+y–3=0
Sol.20 (x1 – x2)2 + (y1 – y2)2 = a2
(x2 – x3)2 + (y2 – y3)2 = b2
& (x3 – x1)2 + (y3 – y1)2 = c2
B (x1, y1)
(1, –10)
c
a
x y 3
2
7x  y  3
=
5 2
C (x2, y2)
 5x + 5y – 15 = 7x – y + 3
 2x – 6y + 18 = 0  x – 3y + 9 = 0
Slope of angle bisectors are
1
&–3
3
& req. lines are passing through (1, –10)
(y + 10) = –3 (x – 1) & (y + 10) =
3x + y + 7 = 0
1
(x–1)
3
& x – 3y = 31
b
A (x3, y3)
2
x1 y1 1
 x2 y 2 1
x3 y 3 1
= (a + b + c) (b + c – a) (c + a – b) (a + b – c)
x1 y1 1
1 x2 y2 1
=
2s = a + b + c
2 x y 1
3
3
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91
Solution Slot – 3 (Mathematics)
Page # 21
(2)2 = 2s (2s – 2a) (2s – 2b) (2s – 2c)
22 . 2 = 24 s(s – a) (s – b) (s – c)
 = 22 = 4
C
Sol.21 AB = C = 3 2
BC = a = 5 2
5 2
4 2
AC = b = 4 2
– p + 11q –197 = ± 140
– P + 11q - 337 ....(ii)
OR –P + 11Q = 57 ....(iii)
(i) & (ii)
or (i) & (iii)
P = 15, 32
p = 20, q = 7
A (15, 32)
A (20, 7)
p + q = 47
p + q = 27
max. (p + q) = 47
I
A(1,1)
B(4, –2)
3 2
Incentre
 5 2.1  4 2.4  5.3 2 5 2.1  4 2  5.3 2 


,
 5 2 4 2 3 2

5
2

4
2

3
2


Sol.23 Origin lies outside
P (cos , sin )
L1 : x + y – 2 = 0
L2 : x – y – 1 = 0
L3 : 6x + 2y –
C
10 = 0
y
 36 2 12 2 


,
 
  I (3, 1)
 12 2 12 2 
L1
C
P
A
L3
5 2
4 2
L2
x
O
I
B
A(1, 1)
Incentre
3 2
B(4, –2)
 5 2.1  4 2.4  5.3 2 5 2.1  4 2(2)  5.3 2 


,


5 2 4 2 3 2
5 2 4 2 3 2


 36 2 12 2 


,
= 
  I (3, 1)
12
2
12
2


Interior angle angle bisector of A is
(y – 1) = 0 (x – 3)  y = 1(|| to x-axis)
perpendicular from C on A I is x = 5
If P lies inside ABC
L1 (d) < 0 & L2(2) < 0 & L3(0) < 0
L1 (P) < 0 & L2(P) < 0 & L3 (P) > 0
cos + sin –2 < 0


sin     <
4

2
  (0, /4)  (3/4, 4)
& cos – sin – 1 < 0
1


cos     <
4
2

Sol.22 Area ABC = 70 sq. units
 35 39 
,
 , sq. units
M 
 2 2 
7


<+
<
4
4
4
35 

39

= – 5 x 
2 
2

 3 
 ....(ii)
  0,
 2 
y–
10x + 2y = 214 satisfy (p, q)
10p + 2q = 214
....(i)
&
6 cos + 2 sin –
3
p
Now
q
1
12 12 1
...(i)
10
10 > 0
1
cos +
= 170
23 20 1
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
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10
10
sin >
2 10
92
Solution Slot – 3 (Mathematics)
Page # 22
sin ( + ) >
1
2
MN = b – a = 1 +
5

< ( + ) <
6
6
Area DMN =
 3 

 [0, 2] &   0,
 2 
&
21
25
=
& DP = 3
4
4
1 25
75
×
×3=
sq. units
2
4
8
area above x-axis = 75 –
5

–<<
–
6
6
=
525
sq.units
8
75
8
 525 + 8 = 533
tan = 3


 6

 
    < 0 { > 0 0  < 
6

 6

Sol.25 B median 2x + y – 3 = 0
angle bisector of C 7x – 4y –1 = 0
A(–3, 1)
From (i) & (ii) & (iii)
 5

   0 <  < 5 – tan–1 3
  0,
 6

6
D
Sol.24 Area of ABCD = (AB) × (AD)
AB =
122  92 =15
AD =
32  42 = 5
B
Let C on the line 7x – 4y – 100
y
B(6, 10)
C
 7  1 

C  ,
4 

D is mid point of AC lie median
C(9, 6)
A(–6, 1)
N
P
O
x
M
D(–3, –3)
ABCD = 15 × 5 = 75 sq. units
Let N (a, 0)
mAD = mAN

13
0 1
=
63
a6
7  1 

3  1

4 

,
D  2
2





3
3  7
 +
2 
–3=0
 2 
8
C (3, 5) & D(0, 3)
(C) line AC is y – 3 = 0

4
1
21
=–
a = –
3
a6
4
Let M (b, 0)
mCD = mDM 
 = 3
– 48 + 8 + 3 + 7 = 0
2
(x – 0)
3
 2x –3y6 + 9 = 0 (Q)
(P) will not a side Q (It's given median)
63
03
3
3
=

=
93
b3
4
b3
(A) Line AB A (–3, 1) satisfy (R) 4x + 7y + 5 = 0
& (B) Line BC is only (S) 18x – y – 49 = 0
b = 1
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93
Solution Slot – 3 (Mathematics)
Page # 23
HINTS & SOLUTIONS
EXERCISE – IV
Sol.1 Centroid  G
Orthocentre  H
Circumcentre  s
(b)
L1 : 2x + y = 0
L2 : x – y + 2 = 0
A
A
L2 = 0
L1 = 0
H(2, 3)
L2 = 0
L1 = 0
B
B
(a)
Altitudes from B & C
Passing through H(2, 3)
LC : x – 2y = – 4 & LB : x + y = 5
Intersection of LB & L1
x = – 5 & y = 10
B(–5, 10)
Intersection of LC & L2
y = 2 & x= 0 C(0, 2)
Line BC is
C
L=0
C
L=0
L1 : 2x – y = 0
L2 : x + y = 3 & G(2, 3)
A
(y – 2) =
L2 = 0
10  2
(x)
5
L1 = 0
G(2, 3)
B
L=0
C
y intercept is (If x = 0)
y=2
(c)
L1 : x + y – 1 = 0
L2 : 2x – y + 4 = 0
A
A (1, 2)
AG
2
=
Let D (x1, y1)
GD
1
L2 = 0
L1 = 0
2x  1
2y1  2
 1
=2&
=3
2 1
2 1
E
F
S(2, 1)
5 7
5
7
x1 =
& y1 =
D  , 
2
2
2 2
Let B (a, 2a) & C(b, 3 – b)
D is mid point of BC
ab
5
2a  3  b
7
=
&
=
2
2
2
2
&
a+b=5
2a –b = 4
3a = 9
Line BC is y – 1 =
5x – y = 9
a = 3 B  (3, 6)
b = 2 C  (2, 1)
6 1
(x – 2)
32
Slope is 5
(x, y) B
L=0
C(x2, y1)
Intersection point A (–1, 2)
Line SE
x + 2y – 4 = 0
& Line SF x – y – 1 = 0
B, C is image of A.w.r. to., SF & SE
respectively.
2(1  2  1)
x1  1
y 2
= 1
=
12  (1)2
1
1
x1 = – 1 + 4 = 3 & y1 = 2 – 4 = – 2
 B (3, –2)
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Solution Slot – 3 (Mathematics)
Page # 24
2(1  4  4)
x2  1
y 2
= 2
=–
1
2
12  22
x2 = – 1 +
2
4
3
=
, y2 = 2 +
= 14
5
5
5
Median BE is y + 1 = –
5
(x – 2)
2
 5x + 2y = 8
& CF is (y – 1) = –
1
x
4
 x + 4y = 4
  3 14 
,

 C 
 5 5 
Sol.3 Angle bisector AC is ax + 2y + c = 0
Line BC
14
2
L : (y + 2) = 5
(x – 3)
3
3
5
y+2=
AB =
72  242 = 25
AC =
92  122 = 15
AB
BD
25
5
=
=
=
AC
CD
15
3
4
(x – 3)
3
A(–8, 5)
x intercept is (put y = 0)
x=–
23
+3
4
 x=
3
2
Sol.2 Perpendicular of side AB & AC are
x–y–4=0
2x – y – 5 = 0
15
25
5
A(–2, 3)
3
B(–15, –19)
D
C (1, –7)
D is divdes BC in the ratio 5 : 3
E
F
  45  5  57  35 
,

Coordinate of D 
8
8


G
3 5
O  , 
2 2
C(x2, y2)
B(x1, y1)
 23 


 D   5,
2 

line OE should be
1
2x – y –
= 0
2
2x – y – C1 = 0
4x – 2y – 1 = 0
Line of should be
x–y+1=0
Image of A w.r.t. to OE & OF is C & B
respectively
2(2  3  1)
(x1  2)
(y1  3)
=
=
=4
1
1
12  12
 x1 = 2, y1 = –1
 y–5=
33
(x + 3)
2 .3
 2y – 10 = – 11x – 28
 11x + 2y + 78 = 0
a = 11, c = 78
B (2, –1)
(x  2) (y2  3) 2(8  6  1) 3
& 2
=
=
=
4
2
16  4
2
 x2 = 4
23
5
Line AD is y – 5 = 2
(x + 8)
58
& y2 = 0
 C(4, 0)
Sol.4 L1 : 4x +3y – 12 = 0
L2 : 4x + 3y – 3 = 0
x 2
y7
=
=a
cos 
sin 
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Solution Slot – 3 (Mathematics)
Page # 25
xya
Q
2
3
P
a
3
a
A
(–2, –7)
L1
 7x  y  a 

= ± 
50


5x – 5y + 5a = ± (7x – y + a)
passing through (1, 2)
– 5 + 5a = ± (5 + a)
– 5 + 5a = 5 + a or – 5 + 5a = – 5 – 9
5a = 10 or a = 0
L2
a=
P(a cos – 2, a sin – 7) lies on L2
4a cos – 8 + 3a sin – 21 – 3 = 0
a (4 cos + 3 sin) = 32
....(i)
x 2
y7
=
= (a + 3)
cos 
sin 
Q ((a + 3) cos – 2, (a + 3) sin – 7) lies on L1
4 (a + 3) cos – 8 + 3 (a + 3) sin – 21 = 0
5
2
 5
A should be (0, 0) or  0, 
 2
Sol.6 AB1 : x +y – 5 = 0  m2 = – 1
BC2 : x + 7y – 7 = 0  m1 = –
32
32
32 + 3 .
= 41  a =
a
3
From (i)
4 cos + 3 sin = 3
1
7
A


)  3.2 tan
2
2 =3

1  tan2
2
4(1  tan2
 7tan2
tan


– 6 tan – 1 = 0
2
2
AC3 : 7x + y + 14 = 0  m3 = –7
 1
2  
 4
–7
 tan =
or tan =
=
2
2
24
11
  1
1

 7 
2.1
 = 90º  poarallel to y-axis
Lines x = –2  x + 2 = 0
–
1
>–1>–7
7
m1 =
1
, m2 = –1, m3 = –7
7
1
1 6 3
tan B = 7
= =
1
8 4
1
7
7
&y+7=–
(x + 2)
24
7x + 24y + 154 = 0
Sol.5 Sides parallel to
y = x + 2, y = 7x + 3
A lie on y-axis
let (0, a)
Sides which passing
through are
B
y=x+a
& y = 7x + a
or x – y + a = 0
& 7x – y + a = 0
diagonals should be
(angle bisectors)
C
B


1
= 1 or tan
=–
2
2
7
A(0, a)
O (1, 2) D
C
B  acute <

4
tan A =
6
3
1  7

=
=
A  acute <
17
8
4
4
tan C =
1  7
<0
17
ABC is isosceles traingle
Make constant positive sign in lines
Bisector A
–x–y+5=0
7x + y + 14 = 0
(–1) (7) + (–1) (1) < 0
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96
Solution Slot – 3 (Mathematics)
Page # 26
acute angle bisector of A with (+) sign.
x  y  5
2
Sol.8 Image of A w.r.t to x – 1 = 0 is P
7x  y  15
=
x –1 = 0
B
 12x + 6y = 11
5 2
Bisector B
– x – y + 5 = 0 & – x – 7y + 7 = 0
(–1) (–1) (–7) > 0
acute angle bisector of B with (–) sign
y=x–1
Q
I(1, 0)
P
x  y  5
2
(x  7y  7)
A(4, –1)
=–
5 2
C
 6x + 12y = 32  3x + 6y = 16
Bisector-C
–x – 7y + 7 = 0
7x + y + 14 = 0
(–1) (7) + (–7) (1) < 0
External bisector of C is acute with (+) sign
x  7y  7
5 2
7x  y  14
=+
 8x+8y+7=0
5 2
Sol.7 (PM)2 = (OP)2
2(4  1)
x4
y 1
=
= 2
=–6
1
0
1  O2
P(–2, –1)
Image of A w.r. to x – y – 1 = 0 is Q
2(4  1  1)
xy
y 1
=
=–
= –4
12  (– 1)2
1
1
x = 0, y = 3 Q (0, 3)
Line BC is same PQ
4
x  2x – y + 3 = 0
2
 B (1, 5) & C (–4, –5)
y–3=
M
y – 2x + 1= 0
Line AB is y + 1 =
6
(x – 4)
3
 2x + y – 7 = 0
O
(0, 0)
P (h, k)
L1
Line AC is y + 1 =
4
(x – 4)
8
 x – 2y = 6 = 0
2
 k  2h  1 

 = ( (h  0)2  (k  0)2 )
5


(k – 2h + 1)2 = 5 (h2 + k2)
(y – 2x + 1)2 = 5 (x2 + y2)
Locus
x2 + 4y2 + 4xy + 4x – 2y – 1 = 0
Locus C at by y= 2x at Q & R
(1)2 = 5 (x2 + 4x2)
Sol.9 Line
y – 2 = m (x + 1)
m2 
,0 
B (0, m + 2), A 
 m

y
P
(–1, 2)
B(a, m + 2)
1
2
 x=±
 y=±
5
5
Q(h, k)
1 2
 1 2

Q  ,  & R  ,
5 5
 5 5 
2
QR =
2
2
4
   
5
5
x
O
A
m2 
,0 

 m

2
=
Mid point of QR is (0, 0)
Let
5
PA

=
PB
1
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97
Solution Slot – 3 (Mathematics)
Page # 27
P divides AB externally in ratio  : 1
2=
(m  2)  0)
2
 =
 1
m
Sol.11 Let line y = mx have {y = m1x & y = m2x}
distance from (x1, y1) is 
=
| mx1  y1 |
m2  1
Q divides AB internally in ratio  : 1
h=
2m2 + 2 = m2x12 + y12 – 2m1x1y1
m2(x12 – 2) – 2mx1y1 + y12
m  2
2(m  2)
&k=
(m  2)
m2
m
m
m
m
h=–
 2 (m2 + 1) = (mx1 – y1)2
m1
– d2 = 0
m2
(m  2)
(m  2)
&k=–2
m2
m2
m1 + m2 =
k = –2 (–h)
mx1y1
x12
{m1 + m2 =
2

y12  2
m1m2 =
k = 2h
x12
2

{m1m2 =
2h
b
a
b
Locus y = 2x
2
y
 
x
Sol.10 AD : x + y = 0
BE : x – 4y = 0
2x1y1  y 
(y12  2 )


– 2
+
=0
x1  2  x 
x12  2
x2 (y12 – 2) – 2x1y1xy + (x12 – 2) y2 = 0
CF : 2x – y = 0
Sol.12
A(t, –t)
 2v
2q 

,
 A
 2p  1 2p  1 
E
E
x+2y=q
 3p  q q  3 

,
C 
 p 1 p 1
x
–
y
=
3
P(2, 3)
F
D
F
O
(0,0)
(1, –2)
2x
B
+y
D
=0
B
C
Line AC is  BE passing A
4x + y – 3t = 0
(a)
d
q
3p  q
+
+ 1 = 3.2
2p  1
p 1
Line AB is  CF passing A
x + 2y + 1 = 0
B & C are intersection point of AB & BE
and AC × CF respectively
 2t  t 
t 
,
 & C  ,t
B 
 3 6 
3 
Let centroid is (h, k)
3h = t –
2t
t
t
+
& 3k = – t –
+t
3
3
6
5t
t
3h =
& 3k = –
6
6
3h = 5 (–3k)  h = – 5k  x + 5y = 0
 q 
3(p  q)  (q  3)
 +
– 
=5
(p  1)
 2p  1 
 q 
q  3
 + 
 = 2
– 
2
p

1


p 1
....(i)
 2q 
q  3
 + 
 – 2 = 3.3
& 
 2p  1 
p 1
 q 
q  3
 + 
 = 11 ....(ii)
2 
2
p

1


p 1
Solve (i) & (ii)
q
q3
=3&
=5
2p  1
p 1
q = 5p + 8 = 6p – 3
p = 11 & q = 63  p + q = 74
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98
Solution Slot – 3 (Mathematics)
Page # 28
(b)
P(2, 3) orthocentre
mBP mAC = –1
5   1 
. 
 = –1
1  p
(b)
p = 5
PD  AC
equation of PD x + y = 
Solve the 2x + y = 0
a (–5, 10) satisfy BC
–5 + 10p = q
q = 45
p + q = 50
(c)
Line PD
 4p  q  1 q  2p  5 

,
 x + y = 5 & D 
2(P  1) 
 2(P  1)
 PD  BC dis mid point of BC satisfy line PF
2P – q –1 – 4q + 8p – 4 + 16 P – 8 = 0
26p – 5q = 13
....(ii)
Solve (i) & (ii) & get P = 8 q = 39
p + q = 47
=
AB =
9  81 = 3 10
  3 18 



5 
 5
p=
=
(c)
Sol.13 2x2 + 3xy – 2y2 – 10x + 15y –28 = 0
& y = 3x
a + b   lines
21
=
32  12
r=

AB = 3 10
s
2
AC =
=
x + 2y – 7 = 0
 1 18 
 ,

C  5 5 
2x – y + 4 = 0
2
BC =
2.S = 3 10 +
=
(5)2
2
 21 
 42 

 

 5 
 5 
3 5
21 5
+
5
5
15 10  24 5
5
S=
3(5 10  8 5 )
10
 x =
5x + 1 = 0
(a)
36  9
=
45
3 5
=
5
5
B
(4, 12)
1
18
y=
5
5
& x + 6x – 7 = 0  x = 1, y = 3
& 2x – 3x + 4 = 0 x = 4, y = 12
2
6
3
   
5
5
P
(x + 2y) (2x – y) + 15y – 25 = 0
(x + 2y – 7) (2x – y + 4) = 0
x + 2y – 7 = 0
4x – 2y + 8 = 0
5 10
21
1
63
× 3 10 ×
=
5
10
2
10
A(1, 3)
y=–3
1
(AB) × p
2
Slopes 3, 2, –
r=
1
2
21
=
m1, m2, m3
m1  m2
32
1
tanB = 1  m m =
=
16
7
1 2
cot B = 7
tan C = N.D. cot C = 0
cot A + cotB + cot C = +
10
63
×
3(5 10  8 5 )
10
1
50
+7+0=
7
7
=
=
(8 5  5 10 )
(5 10  5 10 )
×
(8 5  5 10 )
21(8 5  5 10 )
21(8 5  5 10 )
=
5(64  50)
64.5  25.10
321
3
(8 5 – 5 10 ) =
(8 5 –5 10 )
5  14
10
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99
Solution Slot – 3 (Mathematics)
Sol.14 Let chorcd is y = mx + c

y  mx
=1
c
honoginizatoin with
3x2 – y2 – 2x + 4y = 0
3x2
 y  mx 
 y  mx 
 +4y 
 =0
– 2x 
C
c




–
y2

2m    2 4m 
4


 +
 xy + y2   1   = 0
c   c
c 
c

Page # 29
Sol.15 Line passes through (1, 0)
y = m (x – 1)  y = mx – x 
mx  y
=1
m
Homogination with x2 + y2 + 6x–10y + 1=0
 mx  y 
 mx  y 
 – 10y 

x2 + y2 + 6x 
 m 
 m 
2

 x2  3

represent  lines
 3+
2m
4
–1+
=0
c
c
1+
m2
=0c=–m–2
c
 mx  y 
 =0
+ 
 m 
Coeff. of x2 = 1 + 6 + 1 = 8
& coeff of y2 = 1 +
1
10
+
m
m2
If equatoin (i) subtends right angle of origin
1
10
+ 2 =0
m
m
put in equation of chord
 (y + 2) = m (x – 1)
passes through always (1, –2)
Homogenization with
3x2 + 3y2 – 2x + 4y = 0
8+1+
 y  mx 
 y  mx 
 + 4y 
=0
3x2 + 3y2 – 2x 
c
c




1
 1

  9    1 = 0
m
m

 

m    2  4m 
y


 xy + y2  3   = 0
 x2  3  2  + 
c 
c
c



Represent  lines
 3+
2m
4
+3+
=0
c
c
6c + 2m + 4 = 0
3c + m + 2 = 0
c=–
m
2
–
3
3
Put in eqaution of chord
y = mx –
 1 
1


 
2  + 10 m + 9 = 0
 
m 
m=–
1
, m = –1
9
lines are
y=–
1
(x – 1)
9
x + 9y – 1 = 0
or y = – 1(x – 1)
or x + y – 1 = 0
Sol.16 y2 – 4y + 3 = 0
(y – 3) (y – 1) = 0
y = 1
L1
 y=3
L2
m
2
–
3
3
2
1


y   = m x  
3
3


y
L2
D
C
always passes through
L3
1 2
 , 
3 3
....(i)
L4
2=h
A
Yes given result hold with
3x2 + 3y2 – 2x + 4y = 0
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L1
B
O
x
100
Solution Slot – 3 (Mathematics)
Page # 30
x2 + 4xy + 4y2 – 5x – 10 y + 4 = 0
(x + 2y + A) (x + 2y + B) = 0
 x + 2y + 4 = 0
L3
x + 2y + 1 = 0
L4
A(–6, 1), B (–3, 1)
C(–7, 3), D(–10, 3)
AB = 3, hight h = 2
Area of parallelogram = base × hight
= 3 × 2 = 6 sq. units
Diagonals
12  q2 =
AC =
7 
 21 7 
 21
,
,
 &  is  

A is 
10
10
10
10




Equation of side is || to 2x + y +  = 0
passes through C
 = – 5
2x – y – 5 = 0
equation of side is || to 3x + y – 7 = 0 &
  21  7 
,

passes through B 
 10 10 
5
3x + y +  = 0
2
BD =
2
7 2
=
 = 47
3x + y + 7 = 0
53
combined equation is
y2
– 16y –
+ x + 12y – 35 = 0
(2x – y + A) (3x + y + B) = 0
(2x – y + 5) (3x + y – 7) = 0
(2x – y – 5) (3x + y + 7) = 0
6y2 – xy – y2 – x – 12y – 35 = 0
Aliter :
B
P
5=
0
2x–y2+5=0
2x
–y
+
Sol.17
6x2
D
3x+y–7=0
L2
L1
3x+y–7=0
A
P2
C
O
A
P1
L4
3x+y+d
B
L3
2x–y+c=0
C
D
diagonal AC
L1 L2 – L3 L4 = 0
Q
Passing through (0, 0)
– 35 – cd = 0  cd = – 35 & diagonal BD
p1 =
5
L1 L4 – L2 L3 = 0
= 3
5
passing though (0, 0)
5d – (–7c) = 0
| 7 |
p2 =
10
 5d + 7c = 0
7
=
  35 
 + 7c = 0
5 
 c 
10
Line CD
c2 = 25  25 c = ± 5 & d = ± 7
x0
y0
=
=±
2
1
5
5
7
for required line c = – 5, d = + 7
10
(2x – y – 5) (3x + y +7) = 0
6x2 – xy – y2 – x – 12y – 35 = 0
7 
 21
,

Point A or B  
10
10


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101
Solution Slot – 3 (Mathematics)
Page # 31
HINTS & SOLUTIONS
EXERCISE – V
Sol.1 (a)

(c) distance d(P, Q) = |x1 – x2| + |y2 – y2|,
P(x, y), Q(x2, y2)
O (0, 0), A(3, 2), & Let P(h, k)
Sides are 2
 is equlateral
Incentre is centroid
y
A(3, 2)
(1, 3 )
x
2
2
(0, 0)
(0, 0)O
2
(2, 0)
0 2 1 0  0  3 


,
Incentre 

3
3



 1 
1,

3

(b)
 1 

S 1,
3

mPS =
2
2 1
1
=
=
13
9
9
2
2
2
P(2, 2)
P lies in Ist Quadrant
d(P, O) = d (P,A)
|h| + |k| = |h – 3| + |k –2|
h1 k > 0
h + k = |h – 3| + |k–2|
h1 k > 0
h + k = |h – 3| + |k – 2|
Case-I
0 < h < 3, & 0 < k < 2
h + k = 3 – h + 2 –k

2h + 2k = 5

2x + 2y = 5
For
0<x<3
0<y<2
accept
Case-II
0<h<3&k2
h+k=3–h+k–2

2h = 1
2x = 1 accept for y  2
Case-III
h3&0<k<2
h+k=h–3+2–k

2k + 1 = 0
2y + 1 = 0
Case-IV
h3&k2
h+k=h–3+k–2
N.S. reject
3
Q(6, –1)
S
 13 
,1

 2 
Infinite ray
R(7, 3)
A(3, 2)
2
finite length
passes through (1, –1)
O
2
y+1=
(x – 1)
9
2x + 9y + 7 = 0
1
2
3
Set = {(x, y) : 2x + 2y = 5 if 0 < x < 3 and
& 0 < y <2 or 2x = 1 if y  2}
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102
Solution Slot – 3 (Mathematics)
Page # 32
Sol.2 (i)
Reflection about y = x – 1
Solo.3 (a)
1
y = mx +
x
(x1, y1)
y
–
1
=
(4, 1)
y = nk
y = nx + 1
O
y = mx
2(4  1  1)
x1  4
y 1
= 1
=
12  (1)2
1
1
x1 = 4 – 2, y1 = + 1 + 2
x1 = 2, y1 = 3 
(2, 3)
(ii)
One unit in + x-axis direction new
position of point (3, 3)
(c1  c2 )(d1  d2 )
m1m2
Area =
=
(1  0)(1  0)
mn
=
1
|m  n|
(b)
(2, 3)
x=
(3, 3)
3x + 4y = 9, y = mx + 1
3x + 4mx + 4 = 9
5
5 is divisible by (3 + 4m)
(3  4m)
4m + 3 = ± 1 or 4m + 3 = ± 5
m=–
Sol.4 (a)
(iii)
Rotation through an angle /4
about the origin in the anticlockwise
direction
Finally coordinate of point
1
1
, – 1 or m = , – 2 Two values
2
2
PQ lies on x-axis
3 3
=
3
PQR = 120º
mQR =
3
R(3, 3 3 )
(0, 3 2 )
(3, 3)

4
3
D
6
1
3 2
3
P(–1, 0)
60º
60º
60º
Q(0, 0)
(0, 3 2 )
inclination of required angle bisector is 120º.
(4, 1)  (2, 3)  (3, 3)  (0, 3 2 )
Equation is y = – 3 x 
3x+y=0
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103
Solution Slot – 3 (Mathematics)
Page # 33
Aliter :
PQ = 1
P(r1 cos , r1 sin )
63 03 3 3 3 3

 

QR = 6 D   6  1 , 6  1    7 , 7 

 

r1
Q
L1
r2
PD
1
=
RD
6
(r2 cos ,r2 sin ) Q
mQD = –
3
y – O = – 3 (x – o) 
(b)
3x+y=0
L2
4r1 cos + 2r1sin = 9
2(r2) cos + (–r2) sin + 6 = 0
9
r1
3
2
(r2 ) = 6 = 4
L1 : 4x + 2y = 9
L2 : 2x + y + 6 = 0
L1
L2
{In opposite sides r2 is negative}
 2
 0, 
 2
P
9
2
y = |x| – 1
|x| – y = 1
&
y = – |x| + 1 |x| + y = 1
y
x
O9
3
(–3, 0)
9 
 ,0 
4 
(c)
1
4
6
1
x
O
Q
1
1
(0, –6)
Perpendicular distance
Anyline passing through origin
9
:3 3:4
4
1

Area = 4  .1.1  = 2 sq. units
2

(d)
9
:6 3:4
2
x+y=1
x+y=3
L1 is || to 2x – y = 5
L2 is || to 3x + y = 5
Let line L = y = mx
9
2 5
y
6
:
5

3:4
L
L2
Aliter :
Q
Let a line passing through (0, 0) & slope
tan
x 0
y0
=
=r
cos 
sin 
R(h, k)
O
P
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
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L1
x
104
Solution Slot – 3 (Mathematics)
Page # 34
m 
 1
,

 P
1  m 1  m 
3m 
 3
,

Q 
1  m 1  m 
&

L1 : 2x – y =
2m
1m
L2 : 3x + y =
9  3m
1m
& b2 – 4ac  0
(y – 10)2 – (8)2  0  (y – 2) (y – 18)  0
3
2
2
Intersection point R h =
11  2m
5(1  m)
12  9m
5(1  m)
18
y  [18, )
ymin = 18
Sol.5 x2 – y2 + 2y –1 = 0
x2 (y –1)2 = 0
(x + y –1) (x – y + 1) = 0
x+y=1
& x – y +1
&
k=

m=

55k – 25hk – 99 + 45h
= 60h – 25hk – 24 + 10k
15h – 45k + 75 = 0
x – 3y + 5 = 0
Let line is
(y – 2) = m (x – 8)
mx – y = 8m – 2


(e)
10
11  5h
12  5k
=
5h  2
5k  9
y
(0, 3) C
2
B
A
(0, 1)
(2, 1)
2
1
y
x
1
(92 – 8m)
1
Q
angle bisector are
y=1&x=0
A (0, 1), B(2, 1), C(0, 3)
(8, 2)
area ABC =
x
O
P
 8m  2 
,0 

 m

Sol.6 BC = 2 – k – k = 2 – 2k = 2 (1 – k)
AB = (1 – k)
(8m  2)
+ (2 – 8m)
m
y
=
x
y = OP + OQ =
1
. 2. 2 = 2 sq. units
2
my = – 8m2 + 10m – 2
8m2 + m (y – 10) + 2= 0
m should be negative & poroduct of roots > 0

sum of roots < 0
(y  10)
<0 
8
y
 1 
(8m  2)(1  m)
y = (8m – 2) 
 =
m

1


m
A
(1, 1)
P(h, k)
y = k (k, k)
M
C (2 – k, k)
B
y > 10
x
x+y=2
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
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105
Solution Slot – 3 (Mathematics)
Page # 35
Sol.9 L1 : (1 + p)x – py + p(1 + p) = 0
L2 : (1 + q)x – qy + q(1 + q) = 0
L3 : y = 0
A(–p, 0), B(–q, 0)
y
Altitude from A
C
q
y=–
(x + p)
1 q
0 A B
&
Altitude from B
L1
p
y=–
(x + q)
1p
1
2(1  k )(1  k) = 4h2
ABC =
2
(1 – k)2 = (2h)2
1 – k = ± 2h
y = ± 2x + 1
Sol.7 (a)
centroid is that point s.t.
OPR = OQR = PQR
y
0 3  6 0  4  0
,

R
3
3


x
L2
P(3,4)
Solving these Altitudes
5
 4
R   3, 
 3
(b)
L3
R
5
 1
1 


= pq 
1

p
1
 q 

 q
p 



1

q
1
 p 

x(q – p) = pq(q – p)
x
x
0
6
Q(6,0)
L1 : y – x = 0
L2 : 2x + y = 0
L3 : y + 2= 0
Statement II is false  (C)
S-I OP = 2 2 , OQ =
x = pq &
y=
q
(pq + p)
1 q
=–
5
q
p(1 + q)
(1  q)
q = – pq
Orthocentre is (h, k)  (pq, –pq)
h = pq &
k = – pq
h – pq = 0
h+k=0
x+y=0
which is straight line
PR
2 2
OP
=
=
RQ
OQ
5
Sol.8 L1 : x + 3y – 5 = 0
L2 : 3x – ky – 1 = 0
L3 : 5x + 2y – 12 = 0
Sol.10 B
1
(A)
5
3
3 k
5
2
1
ML1   3
=0
Let slope of L = m
 12
–3(–36+10) –k (–12+25) + 1(2–15) = 0
78 – 13k – 13 = 0
13k = 65 
k=5
(B)
1
3
=
3
k
or
(C)
k = –9
60º
=±
1  m. 3
on solving m =
k=
6
5
5
6
k = 5, k = –9,k = –
3
3x  y
 1 L1
3,0
equation of line : y –
3
6
5
=
k = –
k
2
5
k  0, k  –9, k  –

(D)

m 3
L
2)
3,–
P(
6
5
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3x+2+3 3 =0