MathCity.org
Merging man and maths
Exercise 4.1 (Solutions)
Page 185
Calculus and Analytic Geometry, MATHEMATICS 12
Available online @ http://www.mathcity.org, Version: 3.0
Distance Formula
Let A( x1 , y1 ) and B( x2 , y2 ) be two points in a
plane and d be a distance between A and B
then
d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2
or
B( x2 , y2 )
A ( x1 , y 1 )
d = ( x1 − x2 ) 2 + ( y1 − y2 ) 2
O(0,0)
See proof on book at page181
Ratio Formula
Let A( x1 , y1 ) and B( x2 , y2 ) be two points in a plane. The coordinates of the point C
dividing the line segment AB in the ratio
k1 : k2 are
 k1 x2 + k2 x1 k1 y2 + k2 y1 
,


k
+
k
k1 + k2 
 1 2
See proof on book at page 182
k2
k1
B( x2 , y2 )
C
A ( x1 , y 1 )
If C be the midpoint of AB i.e. k1 : k2 = 1:1
then coordinate of C becomes
 x1 + x2 y1 + y2 
,


2 
 2
O(0,0)
Question # 1
Describe the location in the plane of the point P ( x, y ) for which
(i)
x>0
(ii)
x > 0 and y > 0
(iii) x = 0
(iv) y = 0
(v)
x < 0 and y ≥ 0
(vi) x = y
(vii) x = − y
(viii) x ≥ 3
(ix) x > 2 and y = 2
(x) x and y have opposite signs.
Solution
(i) x > 0
y
Right half plane
2nd Quadrant 1st Quadrant
(ii) x > 0 and y > 0
The 1st quadrant.
-x
(iii) x = 0
y-axis
x <0
y>0
x>0
y>0
x<0
y<0
x>0
y< 0
x
4th Quadrant
3rd Quadrant
(iv) y = 0
x-axis
-y
FSC-II / Ex. 4.1 - 2
(v) x < 0 and y ≥ 0
2nd quadrant & negative x-axis
(vi) x = y
It is a line bisecting 1st and 3rd quadrant.
x =3
(vii) x = − y
A positive value can’t equal to a negative value, except
number zero, so origin,
( 0,0 ) , is the only point which satisfies x = − y
x = -3
y
x
(viii) x ≥ 3
⇒ ± x ≥ 3 ⇒ x ≥ 3 or − x ≥ 3
⇒ x ≥ 3 or x ≤ −3
which is the set of points lying on right side of the
line x = 3 and the points lying on left side of the
line x = −3 .
(ix) x > 2 and y = 2
The set of all points on the line y = 2 for which x > 2 .
(x) x and y have opposite signs.
It is the set of points lying in 2nd and 4th quadrant.
Question # 2
Find each of the following
(i)the distance between the two given points
(ii)Midpoint of the line segment joining the two points
(a) A ( 3,1) : B ( −2, −4 )
(b) A ( −8,3) ; B ( 2, −1)
1

(c) A  − 5, −  ; B −3 5,5
3

(
)
Solution
(a) A(3,1) ; B(−2, −4)
(i)
AB = (−2 − 3) 2 + (−4 − 1) 2 = (−5) 2 + (−5)2
= 25 + 25 = 50 = 25 × 2 = 5 2
 3 − 2 1 − 4   1 −3 
(ii) Midpoint of AB = 
,
 = , 
2  2 2 
 2
Review:
The midpoint of
A( x1 , y1 ) and B ( x2 , y2 )
 x1 + x2 , y1 + y2  .

2 
 2
is 
(b) A(−8,3) ; B(2, −1)
Do yourself as above.
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 3
1

(c) A  − 5 , −  ; B −3 5 , 5
3

(
( −3
AB =
5+ 5
(i)
)
2
)
1

+ 5 + 
3

2
(2 5 )
=
2
 16 
+ 
 3
2
256
436
4 × 109
2 109
=
=
=
9
9
9
3
 − 5 − 3 5 − 13 + 5   − 4 5 14 3  
7
(ii) Midpoint of AB = 
,
,  = −2 5, 
 =
2
2   2
2  
3

= 20 +
Question # 3
Which of the following points are at a distance of 15 units from the origin?
 15 15 
(b) (10, −10 )
(c) (1,15 )
(d)  , 
(a) 176,7
 2 2
Solution
(
)
(a) Distance of
(
)
176,7 from origin =
=
=
⇒ the point
(
(
)
2
176 − 0 + ( 7 − 0 )
(176 ) + ( 49 )
(176 ) + ( 49 )
2
= 225 = 15
)
176,7 is at 15 unit away from origin.
(b) Distance of (10, −10 ) from origin =
(10 − 0 )
2
+ ( −10 − 0 )
2
= 100 + 100 = 200
= 100 × 2 = 10 2 ≠ 15
⇒ the point (10, −10 ) is not at distance of 15 unit from origin.
(c) Do yourself as above
2
2
 15 15 
 15
  15

(d) Distance of  ,  from origin =  − 0  +  − 0 
 2 2
2
 2

225 225
225 15
=
+
=
=
≠ 15
4
4
2
2
 15 15 
Hence the point  ,  is not at distance of 15 unit from origin.
 2 2
Question # 4
Show that
(i)
the point A(0, 2) , B( 3, −1) and C (0, −2) are vertices of a right triangle.
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 4
(ii)
the point A ( 3,1) , B ( −2, −3) and C ( 2,2 ) are vertices of an isosceles triangle.
(iii) the point A ( 3,1) , B ( −2, −3) and C ( 2,2 ) and D ( 4, −5) are vertices of a
parallelogram. Is the parallelogram a square?
Solution
(i)
Given: A(0, 2) , B( 3, −1) and C (0, −2)
AB =
(
)
2
( 3)
2
3 − 0 + ( −1 − 2 ) =
(0 − 3 )
2
( 0 − 0)
2
(− 3 )
2
+ ( −2 + 1) =
2
+ ( −1)
2
2
⇒ BC = 4
2
+ ( 2 + 2) = 0 + ( 4)
2
2
⇒ CA = 16
= 16 = 4
2
2
2
= 3 +1 = 4 = 2
CA =
+ ( −3 )
⇒ AB = 12
= 3 + 9 = 12
BC =
2
2
2
∵ AB + BC = 12 + 4 = 16 = CA
∴ by Pythagoras theorem A, B & C are vertices of a right triangle.
(ii)
Given: A ( 3,1) , B ( −2, −3) and C ( 2,2 )
2
2
AB =
( −2 − 3) )
BC =
( 2 − (−2) )
CA =
(3 − 2)
2
+ ( −3 − 1) =
2
( −5 )
2
+ ( 2 − (−3) ) =
2
+ (1 − 2 ) =
(1)
2
2
+ ( − 4)
( 4)
2
+ ( −1)
2
+ ( 5)
2
= 25 + 16 = 41
= 16 + 25 = 41
2
= 1+1 = 2
∵ AB = BC ⇒ A , B & C are vertices of an isosceles triangle.
(iii)
Given: A ( 5,2 ) , B ( −2,3) & C ( −3, −4 ) and D ( 4, −5)
AB =
( −2 − 5 )
2
2
+ (3 − 2) =
( −7 )
2
+ (1)
2
= 49 + 1 = 50 = 5 2
BC =
( −3 + 2 )
2
2
+ ( −4 − 3 ) =
( −1)
2
+ ( −7 )
D
C
A
B
2
= 1 + 49 = 50 = 5 2
CD =
( 4 + 3)
2
2
+ ( −5 + 4 ) =
(7)
2
+ ( −1)
= 49 + 1 = 50 = 5 2
DA =
2
2
(5 − 4) + ( 2 + 5) =
= 1 + 49 = 50 = 5 2
∵ AB = CD and BC = DA
2
(1) + ( 7 )
2
2
⇒ A , B , C and D are vertices of parallelogram.
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 5
AC =
Now
( −3 − 5 )
2
2
+ ( −4 − 2 ) =
( −8 )
2
+ ( −6 )
2
= 64 + 36 = 100 = 10
BD =
( 4 + 2)
2
2
+ ( −5 − 3 ) =
( 6)
2
+ ( −8 )
2
= 36 + 64 = 100 = 10
Since all sides are equals and also both diagonals are equal therefore A , B , C , D are
vertices of a square.
Question # 5
The midpoints of the sides of a triangle are (1, −1) , ( −4, −3) and ( −1,1) . Find
coordinates of the vertices of the triangle.
Solution
Let A( x1 , y1 ) , B( x2 , y2 ) and C ( x3 , y3 ) are vertices of triangle ABC , and let
D(1, −1) , E (−4, −3) and F (−1,1) are midpoints of sides AB , BC and CA respectively.
Then
A
 x1 + x2 y1 + y2 
,

 = (1, −1)
2 
 2
D
F
⇒ x1 + x2 = 2 ... (i) and y1 + y2 = −2 ...
(ii)
 x2 + x3 y2 + y3 
,

 = ( −4, −3)
2 
 2
B
C
E
⇒ x2 + x3 = −8... (iii) and y2 + y3 = −6 ...
(iv)
 x3 + x1 y3 + y1 
,

 = ( −1,1)
2 
 2
⇒ x1 + x3 = −2 ... (v) , and y1 + y3 = 2 ...
(vi)
Subtracting (i) and (iii)
x1 + x2
= 2
x2 + x3 = −8 …
−
−
(vii)
= 8
⇒
x1 = 4
Putting value of x1 in (i)
4 + x2 = 2
⇒ x2 = 2 − 4 ⇒
x2 = −2
Putting value of x1 in (v)
4 + x3 = −2
⇒ x3 = −2 − 4 ⇒
−
−
+
(viii)
y1
− y3 = 4
Adding (vi) and (viii)
y1 + y3 = 2
y1 − y3 = 4
x1
− x3 = 10
Adding (v) and (vii)
x1 + x3 = −2
x1 − x3 = 10
2 x1
Subtracting (ii) and (iv)
y1 + y2
= −2
y2 + y3 = − 6 …
x3 = −6
2 y1
= 6
⇒
y1 = 3
Putting value of y1 in (ii)
3 + y2 = −2
⇒ y2 = −2 − 3 ⇒
y2 = −5
Putting value of y1 in (v)
3 + y3 = 2
⇒ y3 = 2 − 3 ⇒
y3 = −1
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 6
Hence vertices of triangle are ( 4,3) , ( −2, −5 ) & ( −6, −1) .
Question # 6
Find h such that the point A
(
)
vertices of a right angle with right angle at the vertex A .
Solution
Since ABC is a right triangle therefore by Pythagoras theorem
2
2
2
AB + CA = BC
⇒  0− 3

(
)
2
2
+ ( 2 + 1)  + 
 
(
B
3, −1 , B ( 0, 2 ) and C ( h, −2 ) are
3−h
)
2
C
A
2
2
2
+ ( −1 + 2 )  = ( h − 0 ) + ( −2 − 2 )

⇒ [3 + 9] + 3 − 2 3h + h 2 + 1 = h 2 + 16
⇒ 12 + 4 − 2 3 h + h 2 = h 2 + 16
⇒ − 2 3 h = h 2 + 16 − 12 − 4 − h 2
⇒ − 2 3h = 0
⇒ h=0 .
Question # 7
Find h such that A ( −1, h ) , B ( 3,2 ) and C ( 7,3) are collinear.
Solution
Points ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) are collinear if
x1 y1 1
x2 y2 1 = 0
x3 y3 1
Since given points are collinear therefore
−1 h 1
3 2 1 =0
7 3 1
⇒ − 1(2 − 3) − h(3 − 7) + 1(9 − 14) = 0 ⇒ − 1(−1) − h(−4) + 1(−5) = 0
⇒ 1 + 4h − 5 = 0
⇒ 4h − 4 = 0
⇒ 4h = 4
⇒ h =1
Question # 8
The points A ( −5, −2 ) and B ( 5, −4 ) are end of a diameter of a circle. Find the centre
and radius of the circle.
Solution
The centre of the circle is mid point of AB
 −5 + 5 −2 − 4   0 −6 
,
i.e. centre ‘C’ = 
 =  ,  = ( 0, −3)
2
2

 2 2 
A
B
C
Now radius = AC
= (0 + 5) 2 + (−3 + 2) 2
= 25 + 1 = 26
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 7
Question # 9
Find h such that the points A ( h,1) , B ( 2,7 ) and C ( −6, −7 ) are vertices of a right
triangle with right angle at the vertex A
Solution
Do yourself as Question # 6
Hint: you will get a equation h 2 + 4h − 60 = 0
Solve this quadratic equation to get two values of h .
Question # 10
A quadrilateral has the points A ( 9,3) , B ( −7,7 ) , C ( −3, −7 ) and
D
D ( −5,5) as its vertices. Find the midpoints of its sides. Show that
H
the figure formed by joining the midpoints consecutively is a
parallelogram.
A
Solution
Given: A ( 9,3) , B ( −7,7 ) , C ( −3, −7 ) and D ( 5, −5 )
Let E , F , G and H be the mid-points of sides of quadrilateral
(
Coordinate of F = (
Coordinate of G = (
Coordinate of H = (
Coordinate of E =
C
G
F
B
E
) ( ) = (1,5)
) ( ) = ( −5,0)
) ( ) = (1, − 6)
) ( ) = (7, −1)
9−7 3+7
,
=
2
2
−7 − 3 7 − 7
,
=
2
2
−3 + 5 −7 − 5
,
=
2
2
9+5 3−5
,
=
2
2
2 10
,
2 2
−10 0
,
2 2
2 −12
,
2 2
14 −2
,
2 2
Now EF =
(−5 − 1) 2 + (0 − 5)2 =
36 + 25 =
61
FG =
(1 + 5) 2 + (−6 − 0) 2 =
36 + 36 =
72 = 6 2
GH =
(7 − 1) 2 + (−1 + 6) 2 =
36 + 25 =
61
HE = (1 − 7) 2 + (5 + 1) 2 = 36 + 36 =
Since EF = GH and FG = HE
Therefore EFGH is a parallelogram.
72 = 6 2
Question # 11
Find h such that the quadrilateral with vertices A ( −3,0 ) , B (1, −2, ) C ( 5,0 ) and
D (1, h ) is parallelogram. Is it a square?
Solution
Given: A ( −3,0 ) , B (1, −2 ) , C ( 5,0 ) , D (1, h )
Quadrilateral ABCD is a parallelogram if
D
AB = CD & BC = AD
when AB = CD
⇒
C
(1 + 3) 2 + (−2 − 0)2 = (1 − 5) 2 + (h − 0) 2
A
B
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 8
⇒ 16 + 4 = 16 + h 2 ⇒ 20 = 16 + h 2
On squaring
20 = 16 + h 2 ⇒ h 2 = 20 − 16 ⇒ h 2 = 4
When h = 2 , then D(1, h) = D (1, 2)
Then
AB =
(1 + 3) 2 + (−2 − 0) 2 =
BC =
(5 − 1)2 + (0 + 2) 2 =
16 + 4 =
CA =
(1 − 5)2 + (2 − 0) 2 =
16 + 4 =
16 + 4 =
20
20
20
DA = (−3 − 1)2 + (−0 − 2)2 = 16 + 4 =
Now for diagonals
AC =
(5 + 3)2 + (0 − 0) 2 =
⇒ h = ±2
20
64 + 0 = 8
BD = (1 − 1) 2 + (2 + 2)2 = 0 + 16 = 4
Since all sides are equal but diagonals AC ≠ BD
Therefore ABCD is not a square.
Now when h = −2 , then D(1, h) = D(1, −2) but we also have B(1, −2)
i.e. B and D represents the same point, which can not happened in quadrilateral
so we can not take h = −2 .
Question # 12
If two vertices of an equilateral triangle are A ( −3,0 ) and B ( 3,0 ) , find the third
vertex. How many of these triangles are possible?
Solution
Given: A ( −3,0 ) , B ( 3,0 )
Let C ( x, y ) be a third vertex of an equilateral triangle ABC .
Then
AB = BC = CA
⇒
(3 + 3)2 + (0 − 0) 2 =
( x − 3) 2 + ( y − 0) 2 =
( x + 3) 2 + ( y − 0) 2
⇒ 36 + 0 = x 2 − 6 x + 9 + y 2 = x 2 + 6 x + 9 + y 2
On squaring
36 = x 2 + y 2 − 6 x + 9 = x 2 + y 2 + 6 x + 9 ............(i )
From equation (i)
x2 + y2 − 6x + 9 = x2 + y2 + 6x + 9
C
A
B
⇒ x + y − 6x + 9 − x − y − 6x − 9 = 0
⇒ − 12 x = 0 ⇒ x = 0
Again from equation (i)
36 = x 2 + y 2 − 6 x + 9
2
2
2
2
⇒ 36 = (0) 2 + y 2 − 6(0) + 9
∵ x=0
⇒ 36 = y 2 + 9 ⇒ y 2 = 36 − 9 = 27
(
) (
⇒ y = ±3 3
)
so coordinate of C is 0,3 3 or 0, −3 3 .
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 9
(
)
And hence two triangle can be formed with vertices A ( −3,0 ) , B ( 3,0 ) , C 0,3 3 and
(
)
A ( −3,0 ) , B ( 3,0 ) , C 0, −3 3 .
Question # 13
Find the points trisecting the join of A ( −1, 4 ) and B ( 6, 2 ) .
Solution
Given: A ( −1, 4 ) , B ( 6, 2 )
1
Let C and D be points trisecting A and B
A
Then AC : CB = 1: 2
 1(6) + 2(−1) 1(2) + 2(4) 
So coordinate of C = 
,

1
+
2
1+ 2 

 6 − 2 2 + 8   4 10 
= 
,
=  , 
3  3 3 
 3
Also AD : DB = 2 :1
 2(6) + 1(−1) 2(2) + 1(4) 
,
So coordinate of D = 

2 +1
2 +1 

 12 − 1 4 + 4   11 8 
= 
,
=  , 
3
3

  3 3
4 10
11 8
Hence  ,  and  ,  are points trisecting A and B .
3 3 
 3 3
1
C
1
B
D
Question # 14
Find the point three-fifth of the way along the line segment from A ( −5,8 ) to B ( 5,3) .
Solution
Given: A ( −5,8 ) , B ( 5,3)
3
Let C ( x, y ) be a required point
A
∵ AC : CB = 3: 2
3(5) + 2(−5) 3(3) + 2(8) 
∴ Co-ordinate of C = 
,
3+ 2
3 + 2 

15 − 10 9 + 16 
5 25
= 
,
=  ,  = (1 , 5 )

5 
 5
5 5 
2
C
B
5
Question # 15
Find the point P on the joint of A (1, 4 ) and B ( 5,6 ) that is twice as far from A as B
is from A and lies
(i) on the same side of A as B does.
(ii) on the opposite side of A as B does.
Solution
Given: A (1, 4 ) , B ( 5,6 )
(i) Let P ( x, y ) be required point, then
1
1
B
A
P
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 10
AB : AP = 1: 2
⇒ AB : BP = 1:1 i.e. B is midpoint of AP
1+ x 4 + y 
,
Then B ( 5,6 ) = 
2 
 2
4+ y
1+ x
⇒ 5 =
and 6 =
2
2
⇒ 10 = 1 + x
and 12 = 4 + y
⇒ x = 10 − 1
,
y = 12 − 4
=9
,
=8
Hence P ( 9,8 ) is required point.
(ii) Since PA : AB = 2 :1
2(5) + 1( x) 2(6) + 1( y ) 
⇒ A (1,4 ) = 
,
2 + 1 
 2 +1
10 + x 12 + y 
= 
,
3 
 3
12 + y
10 + x
⇒ 1=
and 4 =
3
3
⇒ 3 = 10 + x
and 12 = 12 + y
⇒ x = 3 − 10
and y = 12 − 12
= −7
,
= 0
Hence P ( −7,0 ) is required point.
2
1
A
P
B
Question # 16
Find the point which is equidistant from the points A ( 5,3) , B ( 2, −2 ) and C ( 4, 2 ) .
What is the radius of the circumcircle of the ∆ABC ?
Solution
Given: A ( 5,3) , B ( −2,2 ) and C ( 4,2 )
Let D ( x, y ) be a point equidistance from A , B and C then
DA = DB = DC
⇒ DA
2
= DB
2
= DC
2
⇒ ( x − 5) 2 + ( y − 3) 2 = ( x + 2) 2 + ( y − 2)2 = ( x − 4) 2 + ( y − 2)2 …….. (i)
From eq. (i)
( x − 5) 2 + ( y − 3) 2 = ( x + 2)2 + ( y − 2) 2
⇒ x 2 − 10 x + 25 + y 2 − 6 y + 9 = x 2 + 4 x + 4 + y 2 − 4 y + 4
⇒ x 2 − 10 x + 25 + y 2 − 6 y + 9 − x 2 − 4 x − 4 − y 2 + 4 y − 4 = 0
⇒ − 14 x − 2 y + 26 = 0 ⇒ 7 x + y − 13 = 0 ……… (ii)
Again from equation (i)
( x + 2)2 + ( y − 2) 2 = ( x − 4)2 + ( y − 2) 2
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 11
⇒ x 2 + 4 x + 4 + y 2 − 4 y + 4 = x 2 − 8 x + 16 + y 2 − 4 y + 4
⇒ 12 x − 12 = 0 ⇒ 12 x = 12
⇒ x =1
Put x = 1 in eq. (ii)
7(1) + y − 13 = 0
⇒ y −6 = 0
⇒ y = 6
Hence (1,6 ) is required point.
Now radius of circumcircle = DA
=
(5 − 1)2 + (3 − 6)2 =
16 + 9
=
25 = 5 units
Intersection of Median
Let A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) are vertices of triangle.
Intersection of median is called centroid of triangle and can be determined as
 x1 + x2 + x3 y1 + y2 + y3 
,
See proof at page 184


3
3


Centre of In-Circle (In-Centre)
Let A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) are vertices of triangle.
And AB = c , BC = a , CA = b
 ax + bx2 + cx3 ay1 + by2 + cy3 
Then in-centre of triangle =  1
,
a + b + c 
 a+b+c
See proof at page 184
Question # 17
The points ( 4, −2 ) , ( −2,4 ) and ( 5,5 ) are the vertices of a triangle. Find in-centre of
the triangle.
Solution
Let A ( 4, −2 ) , B ( −2,4 ) , C ( 5,5 ) are vertices of triangle then
a = BC =
(5 + 2)2 + (5 − 4) 2 =
b = CA =
(4 − 5)2 + (−2 − 5) 2
c = AB =
2
(−2 − 4) + (4 + 2)
2
49 + 1 =
=
50 = 5 2
1 + 49 =
50 = 5 2
A
=
36 + 36
=
72 = 6 2
c
Now
 ax + bx2 + cx3 ay1 + by2 + cy3 
In-centre =  1
,
B
a + b + c 
 a+b+c
 5 2 ( 4 ) + 5 2 ( −2 ) + 6 2 ( 5 ) 5 2 ( −2 ) + 5 2 ( 4 ) + 6 2 ( 5 ) 
= 
,


5
2
+
5
2
+
6
2
5
2
+
5
2
+
6
2


 20 2 − 10 2 + 30 2 −10 2 + 20 2 + 30 2 
= 
,

16
2
16 2


 40 2 40 2 
= 
,

16
2
16 2 

b
a
C
5 5
=  , 
2 2
Available at http://www.mathcity.org
FSC-II / Ex. 4.1 - 12
Question # 18
Find the points that divide the line segment joining A ( x1 , y1 ) and B ( x2 , y2 ) into four
equal parts.
Solution
Given: A ( x1 , y1 ) , B ( x2 , y2 )
Let C , D and E are points dividing AB into four equal parts.
∵ AC : CB = 1: 3
 1( x ) + 3( x1 ) 1( y2 ) + 3( y1 ) 
 3 x + x 3 y + y2 
⇒ Co-ordinates of C =  2
,
=  1 2, 1


1+ 3
1+ 3
4


 4

Now AD : DB = 2 : 2
1
1
1
1
= 1:1 i.e. D is midpoint of AB .
A
B
E
C
D
 x1 + x2 y1 + y2 
⇒ Co-ordinates of D = 
,
2
2 

Now AE : EB = 3:1
 3( x ) + 1( x1 ) 3( y2 ) + 1( y1 ) 
 x + 3 x2 y1 + 3 y2 
⇒ Co-ordinates of E =  2
,
=  1
,


3 +1
3 +1
4


 4

 3 x + x 3 y + y2   x1 + x2 y1 + y2 
 x + 3 x2 y1 + 3 y2 
Hence  1 2 , 1
,
,
,
and  1


 are the points
4
4
2 
 4
  2
 4

dividing AB into four equal parts.
Error Analyst
1- Sohaib Ali
2- Tahir Waqas
3- Muhammad Imran Khan
4- Zohaib Khokhar
– FAZMIC Sargodha.
– UoS Sargodha.
– FAZMIC Sargodha.
2004-06
2003-05
2004-06
Please report us error at www.mathcity.org/error
Book:
Exercise 4.1 (Page 185)
Calculus and Analytic Geometry Mathematic 12
Punjab Textbook Board, Lahore.
Available online at http://www.MathCity.org in PDF Format
(Picture format to view online).
Updated: September,12,2017.
These resources are shared under the licence AttributionNonCommercial-NoDerivatives 4.0 International
https://creativecommons.org/licenses/by-nc-nd/4.0/
Under this licence if you remix, transform, or build upon the
material, you may not distribute the modified material.
Available at http://www.mathcity.org