Mathematics
Class-X
2015-2016
SOLVED QUESTIONS
Q.1. Find the equation of a line passing through (2, –3) and inclined at an angle of 135º with
positive direction of x-axis.
Solution :
The point is (x1,y1) = (2, –3), slope = m = tan135º = –1,
The equation of a line passing through (x1,y1) with slope m is :
y – y1 = m(x – x1)
Hence, equation of line passing through (2, –3) with slope –1 is given by
y – (–3) = –1(x - 2)
Or, y + 3 = –x + 2
Or, x + y + 1 = 0. [Ans.]
Q.2. Find the equation of the line parallel to 3x + 2y = 8 and passing through the point (0,
1).
Solution :
The equation of line parallel to 3x + 2y = 8 may be written as,
3x + 2y = k -------- (i) we have to find the value of k.
Line (i) passes through (0, 1), therefore, 3(0) + 2(1) = k
Or, k = 2,
Hence, equation of required line is 3x + 2y = 2. [Ans.]
Q.3. Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15
= 0.
Solution :
Do yourself. [Ans. 3x + 5y – 20 = 0.]
Q.4. The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A.
Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 =
0.
Solution :
At x-axis, y-co-ordinate is zero. The line 4x – 3y + 12 = 0, meets x-axis. Hence putting y = 0 in
the equation of line we get,
4x – 3×0 + 12 = 0 Or, 4x = –12 or, x = –3.
Therefore, line 4x – 3y + 12 = 0, meets x-axis at A(–3,0).[Ans.]
Slope m1 of the line 4x – 3y + 12 = 0, = – (coefficient of x /coefficient of y)
= – (4/–3) = 4/3
Let the slope of the required line be m2.
Then m1×m2 = –1
Or, (4/3) ×m2 = –1 => m2 = –3/4
Therefore, line through A(–3,0 ) with slope –3/4 is given by
y – 0 = (–3/4){x – (–3)}
Or, 4y = –3(x + 3) = –3x – 9
Or, 3x + 4y + 9 = 0. [Ans.]
Sudheer Gupta .
Be positive and constructive. 
Page 1
Mathematics
Class-X
2015-2016
Q.5. Write down the equation of the line whose gradient is 3/2 and which passes through P,
where P divides the line segment joining A(–2,6) and B(3, –4) in the ratio 2 : 3.
Solution :
Let the co-ordinates of P be (x,y).
x = {2×3 + 3×(–2)}/(2 + 3) = 0, y = {2×(–4) + 3×6}(2 + 3) = 2,
Hence, P is (0,2), m = 3/2, Using the formula, y – y1= m (x – x1),
The equation of the line is given by, y – 2 = 3/2 (x – 0 )
Or, 2y – 4 = 3x Or, 3x – 2y + 4 = 0. [Ans.]
Q.6. ABCD is a square. The co-ordinates of A and C are (3,6) and (–1,2) respectively. Write
down the equation of BD.
Solution :
Co-ordinate of A and C are (3,6) and (–1,2).
Slope of AC = m1 = (y2 – y1)/(x2– x1) = (2 – 6)/(–1– 3) = 1.
As, diagonals of a square bisect each other at right angle,
let slope of BD be m2, then
m1×m2 = –1 or, m2 = –1/1 = –1
Co-ordinates of mid point of AC are {(x1 + x2)/2,(y1 + y2)/2}
= {(3 – 1 )/2, (6 + 2)/2} = (1,4).
Using the formula y – y1 = m(x – x1) , the equation of BD is
0 + y – 4 = (–1)(x – 1)
or, x + y – 5 = 0 . [Ans.]
Q.7. If 3y – 2x – 4 = 0 and 4y – ax – 2 = 0 are perpendicular to each other, find the value of
a.
Solution :
The given equation of lines are : 3y – 2x – 4 = 0 --------- (i)
and 4y – ax – 4 = 0 ---------- (ii)
m1 = slope of (i) = – coefficient of x/coefficient of y = –(–2)/3 = 2/3,
m2 = slope of (ii) = – coefficient of x/coefficient of y = –(– a)/4 = a/4,
lines (i) and (ii) are perpendicular to each other, therefore, m1×m2 = –1,
or, (2/3)(a/4) = –1 or, a = –6. [Ans.]
Q.8. If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of
p.
Solution :
The given equation of lines are : y = 3x + 7 -------------- (i)
and 2y + px = 3 -------------------- (ii)
Slope of (i) , m1 = 3;
(ii) can be written as y = (– p/2)x + 3/2
Slope of (ii), m2 = – p/2;
As, (i) and (ii) are perpendicular to each other,
Therefore, m1×m2 = – 1
Or, 3×(– p/2) = – 1 => p = 2/3. [Ans.]
Sudheer Gupta .
Be positive and constructive. 
Page 2
Mathematics
Class-X
2015-2016
Q.9. Find the value of k for which the lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are
perpendicular to each other.
Solution :
We have, kx – 5y + 4 = 0 => kx + 4 = 5y => y = (k/5)x + 4/5.
Therefore, m1 = k/5.
And 4x – 2y + 5 = 0 => 4x + 5 = 2y => y = 2x + 5/2.
Therefore, m2= 2.
As the lines are perpendicular to each other, m1m2 = – 1
Or, (k/5). 2 = – 1
Or, k = – 5/2. [Ans.]
Q.10. In the adjoining figure, write:
i. the co-ordinates of A, B and C.
ii. the equation of the line through A and || to BC.
Fig.Q.5(c)/page 477[10 yrs]
Solution :
From the given fig.
i. A = (2, 3), B = (– 1 ,2), C(3, 0) [Ans.]
ii. Slope of BC = [y2 – y1]/[x2 – x1]
= [0 – 2]/[3 – (– 1)]
= – 2/4
= – 1/2 = m1.
Since line is parallel, therefore, m = m1 = – 1/2.
Equation of line through (2, 3) with slope m is y – 3 = m (x – 2)
Or, y – 3 = – 1/2 (x – 2)
Or, 2y – 6 = – x + 2
Or, x + 2y = 8. [Ans.]
Q.11. P(3, 4), Q(7, – 2) and R(– 2, – 1) are the vertices of triangle PQR. Write down the
equation of the median of the triangle, through R.
Solution :
Let S(x, y) be the mid-point of PQ, then SR is the median through R.
Therefore, x = (7 + 3)/2 = 5 and y = (4 – 2)/2 = 1. i.e. S is (5, 1).
Slope of SR, m = (1 + 1)/(5 + 2) = 2/7.
Equation of SR will be y – y1 = m(x – x1)
Or, y – 1 = 2/7(x – 5)
Or, 7y – 7 = 2x – 10
Or, 2x – 7y – 3 = 0 [Ans.]
Q.12. The line joining P (– 4, 5) and Q (3, 2), intersects the y-axis at R. PM and QN are
perpendiculars from P and Q on the x-axis. Find:
i. The ratio PR : RQ,
ii. The co-ordinates of R,
iii. The area of the quadrilateral PMNQ.
Solution :
Sudheer Gupta .
Be positive and constructive. 
Page 3
Mathematics
Class-X
2015-2016
Fig. soln10(b)/page493[10 yrs]
Slope of PQ = (2 – 5)/(3 + 4) = –3/7.
Equation of PQ is y – 5 = – 3/7(x + 4)
Or, 7y – 35 = – 3x – 12
Or, 3x + 7y – 23 = 0 ---------------------- (i)
This passes through y – axis , putting x = 0 in (i) we get,
7y = 23 => y = 23/7 ;
the co-ordinates of R is (0, 23/7).
i. Let the ratio be k : 1
Therefore, 0 = {k×3 + 1×(– 4)}/(k + 1)
Or, 3k – 4 = 0 => k = 4/3 => k : 1 = 4 : 3
Or, PR : RQ = 4 : 3. [Ans.]
ii. Co-ordinates of R is R(0, 23/7). [Ans.]
iii. Area of quadrilateral PMNQ = 1/2×[PM + QN]×MN
= 1/2×(5 + 2)×7
= 1/2×7×7
= 49/2 = 24.5 sq. units. [Ans.]
Q.13. A straight line passes through the points P (– 1, 4) and Q (5, – 2). It intersects the coordinate axes at points A and B. M is the mid point of the segment AB. Find :
i. The equation of the line.
ii. The co-ordinates of A and B.
iii. The co-ordinates of M.
Solution:
We have, P (– 1, 4) and Q (5, – 2) and line passes through P and Q.
Therefore, slope of the line = (y2 – y1)/(x2 – x1) = [– 2 – 4]/[5 – (– 1)] = – 6/6 = – 1.
i. Thus the equation of the line is : y – y1 = m (x – x1)
Or, y – 4 = (– 1) [x – (– 1)]
Or, y – 4 = – x – 1
Or, x + y – 3 = 0 [Ans.]
ii. Putting y = 0, co-ordinates of A are (3, 0) [Ans.]
and putting x = 0, co-ordinates of B are (0, 3). [Ans.]
iii. Co-ordinates of ‘M’, mid-point of AB = [(x1 + x2)/2, (y1 + y2)/2]
= [(3 + 0)/2, (0 + 3)/2]
= (3/2, 3/2). [Ans.]
Q.14. Find the equation of a line passing through the point (– 2, 3) and having x-intercept
of 4 units.
Solution :
Line has x-intercept of 4 units means it passes through (4, 0). Another point is (– 2, 3).
The slope of line, m = (y2 – y1)/(x2– x1) = (0 – 3)/(4 + 2) = – 3/6 = – 1/2.
Therefore, equation of line is : y – y1 = m (x – x1)
Or, y – 3 = – 1/2 (x + 2)
Or, 2y – 6 = – x – 2
Sudheer Gupta .
Be positive and constructive. 
Page 4
Mathematics
Class-X
2015-2016
Or, 2y = – x – 2 + 6 = – x + 4
Or, x + 2y – 4 = 0. [Ans.]
Q.15.
i. Find the equation of a line, which has y-intercept 4 and is parallel to the line 2x – 3y
= 7.
ii. Find the co-ordinates of the point where it cuts the x-axis.
Solution :
i. Equation of line is 2x – 3y = 7 => y = (2/3)x – 7/3.
Therefore slope = 2/3 = m1.
Let slope of the line parallel to given line be m2.
Then m1= m2 = 2/3
Therefore, equation of line is y = mx + c
Or, y = (2/3)x + 4
Or, 3y = 2x + 12
Or, 2x – 3y + 12 = 0 [Ans.]
ii. This line meets the x-axis where y = 0 , 2x – 0 + 12 = 0 => x = – 6.
Therefore, the point is (– 6, 0). [Ans.]
Sudheer Gupta .
Be positive and constructive. 
Page 5