# ICSE Solutions for Chapter 14 Equation of a Line Class 10 Mathematics

Question 1: Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3:5. Find its co-ordinates of point P. Also find the equation of the line through P and parallel to 3x + 5y = 7.

Solution 1: Using section formula, the co-ordinates of the point P are
(3×16 + 5×8)/(3 + 5), (3×(-8) + 5×0)/(3 + 5)}
= (11, -3) = (x1, y1)
3x + 5y = 7
⇒ y = -3/5x + 7/5
Slope of this line = -3/5
As the required line is parallel to the line 3x + 5y = 7.
Slope of the required line = Slope of the given line = -3/5
Thus, the equation of the required line is
y – y1 = m(x – x1)
y + 3 = -3/5(x – 11)
5y + 15 = - 3x + 33
3x + 5y = 18

Question 2: The equation of a line x - y = 4. Find its slope and y-intercept. Also, find its inclination.

Solution 2: Given equation of a line is x – y = 4
⇒  y = x - 4
Comparing this equation with y = mx + c. We have:
Slope = m = 1
y-intercept = c = -4
Let the inclination be θ.
Slope = 1 = tanθ = tan 45˚
∴ θ =  45°

Question 3: Find the equation of the line whose slope is -4/3 and which passes through (-3, 4).

Solution 3: Given, slopes= -4/3
The equation passes through (-3, 4) (x1, y1)
Substituting the values in y – y1 = m(x – x1), we get
y – 4 = -4/3(x + 3)
3y - 12 = -4x - 12
4x + 3y = 0, which is the required equation.

Question 4: Find the slope of the line parallel to AB if :
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)

Solution 4:
(i) Slope of AB = (6 – 4)/(0 + 2) = 2/2 = 1
Slope of the line parallel to AB = Slope of AB =  1

(ii) Slope of AB = (5 + 3)/(-2 – 0) = 8/-2 = -4
Slope of the line parallel to AB = Slope of AB = -4

Question 5: The line 3x/5 – 2y/3 + 1 = 0 contains the point (m, 2m - 1); calculate the value of m.

Solution 5:
The equation of the given line is 3x/5 – 2y/3 + 1 = 0
Putting x = m, y = 2m - 1, we have:
3m/5 – 2(2m – 1)/3 + 1 = 0
3m/5 – (4m – 2)/3 = -1
(9m – 20m + 10)/15 = -1
9m – 20m + 10 = -15
-11m = -25
m = 25/11 = 2 3/11

Question 6: Does the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2)?

Solution 6: The given line will bisect the join of A (5, -2) and B (-1, 2), if the co-ordinates of the mid-point of AB satisfy the equation of the line.
The co-ordinates of the mid-point of AB are
{(5 – 1)/2, (-2 + 2)/2} = (2, 0)
Substituting x = 2 and y = 0 in the given equation, we have:
L.H.S. = 3x - 5y = 3(2) - 5(0) = 6 – 0 = 6 = R.H.S.
Hence, the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2).

Question 7: The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.

Solution 7:
Slope of the line passing through (-4, -2) and (2, -3) = (-3 + 2)/(2 + 4) = -1/6
Slope of the line passing through (a, 5) and (2, -1) = (-1 – 5)/(2 – a) = -6/(2 – a)
Since, the lines are perpendicular.
∴ -1/6  = -1/{-6/(2 – a)}
-1/6 = (2 – a)/6
2 – a = -1
a = 3

Question 8: The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equation of AB and CD. Solution 8: Slope of line AB = tan 45˚ = 1
The line AB passes through P (3, 4). So, the equation of the line AB is given by:
y –  y1 = m(x –  x1)
y – 4 = 1(x – 3)
y – 4 =  x – 3
y = x + 1
Slope of line CD = tan 60° = √3
The line CD passes through P (3,4). So, the equation of the line CD is given by:
y -  y1  m(x - x1)
y – 4 = √3(x - 3)
y – 4 = √3× -3√3
y = √3x + 4 - 3√3

Question 9: The lines represented by 4x + 3y = 9 and px - 6y + 3 = 0 are parallel. Find the value of p.

Solution 9:
4x + 3y = 9
3y = -4x + 9
y = -4/3x + 3
Slope of this line = -4/3
px - 6y + 3 = 0
6y = px + 3
y = P/6x + 1/2
Slope of this line = P/6
Since the lines are parallel, their slopes will be equal.
∴ -4/3 = P/6
-4 = P/2
P = - 8

Question 10: Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x - 8y = -23, and perpendicular to the line 4x - 2y = 1.

Solution 10: 7x + 6y = 71
⇒ 28x + 24 = 284 ...(1)
5x - 8y = -23
⇒ 15x - 24y = -69 …(2)
Adding (1) and (2), we get,
43x = 215
⇒ x = 5
From (2), 8y = 5x + 23
Putting value of x,
25 + 23 = 48 ⇒ y = 6
Thus, the required line passes through the point (5, 6).
4x - 2y = 1
2y = 4x - 1
y = 2x – 1/2
Slope of this line = 2
Slope of the required line = -1/2
The required equation of the line is
y - y1 = m(x - x1)
y – 6 = -1/2(x - 5)
2y - 12 = -x + 5
x + 2y = 17