Newbie

# The equation of a line is 3x + 4y – 7 = 0. Find (i) the slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.

• 0

M.L Aggarwal book Important Question of class 10 chapter Based on Equation of a Straight Line for ICSE BOARD.
You have given an equation of a line.
So , Find the the slope of the line and the equation of a line perpendicular to the given line and passing through the intersection of the given lines.
This is the Question Number 22, Exercise 12.2 of M.L Aggarwal.

Share

1. Given line equation: 3x + 4y – 7 = 0

(i) Slope of the line is given by,

4y = -3x + 7

y = (-3/4) x + 7

Hence, slope (m1) = -3/4

(ii) Let the slope of the perpendicular to the given line be m2

Then, m1 x m2 = -1

(-3/4) x m2 = -1

m2 = 4/3

Now, to find the point of intersection of

x – y + 2 = 0 … (i)

3x + y – 10 = 0 … (ii)

On adding (i) and (ii), we get

4x – 8 = 0

4x = 8

x = 8/4 = 2

Putting x = 2 in (i), we get

2 – y + 2 = 0

y = 4

Hence, the point of intersection of the lines is (2, 4)

The equation of the line having slope m2 and passing through (2, 4) will be

y – 4 = (4/3) (x – 2)

3y – 12 = 4x – 8

4x – 3y + 4 = 0

Thus, the required line equation is 4x – 3y + 4 = 0.

• 1