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.

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# 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.

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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 (m

_{1}) = -3/4(ii) Let the slope of the perpendicular to the given line be m

_{2}Then, m

_{1}x m_{2}= -1(-3/4) x m

_{2}= -1m

_{2}= 4/3Now, 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 m

_{2}and passing through (2, 4) will bey – 4 = (4/3) (x – 2)

3y – 12 = 4x – 8

4x – 3y + 4 = 0

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