This question is Based on Equation of a Straight Line Chapter of M.L Aggarwal book for ICSE BOARD for class 10.

Passing point is given of a perpendicular line. So write down the equation

This is the Question Number 17, Exercise 12.2 of M.L Aggarwal.

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# Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (– 1, – 2).

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Given line: 3x + 8y = 12

8y = -3x + 12

y = (-3/8) x + 12

So, the slope (m

_{1}) = -3/8Let’s consider the slope of the line perpendicular to the given line as m

_{2}Then, m

_{1}x m_{2}= -1-3/8 x m

_{2}= -1m

_{2}= 8/3Now,

The equation of the line perpendicular to the given line and passing through the point (-1, -2) will be

y – y

_{1}= m (x – x_{1})y – (-2) = (8/3) (x – (-1))

y + 2 = (8/3) (x + 1)

3y + 6 = 8x + 8

3y = 8x + 2

Thus, the equation of the required line is 3y = 8x + 2.