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.
Deepak BoraNewbie
Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (– 1, – 2).
Share
Given line: 3x + 8y = 12
8y = -3x + 12
y = (-3/8) x + 12
So, the slope (m1) = -3/8
Let’s consider the slope of the line perpendicular to the given line as m2
Then, m1 x m2 = -1
-3/8 x m2 = -1
m2 = 8/3
Now,
The equation of the line perpendicular to the given line and passing through the point (-1, -2) will be
y – y1 = m (x – x1)
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.