Question 37
Find the equation of the line perpendicular to 3y − 2x = 6 and passing through the point (−2, 1).
Solution
The equation of the given line can be written as:
3y = 2x + 6 ==> y = 2x/3 + 2.
Here, 2/3 is the slope of the given equation.
Assume the slope of the perpendicular line to be M. Product of slopes of 2 perpendicular lines = -1.
2M/3 = -1 ==> M = -3/2.
Equation of the perpendicular line:
y = Mx + c
==> y = -3x/2 + c ==> 2y +3x = 2C.
Putting point -2,1 in the equation as the line passes from this point.
2(1) + 3(-2) = 2 - 6 = -4 = 2C.
C = -2.
The equation will be: 2y + 3x = -4 ==> 3x+2y+4 = 0.
