Question

Find the equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5).

Answer 1

The equation of the line perpendicular to y =  3x + 6 and containing the point (-9,-5) is $y = \frac{-1}{3}x - 8$

Solution:

Given that line perpendicular to y =  3x + 6 and containing the point (-9, -5)

We have to find the equation of line

The slope intercept form is given as:

y = mx + c  ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Let us first find the slope of line

The given equation of line is y = 3x + 6

On comparing the given equation of line y = 3x + 6 with eqn 1, we get,

m = 3

Thus the slope of given equation of line is 3

We know that product of slopes of given line and slope of line perpendicular to given line is equal to -1

Slope of given line $\times$ slope of line perpendicular to given line = -1

$3 \times \text{ slope of line perpendicular to given line }= -1$

$\text{ slope of line perpendicular to given line } = \frac{-1}{3}$

Let us now find the equation of line with slope $m = \frac{-1}{3}$ and containing the point (-9, -5)

Substitute $m = \frac{-1}{3}$ and (x, y) = (-9, -5) in eqn 1

$-5 = \frac{-1}{3}(-9) + c\\\\-5 = 3 + c\\\\c = -8$

Thus the required equation of line is:

Substitute $m = \frac{-1}{3}$ and c = -8 in eqn 1

$y = \frac{-1}{3}x - 8$

Thus the required equation of line perpendicular to given line is found