The equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5) is ![y = \frac{-1}{3}x - 8](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B-1%7D%7B3%7Dx%20-%208)
<em><u>Solution:</u></em>
Given that line perpendicular to y = 3x + 6 and containing the point (-9, -5)
We have to find the equation of line
<em><u>The slope intercept form is given as:</u></em>
y = mx + c ------ eqn 1
Where "m" is the slope of line and "c" is the y - intercept
<em><u>Let us first find the slope of line</u></em>
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 <em>product of slopes of given line and slope of line perpendicular to given line is equal to -1</em>
Slope of given line
slope of line perpendicular to given line = -1
![3 \times \text{ slope of line perpendicular to given line }= -1](https://tex.z-dn.net/?f=3%20%5Ctimes%20%5Ctext%7B%20slope%20of%20line%20perpendicular%20to%20given%20line%20%7D%3D%20-1)
![\text{ slope of line perpendicular to given line } = \frac{-1}{3}](https://tex.z-dn.net/?f=%5Ctext%7B%20slope%20of%20line%20perpendicular%20to%20given%20line%20%7D%20%3D%20%5Cfrac%7B-1%7D%7B3%7D)
Let us now find the equation of line with slope
and containing the point (-9, -5)
Substitute
and (x, y) = (-9, -5) in eqn 1
![-5 = \frac{-1}{3}(-9) + c\\\\-5 = 3 + c\\\\c = -8](https://tex.z-dn.net/?f=-5%20%3D%20%5Cfrac%7B-1%7D%7B3%7D%28-9%29%20%2B%20c%5C%5C%5C%5C-5%20%3D%203%20%2B%20c%5C%5C%5C%5Cc%20%3D%20-8)
<em><u>Thus the required equation of line is:</u></em>
Substitute
and c = -8 in eqn 1
![y = \frac{-1}{3}x - 8](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B-1%7D%7B3%7Dx%20-%208)
Thus the required equation of line perpendicular to given line is found