Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is ![y=\frac{1}{3} x+2](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B3%7D%20x%2B2)
<u>Solution:</u>
Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).
Generic slope intercept form of a line is given by y = mx + c
where m = slope of the line.
Let's first find slope intercept form of 3x + y = -8
3x + y = -8
=> y = -3x - 8
On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3
And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.
Let required slope = x
![\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%3Dx%20%5Ctimes-3%3D-1%7D%20%5C%5C%5C%5C%20%7B%3D%3Ex%3D%5Cfrac%7B-1%7D%7B-3%7D%3D%5Cfrac%7B1%7D%7B3%7D%7D%5Cend%7Barray%7D)
So we need to find the equation of a line whose slope is
and passing through (-3,1)
Equation of line passing through
and having lope of m is given by
![\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)](https://tex.z-dn.net/?f=%5Cleft%28y-y_%7B1%7D%5Cright%29%3D%5Cmathrm%7Bm%7D%5Cleft%28x-x_%7B1%7D%5Cright%29)
![\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20In%20our%20case%20%7D%20x_%7B1%7D%3D-3%20%5Ctext%20%7B%20and%20%7D%20y_%7B1%7D%3D1%20%5Ctext%20%7B%20and%20%7D%20%5Cmathrm%7Bm%7D%3D%5Cfrac%7B1%7D%7B3%7D)
Substituting the values we get,
![\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%28%5Cmathrm%7By%7D-1%29%3D%5Cfrac%7B1%7D%7B3%7D%28%5Cmathrm%7Bx%7D-%28-3%29%29%7D%20%5C%5C%5C%5C%20%7B%3D%3E%5Cmathrm%7By%7D-1%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cmathrm%7Bx%7D%2B1%7D%20%5C%5C%5C%5C%20%7B%3D%3E%5Cmathrm%7By%7D%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cmathrm%7Bx%7D%2B2%7D%5Cend%7Barray%7D)
Hence the required equation of line is found using slope intercept form