Question

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. Show your work.

Answer 1

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is
`y=(1)/(3) x+2`

Solution:

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 *-3=-1} \\\\ {=>x=(-1)/(-3)=(1)/(3)}\end{array}`

So we need to find the equation of a line whose slope is
`(1)/(3)` and passing through (-3,1)

Equation of line passing through
`(x_1 , y_1)` and having lope of m is given by

`\left(y-y_(1)\right)=\mathrm{m}\left(x-x_(1)\right)`

`\text { In our case } x_(1)=-3 \text { and } y_(1)=1 \text { and } \mathrm{m}=(1)/(3)`

Substituting the values we get,

`\begin{array}{l}{(\mathrm{y}-1)=(1)/(3)(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=(1)/(3) \mathrm{x}+1} \\\\ {=>\mathrm{y}=(1)/(3) \mathrm{x}+2}\end{array}`

Hence the required equation of line is found using slope intercept form