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

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


Substituting the values we get,

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