Question

What is the slope-intercept form of the equation of the line that passes through the point (–6, 1) and is perpendicular to the graph of 2x + 3y = –5?

Answer 1

First you want to find the slope of the reference line.  We are given a line in standard form ax+by=c, and we can rearrange this into point slope form, y=mx+b where m=slope.

2x+3y=-5  subtract 2x from both sides

3y=-2x-5

y=(-2x-5)/3

So the slope of the reference line is -2/3

For lines to be perpendicular their slopes must be negative reciprocals of one another, mathematically:

m1*m2=-1  ie:  the product of the slopes is negative one.

In this case our perpendicular line must have a slope of:

m(-2/3)=-1

-2m=-3

m=3/2

m=1.5

So far we now have:

y=1.5x+b, using the point (-6,1) we can solve for b, the y-intercept...

1=1.5(-6)+b

1=-9+b

10=b

So the perpendicular line is:

y=1.5x+10