Math Help

It's been a long time since I've done this, so PLEASE check this out in your textbook. As close as I can remember, and figure out, perpendicular lines have negative reciprocal slopes.

The first thing to do is to put the equation of your line into standard form. x-3y = -2 then becomes:

y = x/3 + 2/3

so the slope of your first line is 1/3.

Iff (not a typo - it stands for "if and only if") my memory is correct, then to be perpendicular, the second line must have a slope of -3 to be a negative reciprocal.

Your second line is now in the form of:

y= -3x + b.

To solve for b, plug in the point (-1,2) from the problem statement:

2 = (-3)(-1) +b
2= 3+b
-1 = b

so the answer is y = -3x-1

I just graphed up the two lines, and they do look perpendicular.

yep you're right