Here is the problem. I got the answer but I would like a better understanding.

Find the equation of a line that passes thru(6,-2) and is perpendicular to equation x-3y+16. I got the final answer being -3x+16. Can someone please explain how to get this? I am so confused!

2 answers

the given line has slope 1/3

so, perpendicular lines will have slope -3

The line you want is thus (using the point-slope form):

y+2 = -3(x-6)
or, if you prefer,
y = -3x + 16
Please Steve, how did you find -3 and 1/3?
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