Question
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!
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!
Answers
Steve
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
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
Jenn
Please Steve, how did you find -3 and 1/3?