Expert: Abe Mantell - 7/9/2009

Question
How do I find the equation of a line that is perpendicular to another line with equation x-3y=5 and passes through the point (4,13)

Answer
Hello Belle,

Rewrite x-3y=5 as -3y=-x+5 then y=(1/3)x-5/3, so the slope is 1/3.
Since perpendicular lines have slopes that are negative reciprocals
of each other, the perpendicular line has a slope of -3.

Now find the equation of the line with slope -3 through the point (4,13).
==> y-13=-3(x-4) ==> y-13=-3x+12 ==> y=-3x+25

OK?

Abe