Question

write the equation of a line that is perpendicular to y=3x-2 that passes trough the point (-9,5)

Answer 1

y= -1/3x + 2

Important takeaway is that the slopes of two perpendicular lines are always negative reciprocals.

Answer 2

$\huge\boxed{y-5&=-\frac{1}{3}(x+9)}$

First, we'll find the slope of the new line. The first line has a slope of $3$. Take the negative reciprocal of this (Flip the numerator and denominator, then multiply by $-1$) to get $-\frac{1}{3}$ for the new slope.

Then, we'll use the point-slope form to make the new equation, where $m$ is the slope and $(x_1,y_1)$ is a point on the line:

$\begin{aligned}y-y_1&=m(x-x_1)\\y-5&=-\frac{1}{3}(x-(-9))\\y-5&=-\frac{1}{3}(x+9)\end{aligned}$