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&=-(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
`-(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&=-(1)/(3)(x-(-9))\\y-5&=-(1)/(3)(x+9)\end{aligned}`