What is the slope of any line perpendicular to the line passing through (0,2) and (3,15)?

Aug 2, 2018

The slope of any line passing through the given two points is $- \frac{3}{13}$.

Explanation:

First find the slope of the line passing through the two given points.

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$,

where:

$m$ is the slope, and $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are the two given points. I'm going to use point $\left(0 , 2\right)$ as $\left({x}_{1} , {y}_{1}\right)$ and $\left(3 , 15\right)$ as $\left({x}_{2} , {y}_{2}\right)$. You could do the reverse and you would get the same slope.

Plug in the known values and solve for slope.

$m = \frac{15 - 2}{3 - 0}$

$m = \frac{13}{3}$

The perpendicular slope of any line is the negative reciprocal of the original slope, so that ${m}_{1} {m}_{2} = - 1$.

$\frac{13}{3} \times - \frac{3}{13} = - \frac{39}{39} = - 1$. The perpendicular slope is $- \frac{3}{13}$.

graph{(y+13/3x)(y-3/13x)=0 [-10, 10, -5, 5]}