# How do you find the slope of a line perpendicular to V(3, 2), W(8, 5)?

May 20, 2017

See a solution process below:

#### Explanation:

First, find the slope of the line V-W. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$

Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.

Substituting the values from the points in the problem gives:

$m = \frac{\textcolor{red}{5} - \textcolor{b l u e}{2}}{\textcolor{red}{8} - \textcolor{b l u e}{3}} = \frac{3}{5}$

Now, let's call the slope of the perpendicular line ${m}_{p}$

The slope of a perpendicular line is the negative inverse of the slope of the line it is perpendicular to, or:

${m}_{p} = - \frac{1}{m}$

Substituting for $m$ gives:

${m}_{p} = - \frac{1}{\frac{3}{5}}$

${m}_{p} = - \frac{5}{3}$

The slope of the line perpendicular to V-W is $- \frac{5}{3}$

May 20, 2017

$m = - \frac{5}{3}$

#### Explanation:

$\text{to calculate the slope of the line VW use the "color(blue)"gradient formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 coordinate points}$

$\text{the points are } \left({x}_{1} , {y}_{1}\right) = \left(3 , 2\right) , \left({x}_{2} , {y}_{2}\right) = \left(8 , 5\right)$

$\Rightarrow {m}_{\textcolor{red}{V W}} = \frac{5 - 2}{8 - 3} = \frac{3}{5}$

$\text{ the slope of a line perpendicular to VW is}$

${m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{m} _ \left(\textcolor{red}{V W}\right)$

$\Rightarrow {m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{\frac{3}{5}} = - \frac{5}{3}$