# What is the equation of the line that passes through (5,7) and is perpendicular to the line that passes through the following points: (1,3),(-2,8) ?

Jan 31, 2017

$\left(y - \textcolor{red}{7}\right) = \textcolor{b l u e}{\frac{3}{5}} \left(x - \textcolor{red}{5}\right)$

Or

$y = \frac{3}{5} x + 4$

#### Explanation:

First, we will find the slope of the perpendicular line. 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 two points from the problem gives:

$m = \frac{\textcolor{red}{8} - \textcolor{b l u e}{3}}{\textcolor{red}{- 2} - \textcolor{b l u e}{1}}$

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

A perpendicular line will have a slope (let's call it ${m}_{p}$) which is the negative inverse of the line or ${m}_{p} = - \frac{1}{m}$

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

Now that we have the slope of the perpendicular line and one point we can use the point-slope formula to find the equation. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the perpendicular slope we calculated and using the point from the problem gives:

$\left(y - \textcolor{red}{7}\right) = \textcolor{b l u e}{\frac{3}{5}} \left(x - \textcolor{red}{5}\right)$

Or, if we solve for $y$:

$y - \textcolor{red}{7} = \left(\textcolor{b l u e}{\frac{3}{5}} \times x\right) - \left(\textcolor{b l u e}{\frac{3}{5}} \times \textcolor{red}{5}\right)$

$y - \textcolor{red}{7} = \frac{3}{5} x - 3$

$y - \textcolor{red}{7} + 7 = \frac{3}{5} x - 3 + 7$

$y - 0 = \frac{3}{5} x + 4$

$y = \frac{3}{5} x + 4$