# What is the equation of the line that is perpendicular to the line passing through (-8,10) and (-5,12) at midpoint of the two points?

Aug 28, 2017

See a solution process below:

#### Explanation:

First, we need to find the midpoint of the two points in the problem. The formula to find the mid-point of a line segment give the two end points is:

$M = \left(\frac{\textcolor{red}{{x}_{1}} + \textcolor{b l u e}{{x}_{2}}}{2} , \frac{\textcolor{red}{{y}_{1}} + \textcolor{b l u e}{{y}_{2}}}{2}\right)$

Where $M$ is the midpoint and the given points are:

$\left(\textcolor{red}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ and $\left(\textcolor{b l u e}{{x}_{2}} , \textcolor{b l u e}{{y}_{2}}\right)$

Substituting gives:

$M = \left(\frac{\textcolor{red}{- 8} + \textcolor{b l u e}{- 5}}{2} , \frac{\textcolor{red}{10} + \textcolor{b l u e}{12}}{2}\right)$

$M = \left(- \frac{13}{2} , \frac{22}{2}\right)$

$M = \left(- 6.5 , 11\right)$

Next, we need to find the slope of the line containing the two points in the problem. 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:

$m = \frac{\textcolor{red}{12} - \textcolor{b l u e}{10}}{\textcolor{red}{- 5} - \textcolor{b l u e}{- 8}} = \frac{\textcolor{red}{12} - \textcolor{b l u e}{10}}{\textcolor{red}{- 5} + \textcolor{b l u e}{8}} = \frac{2}{3}$

Now, let's call the slope of the perpendicular line ${m}_{p}$. The formula for finding ${m}_{p}$ is:

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

Substituting gives: ${m}_{p} = - \frac{1}{\frac{2}{3}} = - \frac{3}{2}$

We can now use the point-slope formula to find an equation for the perpendicular line going through the midpoint of the two points given in the problem. The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

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

Substituting the slope we calculated and the values from the mid-point we calculated gives:

$\left(y - \textcolor{b l u e}{11}\right) = \textcolor{red}{- \frac{3}{2}} \left(x - \textcolor{b l u e}{- 6.5}\right)$

$\left(y - \textcolor{b l u e}{11}\right) = \textcolor{red}{- \frac{3}{2}} \left(x + \textcolor{b l u e}{6.5}\right)$

If necessary, we can solve for $y$ to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y - \textcolor{b l u e}{11} = - \frac{3}{2} x + \left(- \frac{3}{2} \times \textcolor{b l u e}{6.5}\right)$

$y - \textcolor{b l u e}{11} = - \frac{3}{2} x - 9.75$

$y - \textcolor{b l u e}{11} + 11 = - \frac{3}{2} x - 9.75 + 11$

$y - 0 = - \frac{3}{2} x + 1.25$

$y = \textcolor{red}{- \frac{3}{2}} x + \textcolor{b l u e}{1.25}$