# What is the equation of the line that is perpendicular to the line passing through (-5,12) and (4,-3) at midpoint of the two points?

Nov 5, 2017

See a solution process below:

#### Explanation:

First, we need to find the mid-point and of the two points and the slope of the line going through the two points.

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 the values from the points in the problem gives:

$M = \left(\frac{\textcolor{red}{- 5} + \textcolor{b l u e}{4}}{2} , \frac{\textcolor{red}{12} + \textcolor{b l u e}{- 3}}{2}\right)$

$M = \left(\frac{\textcolor{red}{- 5} + \textcolor{b l u e}{4}}{2} , \frac{\textcolor{red}{12} - \textcolor{b l u e}{3}}{2}\right)$

$M = \left(- \frac{1}{2} , \frac{9}{2}\right)$

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:

$m = \frac{\textcolor{red}{- 3} - \textcolor{b l u e}{12}}{\textcolor{red}{4} - \textcolor{b l u e}{- 5}} = \frac{\textcolor{red}{- 3} - \textcolor{b l u e}{12}}{\textcolor{red}{4} + \textcolor{b l u e}{5}} = - \frac{15}{9} = - \frac{5}{3}$

Let's call the slope of a perpendicular line: ${m}_{p}$

The formula for the slope of a perpendicular line is:

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

Substituting gives:

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

Now that we have the slope of the perpendicular line and a point on the line (the midpoint of the line segment) we can use the point-slope formula to write an equation for the line. 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 and the values from the mid-point gives:

$\left(y - \textcolor{b l u e}{\frac{9}{2}}\right) = \textcolor{red}{\frac{3}{5}} \left(x - \textcolor{b l u e}{- \frac{1}{2}}\right)$

$\left(y - \textcolor{b l u e}{\frac{9}{2}}\right) = \textcolor{red}{\frac{3}{5}} \left(x + \textcolor{b l u e}{\frac{1}{2}}\right)$

Nov 5, 2017

$3 x - 5 y = - 23$

#### Explanation:

Given the points:
color(white)("XXX")(-5,12)" and "(4,-3)

a. The slope between the given points is
$\textcolor{w h i t e}{\text{XXX}} m = \frac{\Delta y}{\Delta x} = \frac{12 - \left(- 3\right)}{- 5 - 4} = \frac{15}{- 9} = - \frac{3}{5}$

b. The slope of any line perpendicular to this is
$\textcolor{w h i t e}{\text{XXX}} \hat{m} = - \frac{1}{m} = \frac{5}{3}$

c. The midpoint between the given points is
$\textcolor{w h i t e}{\text{XXX}} \left(\hat{x} , \hat{y}\right) = \left(\frac{- 5 + 4}{2} , \frac{12 + \left(- 3\right)}{2}\right) = \left(- \frac{1}{2} , \frac{9}{2}\right)$

d. The equation, in slope-point form, for the perpendicular through the midpoint is
$\textcolor{w h i t e}{\text{XXX}} y - \hat{y} = \hat{m} \left(x - \hat{x}\right)$

$\textcolor{w h i t e}{\text{XXX}} y - \frac{9}{2} = \frac{3}{5} \left(x - \left(- \frac{1}{2}\right)\right)$

e. Converting to standard form:
after multiplying both sides by $10$ to get rid of the fractions:
$\textcolor{w h i t e}{\text{XXX}} 10 y - 45 = 6 x + 1$

rearranging into the standard form: $A x + B y = C$
$\textcolor{w h i t e}{\text{XXX}} 6 x - 10 y = - 46$
and simplifying by dividing all terms by $2$
$\textcolor{w h i t e}{\text{XXX}} 3 x - 5 y = - 23$