Question

How do I find the equation of the perpendicular bisector of the line segment whose endpoints are $\left(-4,8\right)$$\left(-5,3\right)$ and $\left(-6,-2\right)$ using the Midpoint Formula?

Answer 1

Step-1: Mid Point of the line segment:

A midpoint $M$ of a line segment with endpoints $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is defined as :

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Here the endpoints of the line segment are $\left(-4,8\right)$ and $\left(-6,-2\right)$.

So mid point of this line segment is :

$\begin{array}{rcl}M& =& \left(\frac{-4-6}{2},\frac{8-2}{2}\right)\\ & \Rightarrow & M=\left(-5,3\right)\end{array}$

As midpoint bisects the line segment.

So it will lies on perpendicular bisector.

Step-2: Slope of the Perpendicular bisector:

Slope of the given line is :

$\begin{array}{rcl}m& =& \frac{-2-8}{-6+4}\\ & \Rightarrow & m=\frac{-10}{-2}\\ & \Rightarrow & m=5\end{array}$

Slope of two perpendicular lines are negative reciprocal of each other.

So the slope of perpendicular bisector is :

$m=-\frac{1}{5}$.

Step-3: Equation of perpendicular bisector:

By using point slope form the equation of a line is :

$y-y_1=m\left(x-x_1\right)$.

Here the point $(-5,3)$ lies on perpendicular bisector with slope $-\frac{1}{5}$.

So the equation of perpendicular bisector is :

$\begin{array}{rcl}y-3& =& -\frac{1}{5}\left(x+5\right)\Rightarrow 5\left(y-3\right)=-1(x+5)\\ & \Rightarrow & 5y-15=-x-5\\ & \Rightarrow & 5y+x=-5+15\\ & \Rightarrow & 5y+x-10=0\end{array}$

Therefore, the required equation of perpendicular bisector is $\mathbf{5}\mathit{y}\mathbf{+}\mathit{x}\mathbf{-}\mathbf{10}\mathbf{=}\mathbf{0}$.