Chapter 13.5, Problem 13WE

To determine

To show: that $P\left(2,6\right)$ is on the perpendicular bisector of $\overline{AB}$

Explanation of Solution

Given information: The points are $A\left(0,0\right),B\left(8,4\right)$

Proof:

Method 1:

Midpoint of $AB$ is $M$

  $\begin{array}{l}M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ M=\left(\frac{0+8}{2},\frac{0+4}{2}\right)\\ M=\left(\frac{8}{2},\frac{4}{2}\right)\\ M=\left(4,2\right)\\ slope\left(m_1\right)=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{8-0}=\frac{4}{8}=\frac{1}{2}\\ slope\left(m_2\right)=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{2-4}=\frac{4}{-2}=-2\\ m_1\times m_2=\frac{1}{2}\times -2=-1\end{array}$

This proves that $P$ is on perpendicular bisector

Method 2:

  $\begin{array}{l}PA=\sqrt{{\left(6-0\right)}^2+{\left(2-0\right)}^2}\\ PA=\sqrt{36+4}=\sqrt{40}\\ PB=\sqrt{{\left(8-2\right)}^2+{\left(6-4\right)}^2}\\ PB=\sqrt{36+4}=\sqrt{40}\\ PA=PB\end{array}$