Chapter 13.7, Problem 30WE

To determine

To find : an equation of line described

Answer to Problem 30WE

The equation is $4x-3y+8=0$

Explanation of Solution

Given information :

Perpendicular bisector of segment joining $\left(-3,7\right)$ and $\left(5,1\right)$

Formula used :

  $y-y_1=m\left(x-x_1\right)$

Calculation :

The point of bisection

  $\begin{array}{l}=\left(\frac{-3+5}{2},\frac{7+1}{2}\right)\\ =\left(1,4\right)\end{array}$

The slope of segment

  $\begin{array}{l}=\frac{1-7}{5-\left(-3\right)}\\ =\frac{-6}{8}\\ =-\frac{3}{4}\end{array}$

Therefore, the slope is $-\frac{3}{4}$ .

The required line has slope

  $\begin{array}{l}{m}_2=-\frac{1}{m_1}=-\left(-\frac{4}{3}\right)\\ m_2=-\frac{1}{m_1}=\frac{4}{3}\end{array}$

Using $\left(1,4\right)$ , the equation is

  $\begin{array}{l}y-4=\frac{4}{3}\left(x-1\right)\\ y=\frac{4}{3}x-\frac{4}{3}+4\\ y=\frac{4}{3}x+\frac{8}{3}\\ 3y=4x+8\\ 4x-3y+8=0\end{array}$

Hence, the equation is $4x-3y+8=0$ .