Answer
$x-3y=-9$
Work Step by Step
Using $\left( \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right)$ or the Midpoint Formula, then the midpoint of the line segment joining the points $
(5,8) \text{ and } (7,2)
$ is
\begin{array}{l}\require{cancel}
\left( \dfrac{5+7}{2},\dfrac{8+2}{2} \right)
\\\\
\left( \dfrac{12}{2},\dfrac{10}{2} \right)
\\\\
\left( 6,5 \right)
.\end{array}
Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, then \begin{array}{l}\require{cancel}
m=\dfrac{8-2}{5-7}
\\\\
m=\dfrac{6}{-2}
\\\\
m=-3
.\end{array}
Taking the negative reciprocal of $m$, then the slope of the perpendicular bisector is $m_p=
\dfrac{1}{3}
.$
Using $
(6,5)
$ and $m_p=
\dfrac{1}{3}
,$ the equation of the perpendicular bisector is
\begin{array}{l}\require{cancel}
y-5=\dfrac{1}{3}(x-6)
\\\\
3(y-5)=1(x-6)
\\\\
3y-15=x-6
\\\\
-x+3y=-6+15
\\\\
-x+3y=9
\\\\
x-3y=-9
.\end{array}