Answer
$3x-2y=-13$
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 $
(2,3) \text{ and } (-4,7)
$ is
\begin{array}{l}\require{cancel}
\left( \dfrac{2+(-4)}{2},\dfrac{3+7}{2} \right)
\\\\
\left( \dfrac{2-4}{2},\dfrac{3+7}{2} \right)
\\\\
\left( \dfrac{-2}{2},\dfrac{10}{2} \right)
\\\\
\left( -1,5 \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{3-7}{2-(-4)}
\\\\
m=\dfrac{3-7}{2+4}
\\\\
m=\dfrac{-4}{6}
\\\\
m=-\dfrac{2}{3}
.\end{array}
Taking the negative reciprocal of $m$, then the slope of the perpendicular bisector is $m_p=
\dfrac{3}{2}
.$
Using $
(-1,5)
$ and $m_p=
\dfrac{3}{2}
,$ the equation of the perpendicular bisector is
\begin{array}{l}\require{cancel}
y-5=\dfrac{3}{2}(x-(-1))
\\\\
y-5=\dfrac{3}{2}(x+1)
\\\\
2(y-5)=3(x+1)
\\\\
2y-10=3x+3
\\\\
-3x+2y=3+10
\\\\
-3x+2y=13
\\\\
3x-2y=-13
.\end{array}
