Answer
$\text{a) Distance: }
10
\text{ units}\\\text{b) Midpoint: }
\left( -1, -1 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the Distance Formula and the Midpoint Formula to find the distance and the midpoint of the given points $\left(
-4,3
\right)$ and $\left(
2,-5
\right).$
$\bf{\text{Solution Details:}}$
With the given points, then $x_1=
-4
,$ $x_2=
2
,$ $y_1=
3
,$ and $y_2=
-5
.$
Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
,$ then
\begin{array}{l}\require{cancel}
d=\sqrt{(-4-2)^2+(3-(-5))^2}
\\\\
d=\sqrt{(-4-2)^2+(3+5)^2}
\\\\
d=\sqrt{(-6)^2+(8)^2}
\\\\
d=\sqrt{36+64}
\\\\
d=\sqrt{100}
\\\\
d=\sqrt{(10)^2}
\\\\
d=10
.\end{array}
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 with endpoints given above is
\begin{array}{l}\require{cancel}
\left( \dfrac{-4+2}{2}, \dfrac{3+(-5)}{2} \right)
\\\\=
\left( \dfrac{-4+2}{2}, \dfrac{3-5}{2} \right)
\\\\=
\left( \dfrac{-2}{2}, \dfrac{-2}{2} \right)
\\\\=
\left( -1, -1 \right)
.\end{array}
Hence,
$
\text{a) Distance: }
10
\text{ units}\\\text{b) Midpoint: }
\left( -1, -1 \right)
.$
