Answer
$\text{a) Distance: }
3\sqrt{41}
\text{ units}\\\text{b) Midpoint: }
\left( 0, \dfrac{5}{2} \right)$
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 $\left(
-6,-5
\right)$ and $\left(
6,10
\right).$
$\bf{\text{Solution Details:}}$
With the given points, then $x_1=
-6
,$ $x_2=
6
,$ $y_1=
-5
,$ and $y_2=
10
.$
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{(-6-6)^2+(-5-10)^2}
\\\\
d=\sqrt{(-12)^2+(-15)^2}
\\\\
d=\sqrt{144+225}
\\\\
d=\sqrt{369}
\\\\
d=\sqrt{9\cdot41}
\\\\
d=\sqrt{(3)^2\cdot41}
\\\\
d=3\sqrt{41}
.\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{-6+6}{2}, \dfrac{-5+10}{2} \right)
\\\\=
\left( \dfrac{0}{2}, \dfrac{5}{2} \right)
\\\\=
\left( 0, \dfrac{5}{2} \right)
.\end{array}
Hence,
$
\text{a) Distance: }
3\sqrt{41}
\text{ units}\\\text{b) Midpoint: }
\left( 0, \dfrac{5}{2} \right)
.$
