91Ó°ÊÓ

Coordinates are $(−1,−3)$ and$(3,5)$

The distance ($d$) between the points$(x_1,y_1)$ and$(x_2,y_2)$ is given by:

$d=\sqrt{{(x_2−x_1)}^2+{(y_2−y_1)}^2}$

Therefore, the distance ($d$) between the points$(−1,−3)$ and$(3,5)$ is:

$\begin{array}{l}d=\sqrt{{(3−(−1))}^2+{(5−(−3))}^2}\\ =\sqrt{{(3+1)}^2+{(5+3)}^2}\\ =\sqrt{4^2+8^2}\\ =\sqrt{16+64}\\ =\sqrt{80}\\ =\sqrt{16×5}\\ =4\sqrt{5}\\ =4\left(2.236068\right)\\ =8.944272\\ ≈8.94\left(\text{roundedtothenearesthundredth}\right)\end{array}$

Therefore, the distance ($d$) between the points$(−1,−3)$ and$(3,5)$ is $8.94$.

The midpoint formula states that the midpoint of a line segment with endpoints at $(x_1,y_1)$and $(x_2,y_2)$is given by $M=\left(\frac{{\displaystyle x_1+x_2}}{{\displaystyle 2}},\frac{{\displaystyle y_1+y_2}}{{\displaystyle 2}}\right)$.

The midpoint ($M$) of the line segment with endpoints at$(−1,−3)$ and$(3,5)$ is given by:

$\begin{array}{l}M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{−1+3}{2},\frac{−3+5}{2}\right)\\ =\left(\frac{2}{2},\frac{2}{2}\right)\\ =(1,1)\end{array}$

Therefore, the midpoint of the line segment with the endpoints at$(−1,−3)$ and$(3,5)$ is $(1,1)$.