Answer
$$\color {blue}{\bf\text{Yes, the points make a right triangle}}$$
Work Step by Step
For clarity lets start by naming our points:
$A=(-2,-8)$
$B=(0,-4)$
$C=(-4,-7)$
First, we'll use the distance formula to find the lengths of each side:
$d(P,Q)=\sqrt{(x_{P}-x_{Q})^2+(y_{P}-y_{Q})^2}$
$AB= \sqrt{(-2-0)^2+(-8-(-4))^2}$
$AB= \sqrt{(-2)^2+(-4)^2}$
$AB= \sqrt{4+16}$
$AB= \sqrt{20}$
$AC= \sqrt{(-2-(-4))^2+(-8-(-7))^2}$
$AC= \sqrt{2^2+(-1)^2}$
$AC= \sqrt{4+1}$
$AC= \sqrt{5}$
$BC= \sqrt{(0-(-4))^2+(-4-(-7))^2}$
$BC= \sqrt{4^2+3^2}$
$BC= \sqrt{16+9}$
$BC= \sqrt{25}$
$BC= 5$
Now that we have the lengths of the sides,
$AB= \sqrt{20}$, $AC= \sqrt{5}$, $BC= 5$,
we can apply the Pythagorean Theorem $a^2+b^2=c^2$ to see if the sides make a right triangle.
$(\sqrt{20})^2+(\sqrt{5})^2=5^2$
$20+5=25$
Which is $\bf \text{true}$, so:
$$\color {blue}{\bf\text{Yes, the points make a right triangle}}$$
