Answer
$$\color {blue}{\bf\text{Yes, the points make a right triangle}}$$
Work Step by Step
For clarity lets start by naming our points:
$A=(-6,-4)$
$B=(0,-2)$
$C=(-10,8)$
First, we'll use the distance formula to find the lengths of each side:
$d(P,Q)=\sqrt{(x_{P}-x_{Q})^2+(y_{P}-y_{Q})^2}$
$AB= \sqrt{(-6-0)^2+(-4-(-2))^2}$
$AB= \sqrt{(-6)^2+(-2))^2}$
$AB= \sqrt{36+4}$
$AB= \sqrt{40}$
$AB= 2\sqrt{10}$
$AC= \sqrt{(-6-(-10))^2+(-4-8)^2}$
$AC= \sqrt{(4^2+(-12)^2}$
$AC= \sqrt{(16+144}$
$AC= \sqrt{(160}$
$AC= 4\sqrt{(10}$
$BC= \sqrt{(0-(-10))^2+(-2-8)^2}$
$BC= \sqrt{10^2+(-10)^2}$
$BC= \sqrt{100+100}$
$BC= \sqrt{200}$
$BC= 10\sqrt{2}$
Now that we have the lengths of the sides,
$AB= 2\sqrt{10}$, $AC= 4\sqrt{(10}$, $BC= 10\sqrt{2}$,
we can apply the Pythagorean Theorem $a^2+b^2=c^2$ to see if the sides make a right triangle.
$(2\sqrt{10})^2+(4\sqrt{(10})^2=(10\sqrt{2})^2$
$4(10)+16(10)=100(2)$
$40+160=200$
Which is $\bf \text{true}$, so:
$$\color {blue}{\bf\text{Yes, the points make a right triangle}}$$
