Answer
$AB\perp AC$
Work Step by Step
Determine the slope of the line passing through $A(-3,-1)$ and $B(3,3)$:
$$m_{AB}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{3-(-1)}{3-(-3)}=\dfrac{2}{3}.$$
Determine the slope of the line passing through $B(3,3)$ and $C(-9,8)$:
$$m_{BC}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{8-3}{-9-3}=-\dfrac{5}{12}.$$
Determine the slope of the line passing through $C(-9,8)$ and $A(-3,-1)$:
$$m_{CA}=\dfrac{y_A-y_C}{x_A-x_C}=\dfrac{-1-8}{-3-(-9)}=-\dfrac{3}{2}.$$
We notice that we have
$$m_{AB}\cdot m_{CA}=\dfrac{2}{3}\cdot\left(-\dfrac{3}{2}\right)=-1.$$
This means that $AB$ is perpendicular to $AC$, so the triangle $ABC$ is a right triangle ($m\angle A=90$ degrees).
