Answer
Two of the three sides have slopes whose product is $-1$ which means that they are perpendicular to each other.
This implies that the triangle formed by connecting the three vertices is a right triangle.
Work Step by Step
Solve for the slope (let's name it as $m_1$) of the line joining the vertices $A(- 2, 5)$ and $B(1, 3)$:
\begin{align*}
m_1 &= \frac{3-5}{1-(-2)}\\
&=\frac{-2}{1+2}\\
&=-\frac{2}{3}
\end{align*}
Solve for the slope (let's name is $m_2$) of the line joining the vertices $B(1, 3)$ and $C(-1, 0)$:
\begin{align*}
m_2 &= \frac{0-3}{-1-1}\\
&=\frac{-3}{-2}\\
&=\frac{3}{2}\\
\end{align*}
SOlve for the slope (let's name it $m_3$) of the line joining the vertices $C(-1, 0)$ and $A(- 2, 5)$:
\begin{align*}
m_3&= \frac{5-0}{-2-(-1)}\\
&=\frac{5}{-2+1}\\
&=\frac{5}{-1}\\
&=-5\end{align*}
Now we can see $m_1 × m_2 = -\frac{2}{3} \times \frac{3}{2}=-1$.
Since the product of the slopes of the sides $\overline{AB}$ and $\overline{BC}$ is $-1$ then the sides are perpendicular to each other.
Thus the given given vertices are vertices of a right triangle.
