## Precalculus (10th Edition)

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. 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.