Chapter 1.10, Problem 58E

To determine

By using slope, we have to find whether the vertices form a right-angle triangle or not.

Answer to Problem 58E

Yes, the vertices ABC are forming a right-angle triangle.

Explanation of Solution

Given information:

  $\begin{array}{l}A=\left(-3,-1\right)\\ B=\left(3,3\right)\\ C=\left(-9,8\right)\end{array}$

Formula used: -

  $m_{1}x{m}_2=-1$

Where $m_1$ and $m_2$ are slope of perpendicular.

  $Slope=m=\frac{y_2-y_1}{x_2-x_1}$

“To form a right-angle triangle one of the sides of the triangle is perpendicular on the other side of triangle. Therefore, when two sides are perpendicular their product of slope is -1.”

Slope of AB:

  $\begin{array}{l}A=\left(-3,-1\right),B=\left(3,3\right)\\ \\ Slope\left(AB\right)=\frac{3-\left(-1\right)}{3-\left(-3\right)}=\frac{2}{3}\end{array}$

Slope of BC:

  $\begin{array}{l}B=\left(3,3\right),C=\left(-9,8\right)\\ \\ Slope\left(BC\right)=\frac{8-3}{-9-3}=-\frac{5}{12}\end{array}$

Slope of CA:

  $\begin{array}{l}C=\left(-9,8\right),A=\left(-3,-1\right)\\ \\ Slope\left(CA\right)=\frac{-1-8}{-3+9}=-\frac{3}{2}\end{array}$

Product of slope AB and CA:

  $\begin{array}{l}Slope(AB)xSlope(CA)\\ =\boxed{\left(\frac{2}{3}\right)x\left(\frac{-3}{2}\right)=-1}\end{array}$

Hence, the product of AB and CA is -1 therefore, triangle ABC is a right-angle triangle at angle A.