Precalculus, Chapter 1, 1.1, Section 1.1, Problem 32

Plot the points $A = (-6,3), B = (3,-5)$ and $C = (-1,5)$ and form the triangle $ABC$. Verify that the triangle is a right triangle. Determine its area.








$\begin{equation} \begin{aligned} AB =& \sqrt{[3-(-6)]^2 + (-5-3)^2} \\ =& \sqrt{81 + 64} \\ =& \sqrt{145} \\ BC =& \sqrt{(-1-3)^2 + [5-(-5)]^2} \\ =& \sqrt{16 + 100} \\ =& \sqrt{116} \\ AC =& \sqrt{[-1- (-6)]^2 + (5-3)^2} \\ =& \sqrt{25+4} \\ =& \sqrt{29} \\ (AB)^2 =& (AC)^2 + (BC)^2 \\ (\sqrt{145})^2 =& (\sqrt{29})^2 + (\sqrt{116})^2 \\ 145 =& 29 + 116 \\ 145 =& 145 \end{aligned} \end{equation}$


Thus, $\Delta ABC$ is a right triangle.
The area of $\displaystyle \Delta ABC = \frac{1}{2} AC \cdot BC$,


$\begin{equation} \begin{aligned} \Delta ABC =& \frac{1}{2} (\sqrt{29}) (\sqrt{116}) \\ \\ =& \frac{58}{2} \\ \\ =& 29 \text{ square units} \end{aligned} \end{equation}$