Precalculus, Chapter 1, 1.1, Section 1.1, Problem 30

Plot the points $A = (-2,5), B = (12,3)$ and $C = (10,-11)$ and form the triangle $ABC$. Verify that the triangle is a right triangle. Determine its area.








$\begin{equation} \begin{aligned} AB =& \sqrt{[12-(-2)]^2 + (3-5)^2} \\ =& \sqrt{196 + 4} \\ =& \sqrt{200} \\ BC =& \sqrt{(10-12)^2 + (-11-3)^2} \\ =& \sqrt{4+196} \\ =& \sqrt{200} \\ AC =& \sqrt{[10 - (-2)]^2 + (-11-5)^2} \\ =& \sqrt{144+256} \\ =& \sqrt{400} \\ =& 200 \end{aligned} \end{equation}$


$(AC)^2 = (AB)^2 + (BC)^2$, thus $\Delta ABC$ is a right triangle.

The area of $\displaystyle \Delta ABC = \frac{1}{2} AB \cdot BC$. So


$\begin{equation} \begin{aligned} \Delta ABC =& \frac{1}{2} (\sqrt{200}) (\sqrt{200}) \\ \\ =& \frac{200}{2} \\ \\ =& 100 \text{ square units} \end{aligned} \end{equation}$