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}
$