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}
$