College Algebra, Chapter 2, 2.1, Section 2.1, Problem 46

Show that the triangle with vertices $A(6,-7), B(11,-3)$, and $C(2,-2)$ is a right triangle by using the converse of the Pythagorean Theorem. Find the area of the triangle.

By using distance formula,

$
\begin{equation}
\begin{aligned}
d_{AB} &= \sqrt{(-3-(-7))^2 + (11-6)^2} &&& d_{AC} &= \sqrt{(-2-(-7))^2 + (2-6)^2}\\
\\
d_{AB} &= \sqrt{4^2 + 5^2} &&& d_{AC} &= \sqrt{5^2 + (-4)^2}\\
\\
d_{AB} &= \sqrt{16+25} &&& d_{AC} &= \sqrt{25+16}\\
\\
d_{AB} &= \sqrt{41} \text{ units} &&& d_{AC} &= \sqrt{41} \text{ units}\\
\\
\\
d_{BC} &= \sqrt{(-2-(-3))^2 + (2-11)^2}\\
\\
d_{BC} &= \sqrt{1^2 + (-9)^2}\\
\\
d_{BC} &= \sqrt{1+81}\\
\\
d_{BC} &= \sqrt{82} \text{ units}
\end{aligned}
\end{equation}
$

Next, by using the converse of the Pythagorean Theorem if the sides $\sqrt{82}$, $\sqrt{41}$ and $\sqrt{41}$ forms a right triangle.

$
\begin{equation}
\begin{aligned}
(\sqrt{82})^2 &= (\sqrt{41})^2 + (\sqrt{41})^2\\
\\
82 &= 41+41\\
\\
82 &= 82
\end{aligned}
\end{equation}
$

Since both sides are equal, then the three sides form a right triangle.
Thus, the area is $\displaystyle A = \frac{1}{2} bh = \frac{1}{2} (\sqrt{41}) (\sqrt{41}) = 20.5$ square units