Precalculus, Chapter 1, 1.1, Section 1.1, Problem 34

Plot the points $A = (4,-3), B = (4,1)$ and $C = (2,1)$ and form the triangle $ABC$. Verify that the triangle is a right triangle. Determine its area.








$
\begin{equation}
\begin{aligned}

AB =& \sqrt{(4-4)^2 + [1-(-3)]^2}
\\
=& \sqrt{0+16}
\\
=& \sqrt{16}
\\
=& 4
\\
BC =& \sqrt{(2-4)^2 + (1-1)^2}
\\
=& \sqrt{4+0}
\\
=& \sqrt{4}
\\
=& 2
\\
AC =& \sqrt{(2-4)^2 + [1-(-3)]^2}
\\
=& \sqrt{4+16}
\\
=& \sqrt{20}
\\
=& 2 \sqrt{5}
\\
(AC)^2 =& (AB)^2 + (BC)^2
\\
(2 \sqrt{5})^2 =& (4)^2 + (2)^2
\\
(4 \cdot 5) =& 16 + 4
\\
20 =& 20


\end{aligned}
\end{equation}
$


Thus, $\Delta ABC$ is a right triangle.

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


$
\begin{equation}
\begin{aligned}

\Delta ABC =& \frac{1}{2} (4)(2)
\\
\\
=& \frac{8}{2}
\\
\\
=& 4 \text{ square units}

\end{aligned}
\end{equation}
$