College Algebra, Chapter 2, 2.1, Section 2.1, Problem 44

Find the area of the triangle shown in the figure.




$A(-2,1)$
$B(4,1)$
$C(7,4)$

By using distance formula,

$
\begin{equation}
\begin{aligned}
d_{AB} &= \sqrt{(1-1)^2 + (4-(-2))^2}\\
\\
d_{AB} &= \sqrt{0^2 + (6)^2}\\
\\
d_{AB} &= \sqrt{0+36}\\
\\
d_{AB} &= 6 \text{ units}
\end{aligned}
\end{equation}
$

Then,

$
\begin{equation}
\begin{aligned}
d_{AC} &= \sqrt{(4-1)^2 + (7-(-2))^2}\\
\\
d_{AC} &= \sqrt{3^2 + (9)^2}\\
\\
d_{AC} &= \sqrt{9+81}\\
\\
d_{AC} &= \sqrt{90} \text{ units}
\end{aligned}
\end{equation}
$


Lastly,

$
\begin{equation}
\begin{aligned}
d_{BC} &= \sqrt{(4-1)^2 + (7-4)^2}\\
\\
d_{BC} &= \sqrt{3^2 + 3^2}\\
\\
d_{BC} &= \sqrt{9+9}\\
\\
d_{BC} &= \sqrt{18} \text{ units}
\end{aligned}
\end{equation}
$


Now, we can get the area of those triangle by using the Heron's Formula given the length of three sides. Recall that,
$A \sqrt{s(s-a)(s-b)(s-c)}$ where $\displaystyle s = \frac{a+b+c}{2}$
$\displaystyle s = \frac{d_{AB}+d_{AC}+d_{BC}}{2} = \frac{6+\sqrt{90}+\sqrt{18}}{2} = \frac{6+3\sqrt{10}+3\sqrt{2}}{2}$
$\displaystyle =\frac{3}{2} (2 + \sqrt{10}+ \sqrt{2})$ units
Now substitute $s, d_{AB}$ as $a$, $d_{AC}$ as $b$, and $d_{BC}$ as $c$ to the equation of $A$ an we get.
$ A = 9 $ square units