College Algebra, Chapter 2, 2.1, Section 2.1, Problem 36

Draw the parallelogram with vertices $A(1,2), B(5,2), C(3,6)$ and $D(7,6)$ on a coordinate plane. Find the area of the parallelogram.




Recall that the area of the parallelogram is $A = L \times W$ and we know that $AC = BD$ and $AB = CD$
By using distance formula,

$
\begin{equation}
\begin{aligned}
d_{AC} = d_{BD} &= \sqrt{(3-1)^2 + (6-2)^2}\\
\\
&= \sqrt{2^2 + 4^2}\\
\\
&= \sqrt{4+16}\\
\\
&= \sqrt{20} \text{ units}\\
\\
d_{AB} = d_{CD} &= \sqrt{(5-1)^2 + (2-2)^2}\\
\\
&= \sqrt{4^2 + 0^2}\\
\\
&= \sqrt{4^2}\\
\\
&= 4 \text{ units}
\end{aligned}
\end{equation}
$


Therefore, the area of the parallelogram is $A = d_{AC} \cdot d_{AB} = (\sqrt{20})(4)$
$= 8 \sqrt{5}$ square units