Answer
Refer to the image below to see the triangle.
$\triangle{ABC}$ is right since the square of the longest side is equal to the sum of the squares of the two other sides.
$\text{Area}= \frac{13}{2}$ square units
Work Step by Step
Step 1.
Plot the points $A(-2,5)$, $B(1,3)$, and $C(-1,0)$, then connect them to form $\triangle{ABC}$. Refer to the image below.
Step 2.
Find the length of each side using the distance formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ to obtain:
$AB=\sqrt {(-2-1)^2+(5-3)^2}=\sqrt{9+4}=\sqrt {13}$
$BC=\sqrt {(-1-1)^2+(0-3)^2}=\sqrt{4+9}=\sqrt {13}$
$AC=\sqrt {(-2+1)^2+(5-0)^2}=\sqrt{1+25}=\sqrt {26}$
Step 3
A triangle is right if the square of the longest side is equal to the sum of the squared of the two other sides.
Note that:
$\begin{align*}
(\sqrt{26})^2&=(\sqrt{13})^2+(\sqrt{13})^2\\
26&=13+13\\
26&=26
\end{align*}$
Thus, $\triangle{ABC}$ is a right triangle.
Step 4.
The area i$A$ of a triangle is given by the formula $A=\frac{1}{2}bh$ where $b$ is the length of the base and $h$ is the height.
Note that $\triangle{ABC}$ has a height of $\sqrt{13}$ and a base of $\sqrt{13}$.
Thus,
$A=\frac{1}{2}(\sqrt{13})(\sqrt{13})\\
A=\frac{13}{2}$
Therefore, the area of $\triangle{ABC}$ is $\frac{13}{2}$ square units.
