Chapter F - Foundations: A Prelude to Functions - Section F.1 The Distance and Midpoint Formulas - F.1 Assess Your Understanding - Page 7: 29

See graph. $AC^2=AB^2+BC^2$, $\frac{13}{2}\ unit^2$.

Step 1. See graph. Step 2. $AB=\sqrt {(-2-1)^2+(5-3)^2}=\sqrt {13}$, $BC=\sqrt {(-1-1)^2+(0-3)^2}=\sqrt {13}$, $AC=\sqrt {(-2+1)^2+(5-0)^2}=\sqrt {26}$, Step 3. Check $AC^2=AB^2+BC^2$, thus it is a right triangle, Step 4. $Area=\frac{1}{2}(\sqrt {13})(\sqrt {13})=\frac{13}{2}\ unit^2$.