Chapter F - Foundations: A Prelude to Functions - Section F.1 The Distance and Midpoint Formulas - F.1 Assess Your Understanding - Page 7: 31

See graph. $AB^2=AC^2+BC^2$, $26\ unit^2$.

Step 1. See graph. Step 2. $AB=\sqrt {(-5-6)^2+(3-0)^2}=\sqrt {130}$, $BC=\sqrt {(6-5)^2+(0-5)^2}=\sqrt {26}$, $AC=\sqrt {(-5-5)^2+(3-5)^2}=\sqrt {104}=2\sqrt {26}$, Step 3. Check $AB^2=AC^2+BC^2$, thus it is a right triangle, Step 4. $Area=\frac{1}{2}(\sqrt {26})(\sqrt {104})=26\ unit^2$.