Chapter F - Foundations: A Prelude to Functions - Section F.1 The Distance and Midpoint Formulas - F.1 Assess Your Understanding - Page 7: 32

See graph. $AB^2=AC^2+BC^2$, $29\ unit^2$.

Step 1. See graph. Step 2. $AB=\sqrt {(-6-3)^2+(3+5)^2}=\sqrt {145}$, $BC=\sqrt {(3+1)^2+(-5-5)^2}=\sqrt {116}$, $AC=\sqrt {(-6+1)^2+(3-5)^2}=\sqrt {29}$, Step 3. Check $AB^2=AC^2+BC^2$, thus it is a right triangle, Step 4. $Area=\frac{1}{2}(\sqrt {116})(\sqrt {29})=29\ unit^2$.