1. **State the problem:** We are given two vertices of a triangle, $A(2,1)$ and $B(3,-2)$, and the area of the triangle is 5. The third vertex $C$ lies on the line $y = x + 3$. We need to find the coordinates of $C$. 2. **Formula for the area of a triangle given coordinates:** The area of a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$ is given by: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 3. **Apply the formula:** Let $C = (x, y)$ and since $C$ lies on $y = x + 3$, we have $y = x + 3$. Substitute $A(2,1)$, $B(3,-2)$, and $C(x, x+3)$ into the area formula: $$5 = \frac{1}{2} |2(-2 - (x+3)) + 3((x+3) - 1) + x(1 - (-2))|$$ 4. **Simplify inside the absolute value:** $$5 = \frac{1}{2} |2(-2 - x - 3) + 3(x + 3 - 1) + x(1 + 2)|$$ $$5 = \frac{1}{2} |2(-5 - x) + 3(x + 2) + x(3)|$$ $$5 = \frac{1}{2} |-10 - 2x + 3x + 6 + 3x|$$ $$5 = \frac{1}{2} |-4 + 4x|$$ 5. **Multiply both sides by 2:** $$10 = |4x - 4|$$ 6. **Solve the absolute value equation:** $$4x - 4 = 10 \quad \text{or} \quad 4x - 4 = -10$$ For $4x - 4 = 10$: $$4x = 14$$ $$x = \frac{14}{4} = 3.5$$ For $4x - 4 = -10$: $$4x = -6$$ $$x = \frac{-6}{4} = -1.5$$ 7. **Find corresponding $y$ values:** Since $y = x + 3$, - For $x = 3.5$, $y = 3.5 + 3 = 6.5$ - For $x = -1.5$, $y = -1.5 + 3 = 1.5$ 8. **Final answer:** The possible coordinates for $C$ are: $$C_1 = (3.5, 6.5) \quad \text{or} \quad C_2 = (-1.5, 1.5)$$