1. **State the problem:** Find the area of triangle ABC with vertices A(9,9), B(4,-4), and C(-4,1). 2. **Formula used:** The area of a triangle given coordinates $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 3. **Apply the formula:** $$\text{Area} = \frac{1}{2} |9(-4 - 1) + 4(1 - 9) + (-4)(9 + 4)|$$ 4. **Calculate inside the absolute value:** $$= \frac{1}{2} |9(-5) + 4(-8) + (-4)(13)|$$ $$= \frac{1}{2} |-45 - 32 - 52|$$ 5. **Sum the terms:** $$= \frac{1}{2} |-129| = \frac{1}{2} \times 129 = 64.5$$ 6. **Final answer:** The area of triangle ABC is **64.5 square units**.