1. **Problem:** Given points A (1, 3), B (-2, -1), and C (3, -1), find the coordinates of the vertices of triangle ABC and calculate its area. 2. **Vertices:** The vertices are given as A (1, 3), B (-2, -1), and C (3, -1). 3. **Formula for area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$:** $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 4. **Substitute the coordinates:** $$x_1=1, y_1=3; \quad x_2=-2, y_2=-1; \quad x_3=3, y_3=-1$$ 5. **Calculate each term:** $$1((-1) - (-1)) = 1(0) = 0$$ $$-2((-1) - 3) = -2(-4) = 8$$ $$3(3 - (-1)) = 3(4) = 12$$ 6. **Sum the terms:** $$0 + 8 + 12 = 20$$ 7. **Calculate the area:** $$\text{Area} = \frac{1}{2} |20| = 10$$ **Final answer:** The coordinates of the vertices are A (1, 3), B (-2, -1), and C (3, -1). The area of triangle ABC is $10$ square units.