1. **Problem statement:** Given points A(0,1), B(3,4), and C(1,3), we need to show that $\cos \angle ACB = -\frac{4}{5}$ and calculate the area of triangle $\triangle ABC$. 2. **To find $\cos \angle ACB$:** - Use the cosine formula for the angle at point C formed by vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$. - The formula is: $$\cos \theta = \frac{\overrightarrow{CA} \cdot \overrightarrow{CB}}{|\overrightarrow{CA}| |\overrightarrow{CB}|}$$ 3. **Calculate vectors:** - $\overrightarrow{CA} = A - C = (0-1, 1-3) = (-1, -2)$ - $\overrightarrow{CB} = B - C = (3-1, 4-3) = (2, 1)$ 4. **Dot product:** $$\overrightarrow{CA} \cdot \overrightarrow{CB} = (-1)(2) + (-2)(1) = -2 - 2 = -4$$ 5. **Magnitudes:** $$|\overrightarrow{CA}| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ $$|\overrightarrow{CB}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$$ 6. **Calculate cosine:** $$\cos \angle ACB = \frac{-4}{\sqrt{5} \times \sqrt{5}} = \frac{-4}{5}$$ 7. **Therefore, $\cos \angle ACB = -\frac{4}{5}$ as required.** 8. **Calculate area of $\triangle ABC$:** - Use the formula for area given coordinates: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ - Substitute $A(0,1), B(3,4), C(1,3)$: $$= \frac{1}{2} |0(4 - 3) + 3(3 - 1) + 1(1 - 4)|$$ $$= \frac{1}{2} |0 + 3 \times 2 + 1 \times (-3)| = \frac{1}{2} |0 + 6 - 3| = \frac{1}{2} \times 3 = \frac{3}{2}$$ 9. **Final answer:** - $\cos \angle ACB = -\frac{4}{5}$ - Area of $\triangle ABC = \frac{3}{2}$ square units.