1. **Problem statement:** Given triangle ABC with vertices A(-3,7), B(-8,2), and C(4,6), and point D(-2,k) as the midpoint of BC, we need to: 3.1 Determine the gradient of BC. 3.2 Calculate the length of BC. 3.3 Find the value of k. 3.4 Check if AD is perpendicular to BC. --- 2. **Formula and rules:** - Gradient (slope) between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ - Distance between two points is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - Midpoint coordinates between two points are: $$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ - Two lines are perpendicular if the product of their gradients is $-1$. --- 3. **Step-by-step solution:** **3.1 Gradient of BC:** - Points B(-8,2) and C(4,6) - Calculate gradient: $$m_{BC} = \frac{6 - 2}{4 - (-8)} = \frac{4}{12} = \frac{1}{3}$$ **3.2 Length of BC:** - Use distance formula: $$BC = \sqrt{(4 - (-8))^2 + (6 - 2)^2} = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}$$ **3.3 Value of k (midpoint y-coordinate):** - Midpoint D(-2,k) of BC means: $$-2 = \frac{-8 + 4}{2} = \frac{-4}{2} = -2 \quad \text{(x-coordinate matches)}$$ $$k = \frac{2 + 6}{2} = \frac{8}{2} = 4$$ **3.4 Is AD perpendicular to BC?** - Find gradient of AD: - Points A(-3,7) and D(-2,4) $$m_{AD} = \frac{4 - 7}{-2 - (-3)} = \frac{-3}{1} = -3$$ - Check product of gradients: $$m_{AD} \times m_{BC} = (-3) \times \frac{1}{3} = -1$$ - Since product is $-1$, AD is perpendicular to BC. --- **Final answers:** - Gradient of BC: $\frac{1}{3}$ - Length of BC: $4\sqrt{10}$ - Value of $k$: 4 - AD is perpendicular to BC because $m_{AD} \times m_{BC} = -1$