1. **Problem Statement:** Given points A(9, 8), B(12, 4), and C(4, -2), solve the following: (a)(i) Find the gradient of the line through A and B. (a)(ii) Find the equation of the line through C parallel to AB. (b)(i) Calculate the length of line segment AB. (b)(ii) Calculate the length of line segment BC. (c) Show that AB is perpendicular to BC. (d) Calculate the area of triangle ABC. --- 2. **Formulas and Rules:** - Gradient between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ - Equation of a line with gradient $m$ passing through point $(x_0,y_0)$: $$y - y_0 = m(x - x_0)$$ - Distance between two points: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - Two lines are perpendicular if the product of their gradients is $-1$. - Area of triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ --- 3. **Step-by-step Solution:** **(a)(i) Gradient of AB:** $$m_{AB} = \frac{4 - 8}{12 - 9} = \frac{-4}{3} = -\frac{4}{3}$$ **(a)(ii) Equation of line through C parallel to AB:** - Parallel lines have the same gradient, so $m = -\frac{4}{3}$. - Using point C(4, -2): $$y - (-2) = -\frac{4}{3}(x - 4)$$ $$y + 2 = -\frac{4}{3}x + \frac{16}{3}$$ $$y = -\frac{4}{3}x + \frac{16}{3} - 2$$ $$y = -\frac{4}{3}x + \frac{16}{3} - \frac{6}{3} = -\frac{4}{3}x + \frac{10}{3}$$ **(b)(i) Length of AB:** $$d_{AB} = \sqrt{(12 - 9)^2 + (4 - 8)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ **(b)(ii) Length of BC:** $$d_{BC} = \sqrt{(4 - 12)^2 + (-2 - 4)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$ **(c) Show AB is perpendicular to BC:** - Gradient of BC: $$m_{BC} = \frac{-2 - 4}{4 - 12} = \frac{-6}{-8} = \frac{3}{4}$$ - Product of gradients: $$m_{AB} \times m_{BC} = -\frac{4}{3} \times \frac{3}{4} = -1$$ - Since product is $-1$, AB is perpendicular to BC. **(d) Area of triangle ABC:** $$\text{Area} = \frac{1}{2} |9(4 - (-2)) + 12(-2 - 8) + 4(8 - 4)|$$ $$= \frac{1}{2} |9 \times 6 + 12 \times (-10) + 4 \times 4|$$ $$= \frac{1}{2} |54 - 120 + 16| = \frac{1}{2} |-50| = 25$$ --- **Final answers:** - Gradient of AB: $-\frac{4}{3}$ - Equation of line through C parallel to AB: $y = -\frac{4}{3}x + \frac{10}{3}$ - Length of AB: $5$ - Length of BC: $10$ - AB is perpendicular to BC: Yes - Area of triangle ABC: $25$