1. **State the problem:** We have a parallelogram PARM with points P(2,9), R(n,-13), T(10,-3), and M the midpoint of PT. We need to find: 3.1.1 The length of PT in surd form. 3.1.2 The gradient of PT. 3.1.3 The gradient of AR. 3.1.4 The coordinates of M. 3.2 The equation of PM in the form $y=mx+c$. 2. **Given:** - $P=(2,9)$ - $T=(10,-3)$ - $M$ is midpoint of $PT$ - $PT \perp RT$ - $R=(n,-13)$ 3. **Find length of PT:** Length formula: $$\text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Calculate: $$PT = \sqrt{(10 - 2)^2 + (-3 - 9)^2} = \sqrt{8^2 + (-12)^2} = \sqrt{64 + 144} = \sqrt{208}$$ Simplify $\sqrt{208}$: $$208 = 16 \times 13 \Rightarrow \sqrt{208} = \sqrt{16 \times 13} = 4\sqrt{13}$$ So, length of $PT = 4\sqrt{13}$. 4. **Find gradient of PT:** Gradient formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Calculate: $$m_{PT} = \frac{-3 - 9}{10 - 2} = \frac{-12}{8} = -\frac{3}{2}$$ 5. **Find gradient of AR:** Since $PARM$ is a parallelogram, $AR$ is parallel to $PM$. First, find $M$ (midpoint of $PT$): $$M_x = \frac{2 + 10}{2} = 6, \quad M_y = \frac{9 + (-3)}{2} = 3$$ So, $M = (6,3)$. Gradient of $PM$: $$m_{PM} = \frac{3 - 9}{6 - 2} = \frac{-6}{4} = -\frac{3}{2}$$ Since $AR \parallel PM$, $$m_{AR} = -\frac{3}{2}$$ 6. **Find coordinates of M:** Already found: $$M = (6,3)$$ 7. **Find equation of PM:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ Using point $P(2,9)$ and slope $m = -\frac{3}{2}$: $$y - 9 = -\frac{3}{2}(x - 2)$$ Simplify: $$y - 9 = -\frac{3}{2}x + 3$$ $$y = -\frac{3}{2}x + 3 + 9$$ $$y = -\frac{3}{2}x + 12$$ **Final answers:** - Length of $PT = 4\sqrt{13}$ - Gradient of $PT = -\frac{3}{2}$ - Gradient of $AR = -\frac{3}{2}$ - Coordinates of $M = (6,3)$ - Equation of $PM: y = -\frac{3}{2}x + 12$