1. **Stating the problem:** Given a triangle ABC with points E and F on sides AC and CB respectively, and points A, B, D collinear. Given ratios AE : AC = 2 : 3, CF : CB = 3 : 5, and vectors \(\overrightarrow{AC} = 4x\), \(\overrightarrow{AB} = 3y\). Find: (i) \(\overrightarrow{AF}\) (ii) \(\overrightarrow{EF}\) 2. **Expressing \(\overrightarrow{AE}\) and \(\overrightarrow{AF}\):** Since \(AE : AC = 2 : 3\), point E divides AC in ratio 2:3. \[ \overrightarrow{AE} = \frac{2}{3} \overrightarrow{AC} = \frac{2}{3} \times 4x = \frac{8}{3} x \] For F on CB with \(CF : CB = 3 : 5\), F divides CB in ratio 3:5 starting from C. Since \(\overrightarrow{CB} = \overrightarrow{AB} - \overrightarrow{AC} = 3y - 4x\), \[ \overrightarrow{CF} = \frac{3}{5} \overrightarrow{CB} = \frac{3}{5} (3y - 4x) = \frac{9}{5} y - \frac{12}{5} x \] Then, \[ \overrightarrow{AF} = \overrightarrow{AC} + \overrightarrow{CF} = 4x + \left( \frac{9}{5} y - \frac{12}{5} x \right) = \left(4 - \frac{12}{5}\right) x + \frac{9}{5} y = \frac{8}{5} x + \frac{9}{5} y \] 3. **Finding \(\overrightarrow{EF}\):** \[ \overrightarrow{EF} = \overrightarrow{AF} - \overrightarrow{AE} = \left( \frac{8}{5} x + \frac{9}{5} y \right) - \frac{8}{3} x = \left( \frac{8}{5} - \frac{8}{3} \right) x + \frac{9}{5} y = \left( \frac{24 - 40}{15} \right) x + \frac{9}{5} y = -\frac{16}{15} x + \frac{9}{5} y \] 4. **(b)(i) Find h if E, F, D are collinear and \(\overrightarrow{AB} = h \overrightarrow{AD}\):** Since A, B, D are collinear with D beyond B, \(\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = 3y + \overrightarrow{BD}\). Express \(\overrightarrow{BD} = t \overrightarrow{AB} = t (3y)\) for some scalar t, so \[ \overrightarrow{AD} = 3y + 3t y = 3(1 + t) y \] Given \(\overrightarrow{AB} = h \overrightarrow{AD} \Rightarrow 3y = h \times 3(1 + t) y \Rightarrow h = \frac{1}{1 + t} \] Now, E, F, D collinear means vectors \(\overrightarrow{EF}\) and \(\overrightarrow{ED}\) are parallel. Express \(\overrightarrow{ED} = \overrightarrow{AD} - \overrightarrow{AE} = 3(1 + t) y - \frac{8}{3} x\). For \(\overrightarrow{EF} = -\frac{16}{15} x + \frac{9}{5} y\) and \(\overrightarrow{ED} = -\frac{8}{3} x + 3(1 + t) y\) to be parallel, their components must be proportional: \[ \frac{-\frac{16}{15}}{-\frac{8}{3}} = \frac{\frac{9}{5}}{3(1 + t)} \Rightarrow \frac{16/15}{8/3} = \frac{9/5}{3(1 + t)} \] Calculate left side: \[ \frac{16/15}{8/3} = \frac{16}{15} \times \frac{3}{8} = \frac{16 \times 3}{15 \times 8} = \frac{48}{120} = \frac{2}{5} \] Calculate right side: \[ \frac{9/5}{3(1 + t)} = \frac{9}{5} \times \frac{1}{3(1 + t)} = \frac{3}{5(1 + t)} \] Set equal: \[ \frac{2}{5} = \frac{3}{5(1 + t)} \Rightarrow 2 = \frac{3}{1 + t} \Rightarrow 1 + t = \frac{3}{2} \Rightarrow t = \frac{1}{2} \] Then, \[ h = \frac{1}{1 + t} = \frac{1}{1 + \frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \] 5. **(b)(ii) Find \(\frac{\overrightarrow{EF}}{\overrightarrow{ED}}\):** Since \(\overrightarrow{EF} = k \overrightarrow{ED}\) for some scalar k, Using x-components: \[ -\frac{16}{15} = k \times -\frac{8}{3} \Rightarrow k = \frac{16/15}{8/3} = \frac{16}{15} \times \frac{3}{8} = \frac{2}{5} \] Check y-components: \[ \frac{9}{5} = k \times 3(1 + t) = k \times 3 \times \frac{3}{2} = \frac{9}{2} k \Rightarrow k = \frac{9/5}{9/2} = \frac{9}{5} \times \frac{2}{9} = \frac{2}{5} \] Consistent. **Final answers:** (i) \(\overrightarrow{AF} = \frac{8}{5} x + \frac{9}{5} y\) (ii) \(\overrightarrow{EF} = -\frac{16}{15} x + \frac{9}{5} y\) (b)(i) \(h = \frac{2}{3}\) (b)(ii) \(\frac{\overrightarrow{EF}}{\overrightarrow{ED}} = \frac{2}{5}\)