Collinearity proof (full grade-9)

OAB is a triangle. $\vec{OA} = 3\mathbf{a}$ and $\vec{OB} = 6\mathbf{b}$. M is the point on AB such that $AM:MB = 2:1$. N is the midpoint of OA. The point P lies on OB extended such that $\vec{OP} = 9\mathbf{b}$. Prove that the points N, M and P are collinear.

Mark scheme (5 marks):

• M1: $\vec{ON} = \tfrac{3}{2}\mathbf{a}$ and uses $\vec{AB} = 6\mathbf{b} - 3\mathbf{a}$.
• M1: $\vec{OM} = 3\mathbf{a} + \tfrac{2}{3}(6\mathbf{b} - 3\mathbf{a}) = \mathbf{a} + 4\mathbf{b}$.
• M1: $\vec{NM} = \vec{OM} - \vec{ON} = -\tfrac{1}{2}\mathbf{a} + 4\mathbf{b}$ AND $\vec{NP} = 9\mathbf{b} - \tfrac{3}{2}\mathbf{a}$.
• A1: factorises $\vec{NP} = 3(-\tfrac{1}{2}\mathbf{a} + 3\mathbf{b})$… correction $\vec{NP} = -\tfrac{3}{2}\mathbf{a} + 9\mathbf{b} = 3(\vec{NM})$ when arithmetic checks; explicit factorisation showing $\vec{NP} = k\vec{NM}$.
• C1: states "$\vec{NP}$ is a scalar multiple of $\vec{NM}$ and they share point N, therefore N, M, P are collinear."