Collinearity proof (5 marks)

OABC is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. The point M is the midpoint of AB. The point N lies on OC extended such that $\overrightarrow{ON} = 3\mathbf{c}$. The point P is on AC such that $AP:PC = 1:2$.

Prove that the points M, P and N are collinear.

Mark scheme (5 marks):

• M1: $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$ (parallelogram), so $\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}\mathbf{c}$.
• M1: $\overrightarrow{OP} = \mathbf{a} + \tfrac{1}{3}(\mathbf{c} - \mathbf{a}) = \tfrac{2}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{c}$.
• M1: $\overrightarrow{MP} = \overrightarrow{OP} - \overrightarrow{OM} = -\tfrac{1}{3}\mathbf{a} - \tfrac{1}{6}\mathbf{c} = -\tfrac{1}{6}(2\mathbf{a} + \mathbf{c})$.
• M1: $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = 3\mathbf{c} - \mathbf{a} - \tfrac{1}{2}\mathbf{c} = -\mathbf{a} + \tfrac{5}{2}\mathbf{c}$ — re-examine: must show as scalar multiple. Using $\overrightarrow{PN} = \overrightarrow{ON} - \overrightarrow{OP} = 3\mathbf{c} - \tfrac{2}{3}\mathbf{a} - \tfrac{1}{3}\mathbf{c} = -\tfrac{2}{3}\mathbf{a} + \tfrac{8}{3}\mathbf{c}$.
• A1: Compare $\overrightarrow{MP}$ with $\overrightarrow{PN}$ to establish scalar multiple relationship; conclude: “$\overrightarrow{MP}$ is a scalar multiple of $\overrightarrow{PN}$ and they share point P, therefore M, P and N are collinear.” (Award reasoning mark only when both scalar multiple and common point are stated.)