Higher November 2018 Paper 1 Q21
21
\(OAB\) is a triangle.
\(OPM\) and \(APN\) are straight lines.
\(M\) is the midpoint of \(AB\).
\(\overrightarrow{OA} = \mathbf{a} \qquad \overrightarrow{OB} = \mathbf{b}\)
\(OP : PM = 3 : 2\)
Work out the ratio \(ON : NB\) (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3 : 4 | P1 | starts process eg \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) oe |
| P1 | for process to find \(\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}\text{``}(\mathbf{b} - \mathbf{a})\text{''}\) oe \(\left(= \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\right)\) | |
| P1 | for process to find \(\overrightarrow{AP} = -\mathbf{a} + \dfrac{3}{5}\text{``}\left(\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\right)\text{''}\) oe or (indep) for \(\overrightarrow{AN} = -\mathbf{a} + \text{``}k\text{''}\mathbf{b}\) | |
| P1 | process to find “\(k\)” using \(\overrightarrow{AN} = -\mathbf{a} + \text{``}k\text{''}\mathbf{b}\) as a multiple of \(\overrightarrow{AP}\) | |
| A1 | cao |
Alternative
| Answer | Mark | Mark scheme |
|---|---|---|
| 3 : 4 | P1 | for producing \(OM\) to \(C\) such that \(AC\) is parallel to \(OB\) |
| P1 | for process to show that \(MC = OM\), using congruent triangles \(ACM\) and \(BOM\) | |
| P1 | for process to find \(PC\) as a multiple of \(OM/5\ (= 7OM/5)\) | |
| P1 | for process to find \(ON\) as a multiple of \(AC(OB)\ (= 3OB/7)\) using similar triangles \(ACP\) and \(NOP\) | |
| A1 | cao |
Formal geometric reasoning relating to congruent and similar triangles is not required