2019 H2 Maths Paper 2 Q5 — Worked Solution

Part (i)

4 marks

Approach

$X$ is the intersection of the two lines $AC$ and $BD$ . We write $\overrightarrow{OX}$ two ways — once as a point on $AC$ , once as a point on $BD$ — and compare coefficients of $\mathbf{a}$ and $\mathbf{b}$ , since $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel.

Since $X$ lies on $AC$ , for some $t \in \mathbb{R}$ ,

\overrightarrow{OX} = \overrightarrow{OA} + t\overrightarrow{AC}

= \mathbf{a} + t\left\lbrack \left( 2\mathbf{a} + 4\mathbf{b} \right) - \mathbf{a} \right\rbrack

= (1 + t)\mathbf{a} + 4t\mathbf{b}

Since $X$ also lies on $BD$ , for some $s \in \mathbb{R}$ ,

\overrightarrow{OX} = \overrightarrow{OB} + s\overrightarrow{BD}

= \mathbf{b} + s\left\lbrack \left( 5\mathbf{a} + \mathbf{b} \right) - \mathbf{b} \right\rbrack

= 5s\mathbf{a} + \mathbf{b}

Comparing coefficients of $\mathbf{a}$ and $\mathbf{b}$ ,

1 + t = 5s,\quad\quad 4t = 1

Solving: $t = \frac{1}{4}$ , $s = \frac{1}{4}$ .

\therefore\ \overrightarrow{OX} = \frac{5}{4}\mathbf{a} + \mathbf{b}

Part (ii)

6 marks

Approach

$O$ , $X$ , $Y$ collinear means $\overrightarrow{OY}$ is a scalar multiple of $\overrightarrow{OX}$ . Writing $Y$ on $CD$ gives a second expression for $\overrightarrow{OY}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ ; comparing coefficients pins down the scalar.

Since $Y$ lies on $CD$ , for some $\mu \in \mathbb{R}$ ,

\overrightarrow{OY} = \overrightarrow{OC} + \mu\overrightarrow{CD}

= \left( 2\mathbf{a} + 4\mathbf{b} \right) + \mu\left\lbrack \left( 5\mathbf{a} + \mathbf{b} \right) - \left( 2\mathbf{a} + 4\mathbf{b} \right) \right\rbrack

= (2 + 3\mu)\mathbf{a} + (4 - 3\mu)\mathbf{b}

Since $O$ , $X$ , $Y$ are collinear, $\overrightarrow{OY} = \lambda\overrightarrow{OX}$ for some $\lambda \in \mathbb{R}$ , i.e.

(2 + 3\mu)\mathbf{a} + (4 - 3\mu)\mathbf{b} = \frac{5\lambda}{4}\mathbf{a} + \lambda\mathbf{b}

Comparing coefficients of $\mathbf{a}$ and $\mathbf{b}$ ,

2 + 3\mu = \frac{5\lambda}{4},\quad\quad 4 - 3\mu = \lambda

Solving simultaneously, $\mu = \frac{4}{9}$ , $\lambda = \frac{8}{3}$ .

\overrightarrow{OY} = \frac{8}{3}\left( \frac{5}{4}\mathbf{a} + \mathbf{b} \right) = \frac{10}{3}\mathbf{a} + \frac{8}{3}\mathbf{b}

To check your answer: with

\mu = \frac{4}{9}

2 + 3\mu = \frac{10}{3}

and

4 - 3\mu = \frac{8}{3}

, matching the coefficients above. ✓

Since $\overrightarrow{OY} = \lambda\overrightarrow{OX}$ with $\lambda = \frac{8}{3}$ , $Y$ lies on ray $OX$ beyond $X$ , and

OX:OY = 1:\lambda = 1:\frac{8}{3}

\therefore\ OX:OY = 3:8

2019 H2 Maths Paper 2, Question 5

Part (i)

Part (ii)

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