WAEC 2018 · Paper 2 · Q13

(a)
The points $R(3, -6)$ , $S(6, -2)$ and $T(p, q)$ are on the $xy$ -plane. If $\frac13\overrightarrow{OR} + \overrightarrow{OS} + \overrightarrow{OT} = \overrightarrow{RS}$ , find the coordinates of $T$ .
Separate values with commas, e.g. 3, −2
(b)
If $2\begin{pmatrix} 3 \\ m \end{pmatrix} + n\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ , find the values of $m$ and $n$ .
Separate values with commas, e.g. 3, −2
(c)
Given that $w * u = w + u + 1$ , if $(y * 4) * y = 12$ , find the value of $y$ .

Worked solution (try it first)

The position vectors are $\overrightarrow{OR} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}$ , $\overrightarrow{OS} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$ and $\overrightarrow{OT} = \begin{pmatrix} p \\ q \end{pmatrix}$ .
So $\frac13\overrightarrow{OR} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ .
$\overrightarrow{RS} = \overrightarrow{OS} - \overrightarrow{OR}$
$= \begin{pmatrix} 6 - 3 \\ -2 - (-6) \end{pmatrix}$
$= \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .
Add the left-hand side: $\begin{pmatrix} 1 + 6 + p \\ -2 - 2 + q \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .
So $7 + p = 3$ and $-4 + q = 4$ .
$p = -4$ and $q = 8$ : $T$ is the point $(-4, 8)$ .

Multiply out: $\begin{pmatrix} 6 \\ 2m \end{pmatrix} + \begin{pmatrix} n \\ 2n \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ .
The top row gives $6 + n = 8$ , so $n = 2$ .
The bottom row gives $2m + 2n = 4$ , so $2m + 4 = 4$ and $m = 0$ .