WAEC 2017 · Paper 2 · Q13

(a)
The operation $(*)$ is defined on the set of real numbers, $\mathbb R$ , by $x * y = \dfrac{x + y}{2}$ , $x, y \in \mathbb R$ . (i) Evaluate $3 * \frac25$ . (ii) If $8 * y = 8\frac14$ , find the value of $y$ .
Separate values with commas, e.g. 3, −2
(b)
In $\triangle ABC$ , $\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}$ . If $P$ is the midpoint of $\overline{AB}$ , express $\overrightarrow{CP}$ as a column vector.
Show the answer
$\begin{pmatrix} -5 \\ 11 \end{pmatrix}$

Worked solution (try it first)

Put $x = 3$ and $y = \frac25 = 0.4$ into the rule: $3 * \frac25 = \frac{3 + 0.4}{2}$
$= \frac{3.4}{2}$
$= 1.7$ .

$P$ is the midpoint of $AB$ , so $\overrightarrow{AP} = \frac12\overrightarrow{AB}$
$= \frac12\begin{pmatrix} -4 \\ 6 \end{pmatrix}$
$= \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ .
Go from $C$ to $A$ , then from $A$ to $P$ : $\overrightarrow{CP} = \overrightarrow{CA} + \overrightarrow{AP}$ , where $\overrightarrow{CA} = -\overrightarrow{AC}$
$= \begin{pmatrix} -3 \\ 8 \end{pmatrix}$ .
So $\overrightarrow{CP} = \begin{pmatrix} -3 + (-2) \\ 8 + 3 \end{pmatrix}$
$= \begin{pmatrix} -5 \\ 11 \end{pmatrix}$ .