WAEC 2014 · Paper 2 · Q13

(a)
If $\mathbf x = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ and $\mathbf y = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ , find, correct to 1 decimal place, $|\mathbf x + \mathbf y|$ .
(b)
$P(6, 4)$ , $Q(-2, -2)$ and $R(4, -6)$ are the vertices of triangle $PQR$ . (i) Determine the coordinates of $M$ and $S$ , the midpoints of $\overline{PQ}$ and $\overline{PR}$ respectively. (ii) Find $\overrightarrow{QR}$ and $\overrightarrow{MS}$ . (iii) State the relationship between $\overrightarrow{QR}$ and $\overrightarrow{MS}$ . (iv) Find the equation of $\overline{MS}$ .
Show the answer
(i) $M(2, 1)$ , $S(5, -1)$ ; (ii) $\overrightarrow{QR} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ , $\overrightarrow{MS} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ ; (iii) $\overrightarrow{QR} = 2\overrightarrow{MS}$ ; (iv) $2x + 3y = 7$

Worked solution (try it first)

Add the vectors component by component: $\mathbf x + \mathbf y = \begin{pmatrix} -2 + (-3) \\ 4 + 1 \end{pmatrix}$
$= \begin{pmatrix} -5 \\ 5 \end{pmatrix}$ .
Its length is $|\mathbf x + \mathbf y| = \sqrt{(-5)^2 + 5^2}$
$= \sqrt{50}$
$\approx 7.1$ .

Midpoints average the coordinates: $M = \left(\frac{6 + (-2)}{2}, \frac{4 + (-2)}{2}\right)$
$= (2, 1)$ and $S = \left(\frac{6 + 4}{2}, \frac{4 + (-6)}{2}\right)$
$= (5, -1)$ .

A vector from one point to another is "end minus start": $\overrightarrow{QR} = \begin{pmatrix} 4 - (-2) \\ -6 - (-2) \end{pmatrix}$
$= \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ and $\overrightarrow{MS} = \begin{pmatrix} 5 - 2 \\ -1 - 1 \end{pmatrix}$
$= \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ .

$\overrightarrow{QR} = 2\overrightarrow{MS}$ : $QR$ is parallel to $MS$ and twice as long.