WAEC 2018 · Paper 2 · Q12

(a)
If $\mathbf x = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ , $\mathbf y = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ and $\mathbf z = \begin{pmatrix} -4 \\ 13 \end{pmatrix}$ , find scalars $p$ and $q$ such that $p\mathbf x + q\mathbf y = \mathbf z$ .
Separate values with commas, e.g. 3, −2
(b)
Using a scale of 2 cm to 2 units on both axes, draw on a graph paper two perpendicular axes $Ox$ and $Oy$ for $-8 \le x \le 8$ and $-8 \le y \le 8$ respectively. Draw, on the same graph paper, indicating clearly the vertices and their coordinates: (i) the quadrilateral $WXYZ$ with $W(2, 3)$ , $X(4, -1)$ , $Y(-3, -4)$ and $Z(-3, 2)$ ; (ii) the image $W_1X_1Y_1Z_1$ of the quadrilateral $WXYZ$ under an anticlockwise rotation of $90^\circ$ about the origin.
Model answer
A $90^\circ$ anticlockwise rotation about the origin maps $(x, y)$ to $(-y, x)$ . The image (dashed) is $W_1(-3, 2)$ , $X_1(1, 4)$ , $Y_1(4, -3)$ , $Z_1(-2, -3)$ ; note that $W_1$ lands on the same point as $Z$ .

WXYZ (blue) and its image under a 90° anticlockwise rotation about O (red).

Worked solution (try it first)

Write $p\mathbf x + q\mathbf y = \mathbf z$ in columns: $p\begin{pmatrix} 2 \\ 3 \end{pmatrix} + q\begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}$ .
The tops give $2p + 5q = -4$ and the bottoms give $3p - 2q = 13$ .
Multiply the first by 2 and the second by 5: $4p + 10q = -8$ and $15p - 10q = 65$ .
Add them: $19p = 57$ , so $p = 3$ .
Then $2(3) + 5q = -4$ , so $5q = -10$ and $q = -2$ .

Plot $W(2, 3)$ , $X(4, -1)$ , $Y(-3, -4)$ and $Z(-3, 2)$ and join them in order.

A rotation of $90^\circ$ anticlockwise about the origin sends $(x, y)$ to $(-y, x)$ : $W_1(-3, 2)$ , $X_1(1, 4)$ , $Y_1(4, -3)$ and $Z_1(-2, -3)$ .
Plot and join them, labelling each vertex.