NECO 2023 · Paper 2 · Q12

The vectors $\overrightarrow{OX}$ , $\overrightarrow{OY}$ and $\overrightarrow{OZ}$ are $\mathbf{p} = \begin{pmatrix} 10 \\ 1 \end{pmatrix}$ , $\mathbf{q} = \begin{pmatrix} -2 \\ 7 \end{pmatrix}$ and $\mathbf{r} = \mathbf{p} + 3\mathbf{q}$ , where $O$ is the origin. $OZ$ and $XY$ meet at $K$ , where $\overrightarrow{OK} = \alpha\overrightarrow{OZ}$ and $\overrightarrow{XK} = \beta\overrightarrow{XY}$ . Find the:

(i)
equations of the lines $XY$ and $YZ$ ;
Show the answer
$XY$ : $x + 2y = 12$ ; $YZ$ : $2y = 5x + 24$
(ii)
values of $\alpha$ and $\beta$ ;
Separate values with commas, e.g. 3, −2
(iii)
coordinates of $K$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$X(10, 1)$ and $Y(-2, 7)$ .
$\mathbf r = \mathbf p + 3\mathbf q$
$= (10 - 6, 1 + 21)$ , so $Z(4, 22)$ .

(i)

Gradient of $XY$ : $\dfrac{7 - 1}{-2 - 10} = -\dfrac12$ .
So $y - 1 = -\frac12(x - 10)$ , which gives $x + 2y = 12$ .
Gradient of $YZ$ : $\dfrac{22 - 7}{4 + 2} = \dfrac52$ .
So $y - 7 = \frac52(x + 2)$ , which gives $2y = 5x + 24$ .

(ii)

$\overrightarrow{OK} = \alpha(4\mathbf i + 22\mathbf j)$ , so $K = (4\alpha, 22\alpha)$ .
$K$ is on $XY$ : $4\alpha + 2(22\alpha) = 12$ , so $48\alpha = 12$ and $\alpha = \frac14$ .
$\overrightarrow{XK} = (1 - 10, 5.5 - 1)$
$= (-9, 4.5)$ and $\overrightarrow{XY} = (-12, 6)$ .
$(-9, 4.5) = \frac34(-12, 6)$ , so $\beta = \frac34$ .

(iii)

$K = \frac14(4, 22) = (1, 5\frac12)$ .

Report a problem with this question