WAEC 2013 · Paper 2 · Q8

(a)
Given the Cartesian coordinates $A(1, 2)$ , $B(0, 4)$ , $C(-2, -2)$ and $D(-3, 0)$ , on a graph sheet and using a scale of 2 cm to represent 1 unit on both axes, plot the points $A$ , $B$ , $C$ and $D$ .
Model answer
Draw both axes with 2 cm to 1 unit ( $x$ from $-5$ to $2$ and $y$ from $-3$ to $5$ is enough) and mark the four points. The drawing under (c) shows them.
(b)
(i) Join the points to form a quadrilateral. (ii) What type of quadrilateral is formed?
(c)
Using a ruler and a pair of compasses only, construct: (i) the locus $l_1$ of points equidistant from $A$ and $C$ ; (ii) the locus $l_2$ of points equidistant from $\overline{AC}$ and $\overline{BA}$ ; (iii) locate $M$ , the point of intersection of $l_1$ and $l_2$ .
Model answer
$l_1$ is the perpendicular bisector of $AC$ ; $l_2$ is the bisector of $\angle CAB$ . They meet at $M$ , about $(-3.7, 2.4)$ .
(d)
Measure: (i) $\angle MAB$ ; (ii) $|MA|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw the axes with 2 cm to 1 unit on both, and plot $A(1, 2)$ , $B(0, 4)$ , $C(-2, -2)$ and $D(-3, 0)$ .

(b)(i)

(ii)

Compare opposite sides.
From $A$ to $B$ is 1 left and 2 up, and from $C$ to $D$ is also 1 left and 2 up.
From $B$ to $D$ is 3 left and 4 down, and from $A$ to $C$ is also 3 left and 4 down.
Both pairs of opposite sides are equal and parallel, and the corners are not right angles, so $ABDC$ is a parallelogram.

(c)(i)

Points equidistant from $A$ and $C$ lie on the perpendicular bisector of $AC$ .
With centres $A$ and $C$ and the same radius (more than half of $AC$ ), draw arcs on both sides of $AC$ and join the two crossings.
This is $l_1$ .
It passes through the midpoint $(-0.5, 0)$ .

(ii)

Points equidistant from the lines $AC$ and $AB$ lie on the bisector of $\angle CAB$ .
With centre $A$ , draw an arc cutting $AC$ and $AB$ .
From those two points draw equal arcs that cross, and join $A$ to the crossing.
This is $l_2$ .

(iii)

(d)(i)

$\angle CAB \approx 116.6^\circ$ and $l_2$ halves it, so $\angle MAB \approx 58^\circ$ .

(ii)

Measure $MA$ with the ruler: $|MA| \approx 9.5\text{ cm}$ (about 4.76 units at 2 cm to 1 unit).