NECO 2024 · Paper 2 · Q3

(a)(i)
The vertices of $\triangle PQR$ are $P(-3, 8)$ , $Q(4, 3)$ and $R(1, 2)$ . Find the equation of the line $PQ$ (give $y$ in terms of $x$ ).
(a)(ii)
Find the equation of the line $QR$ (give $y$ in terms of $x$ ).
(b)
The interior angles of a hexagon are $(x + 5)^\circ$ , $(2x + 4)^\circ$ , $(x + 10)^\circ$ , $(2x + 2)^\circ$ , $(3x + 2)^\circ$ and $(x + 15)^\circ$ . Find the value of $x$ .

Worked solution (try it first)

(a)(i)

Gradient of $PQ = \frac{3 - 8}{4 - (-3)}$
$= \frac{-5}{7}$
$= -\frac57$ .
Using the point $Q(4, 3)$ : $y - 3 = -\frac57(x - 4)$ .
Multiply by 7: $7y - 21 = -5x + 20$ , so $5x + 7y = 41$ , that is $y = \frac{41 - 5x}{7}$ .
(Check with $P$ : $5(-3) + 7(8) = 41$ ✓.)

(ii)

Gradient of $QR = \frac{2 - 3}{1 - 4}$
$= \frac{-1}{-3}$
$= \frac13$ .
Using $R(1, 2)$ : $y - 2 = \frac13(x - 1)$ , so $3y - 6 = x - 1$ and $x - 3y + 5 = 0$ , that is $y = \frac{x + 5}{3}$ .

(b)

The interior angles of a hexagon add up to $(6 - 2) \times 180^\circ = 720^\circ$ .
So $(x + 5) + (2x + 4) + (x + 10) + (2x + 2) + (3x + 2) + (x + 15) = 720$ .
Collect terms: $10x + 38 = 720$ , so $10x = 682$ and $x = 68.2$ .

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