WAEC 2008 · Paper 2 · Q4

(a)
A straight line passes through the point $P(-1, 3)$ . Another line which passes through $Q(-4, 4)$ intersects the first line at the point $R(k, 5)$ , where $k$ is a constant. If $\angle PRQ = 90^\circ$ , find the values of $k$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Gradient of $PR$ : $\dfrac{5 - 3}{k - (-1)} = \dfrac{2}{k + 1}$ .
Gradient of $QR$ : $\dfrac{5 - 4}{k - (-4)} = \dfrac{1}{k + 4}$ .
$\angle PRQ = 90^\circ$ , so the lines are perpendicular and the product of the gradients is $-1$ : $\dfrac{2}{(k + 1)(k + 4)} = -1$ .
Multiply out: $2 = -(k^2 + 5k + 4)$ , so $k^2 + 5k + 6 = 0$ .
Factorise: $(k + 2)(k + 3) = 0$ .
So $k = -2$ or $k = -3$ .