Determine the

(i) coordinates of M, the midpoint of P and Q.

(ii) gradient of the line drawn through P and Q.

(iii) gradient of the line drawn through M, perpendicular to PQ.

The perpendicular line drawn through M meets the y-axis at R (0, k).

Find k.

(i) (2, – 2) parentheses not required.     (A1)

(ii) gradient of PQ $= \left( {\frac{{ - 5 - 1}}{{0 - 4}}} \right) = \frac{6}{4} = \frac{3}{2}(1.5)$     (M1)(A1)

(M1) for gradient formula with correct substitution

Award (A1) for $y = \frac{3}{2}x - 5$ with no other working

(iii) gradient of perpendicular is $- \frac{2}{3}$     (A1)(ft)     (C4)

[4 marks]

$\left( {\frac{{k + 2}}{{0 - 2}}} \right) =  - \frac{2}{3}$, $k =  - \frac{2}{3}$ or $y =  - \frac{2}{3}x + c$, $c =  - \frac{2}{3}\therefore k =  - \frac{2}{3}$     (M1)(A1)(ft)

Allow ($0, - \frac{2}{3}$)

(M1) is for equating gradients or substituting gradient into $y = mx + c$     (C2)

[2 marks]

There were some good answers, but many candidates showed a shaky understanding of coordinate geometry and some difficulty in dealing with negative numbers. Evidently a favourite question for some centres that consistently scored high marks here.

a) This was done quite well by most candidates with the main errors being reversal of the x, y values in the formula and using the negative, rather than the negative reciprocal for the perpendicular.

There were some good answers, but many candidates showed a shaky understanding of coordinate geometry and some difficulty in dealing with negative numbers. Evidently a favourite question for some centres that consistently scored high marks here.

b) Poorly answered though many candidates did gain a mark by substituting the correct value for gradient into $y = mx + c$.