Calculate the values of s and t.

Find the equation of the straight line perpendicular to AB, passing through the point M.

$s = 6$     (A1)

$t = - 2$     (A1)     (C2)

[2 marks]

${\text{gradient of AB}} = \frac{{ - 2 - 8}}{{ - 2 - 6}} = \frac{{ - 10}}{{ - 8}} = \frac{5}{4}$     (A1)(ft)

(A1) for gradient of AM or BM $= \frac{5}{4}$

${\text{Perpendicular gradient}} = - \frac{4}{5}$     (A1)(ft)

Equation of perpendicular bisector is

$y = - \frac{4}{5}x + c$

$3 = - \frac{4}{5}(2) + c$     (M1)

$c = 4.6$

$y = -0.8x + 4.6$

or $5y = -4x + 23$     (A1)(ft)     (C4)

[4 marks]

(a) It was surprising how many errors were made in finding the values for s and t

(b) The candidates had difficulty in finding the equation of a straight line. They could find the gradient of the line AB and a number could give the gradient of the perpendicular line but most did not substitute the correct midpoint to find the equation of the line.