Write down the coordinates of M.

Calculate the gradient of the line AB.

Find the equation of the line parallel to AB that passes through M.

\(\left( {\frac{{2 + 4}}{2},\frac{{8 + 2}}{2}} \right)\)     (M1)

Note: Award (M1) for a correct substitution into the midpoint formula.

\( = (3, 5)\)     (A1)     (C2)

Note: Brackets must be present for final (A1) to be awarded.

Note: Accept \(x = 3\), \(y = 5\) .

[2 marks]

\(\frac{{8 - 4}}{{2 - 14}}\)     (M1)

Note: Award (M1) for correctly substituted formula.

\( =  - \frac{1}{3}\) \(\left( {\frac{{ - 4}}{{12}}, - 0.333} \right)\)    \(( - 0.333333 \ldots )\)     (A1)     (C2)

[2 marks]

\((y - 5) =  - \frac{1}{3}(x - 3)\)     (M1)(A1)(ft)

OR

\(5 =  - \frac{1}{3}(3) + c\)     (M1)

\(y =  - \frac{1}{3}x + 6\)     (A1)(ft)     (C2)

Notes: Award (M1) for substitution of their gradient into equation of line with their values from (a) correctly substituted.

Accept correct equivalent forms of the equation of the line. Follow through from their parts (a) and (b).

[2 marks]

Overall, there was a very good response to parts (a) and (b) with only a few candidates giving an incorrect expression for the gradient in part (b).

Overall, there was a very good response to parts (a) and (b) with only a few candidates giving an incorrect expression for the gradient in part (b). Occasionally, the final mark in part (b) was lost because the negative sign was dropped by some candidates.

Many able candidates recognized they needed to do something with the equation \(y = mx + c\) in part (c). Weaker candidates clearly showed a lack of understanding of an equation of a line and either simply gave a numerical answer for this part of the question or tried to use the coordinates of \(M\) into what they believed was the required equation of the straight line. A popular incorrect answer seen was \(y = 3x + 5\) .