Write down the coordinates of C, the midpoint of line segment AB.

Find the gradient of the line DC.

Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.

(1, −2)    (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.

Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.

[2 marks]

\(\frac{{1 - \left( { - 2} \right)}}{{ - 3 - 1}}\)    (M1)

Note: Award (M1) for correct substitution, of their part (a), into gradient formula.

\( =  - \frac{3}{4}\,\,\,\left( { - 0.75} \right)\)     (A1)(ft)  (C2)

Note: Follow through from part (a).

[2 marks]

\(y - 1 =  - \frac{3}{4}\left( {x + 3} \right)\)  OR  \(y + 2 =  - \frac{3}{4}\left( {x - 1} \right)\)  OR  \(y =  - \frac{3}{4}x - \frac{5}{4}\)      (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

OR

\(1 =  - \frac{3}{4} \times  - 3 + c\)  OR  \( - 2 =  - \frac{3}{4} \times 1 + c\)     (M1) 

Note: Award (M1) for correct substitution of their part (b) and a given point.

\(3x + 4y + 5 = 0\)  (accept any integer multiple, including negative multiples)    (A1)(ft) (C2)

Note: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be \(\frac{5}{0}\), award at most (M1)(A0) for either \(x =  - 3\) or \(x + 3 = 0\).

[2 marks]