The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is 90°. The coordinates of A, B and C are (2, 6), (5, −3) and (8, 3) respectively. [Figure of trapezium ABCD with coordinates labeled]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [1] Gradient of AB (mAB) = (-3-6)/(5-2) = -9/3 = -3. oe
2. [1] Since angle BAD is 90°, AD is perpendicular to AB. Gradient of AD (mAD) = 1/3. use of m₁m₂ = -1 with grad AB
3. [1] Equation of AD: y – 6 = ⅓(x – 2) or 3y = x + 16. co – OK unsimplified
4. [1] Since DC is parallel to AB, the gradient of CD (mCD) = -3. Equation of CD: y – 3 = -3(x – 8) or y = -3x + 27. OK unsimplified. Follow through on m of AB. Reasonable algebra leading to x = or y = with AD and CD
5. [1] Solve simultaneous equations for AD and CD to find coordinates of D. Reasonable algebra leading to x = or y = with AD and CD
6. [1] Coordinates of D are (6.5, 7.5). co
7. [1] Use of vectors or midpoint formula to find coordinates of E. For parallelogram ABCE, vector AB = vector EC. May be implied
8. [1] Coordinates of E are (5, 12). Length of BE = √((5-5)²+(12-(-3))²) = √(0²+15²) = 15. co