Solutions to this question by accurate drawing will not be accepted. The points A and B are (-8, 8) and (4, 0) respectively.

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

1. [B2] y-8 = -8/12 (x-(-8)) oe isw or y[-0] = -8/12 (x-4) oe isw or 3y = -2x+8 oe isw
2. [B1] for M_AB = -8/12 oe
3. [M1] (-8-4)² + (8[-0])² oe any valid method
4. [A1] √208 isw or 4√13 isw or 14.4222051... rot to 3 or more sf implies M1 provided nfww
5. [B1] [coordinates of D =] (-2, 4) soi If coordinates of D not stated then a calculation for m_CD or a relevant length with the coordinates clearly embedded must be shown to imply B1
6. [M1] Gradient methods: m_CD = (7-their4)/(0-their(-2)) their (3/2) or Length of sides methods: finds or states AC² = 65 or AC = √65 or AC² = (-8-0)²+(8-7)² oe or CD² = their13 or CD = their√13 or CD² = (0-their(-2))²+(7-their4)² oe or AD² = their52 or AD = their2√13 or AD² = (-8-their(-2))²+(8-their4)² or uses a valid method with their coordinates of D to find the exact area of the triangle and equates to 1/2 (AD)(CD)sin(ADC)
7. [A1] states -3/2 x 2/3 = -1 oe or 3/2 is the negative reciprocal of -2/3 oe or finds the equation of the perpendicular bisector of AB as y = 3/2 x + 7 independently of C and states that C lies on this line. applies Pythagoras to confirm, using integer values, that 65 = 13 + 52 or finds e.g. AC = √65 using (2√13)²+(√13)² or solves 1/2 (2√13)(√13)sin ADC = 13 or (√65)² = (2√13)² + (√13)² -2(2√13)(√13) cos ADC to show ADC is a right angle
8. [B1] (-4, 1) or -4i+j condone coordinates
9. [B2] Full valid method e.g. for showing that e.g. CB = (4, -7) and AE = (4, -7) oe
10. [B1] for incomplete method e.g. for stating that CB = (4, -7) or AC = (-8, 8) = EB or just showing that one pair of opposite sides is parallel or has the same length or just showing that length DC = length DE or just showing that C, D and E are collinear