Three points have coordinates A (0, 7), B (8, 3) and C (3k, k). Find the value of the constant k for which

Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

1. [B1] m of AB is -1/2 oe.
2. [M1] Eqn of AB is y = -1/2x + 7 Using A,B or C to get an equation
3. [M1] Let x = 3k, y = k Using C or A,B in the equation
4. [A1] k = 2.8 oe
5. OR
6. [M1A1] (7-k)/(0-3k) = (3-k)/(8-3k) Using A,B & C to equate gradients
7. [DM1A1] →20k = 56 → k = 2.8 Simplifies to a linear or 3 term quadratic = 0.
8. OR
9. [M1A1] (7-k)/(0-8) = (7-3)/(0-8) Using A,B and C to equate gradients
10. [DM1A1] →20k = 56 → k = 2.8 Simplifies to a linear or 3 term quadratic = 0.
11. [B1] M(4, 5) anywhere in (ii)
12. [M1] Perpendicular gradient = 2. Use of m₁m₂=-1 soi
13. [M1] Perp bisector has eqn y-5=2(x-4) Forming eqn using their M and their "perpendicular m"
14. [A1] Let x = 3k, y = k k = 3/5 oe
15. OR
16. [M1A1] (0 – 3k)² + (7 – k)² = (8 – 3k)² + (3 – k)² Use of Pythagoras.
17. [DM1A1] -14k +49 = 73 – 54k → 40k = 24 → k = 0.6 Simplifies to a linear or 3 term quadratic = 0.