A-LevelMathematicsCoordinate geometry9709/12 · Feb/Mar 2016 · Q58 Marks

Two points have coordinates A (5, 7) and B (9, −1).

(i)Find the equation of the perpendicular bisector of AB.[3]
(ii)The line through C (1, 2) parallel to AB meets the perpendicular bisector of AB at the point X. Find, by calculation, the distance BX.[5]

Mark scheme

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

[B1] Mid-point of AB = (7, 3) soi
[M1] Gradient of AB = (-1-7)/(9-5) = -8/4 = -2. Gradient of perpendicular bisector = -1/(-2) = 1/2 (Use of m₁m₂ = -1 soi)
[A1] Equation of perpendicular bisector is y - 3 = 1/2 (x - 7)
[M1] Equation of CX is y − 2 = -2(x-1) (Using their original gradient and (1,2))
[DM1] Solve simultaneously dependent on both previous M's, e.g. 1/2 x - 1/2 = -2x + 4
[A1] x = 9/5, y = 2/5 (Coordinates of X)
[M1] BX² = (9-9/5)² + (-1-2/5)² = (36/5)² + (-7/5)² = 7.2² + 1.4² soi
[A1] BX = 7.33

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Topic: Coordinate geometry, A-Level Mathematics (syllabus 9709). Structured question, 8 marks.
Source: A-Level Mathematics, Feb/Mar 2016 Paper 1 Variant 2, question 5.

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