A-LevelMathematicsCoordinate geometry9709/12 · May/June 2017 · Q26 Marks

The point A has coordinates (−2, 6). The equation of the perpendicular bisector of the line AB is 2y = 3x + 5.

Mark scheme

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

Gradient = 1.5 Gradient of perpendicular = -2/3
[B1]
Equation of AB is y-6 = -2/3(x+2) Or 3y+2x=14 oe
[M1 A1] Correct use of straight line equation with a changed gradient and (–2, 6), the (-(-2)) must be resolved for the A1 ISW. Using y = mx + c gets A1 as soon as c is evaluated.
Simultaneous equations → Midpoint (1, 4)
[M1] Attempt at solution of simultaneous equations as far as x =, or y =.
Use of midpoint or vectors → В (4, 2)
[M1A1] Any valid method leading to x, or to y.

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Topic: Coordinate geometry, A-Level Mathematics (syllabus 9709). Structured question, 6 marks.
Source: A-Level Mathematics, May/June 2017 Paper 1 Variant 2, question 2.

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