4024/12 May/Jun 2023, Question 24: The Perpendicular Bisector
Cambridge O Level Mathematics 4024 · Paper 12 (Non-calculator) · Non-calculator · 5 marks
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The question, in short
Given points A (3, 11) and B (−5, −5) and line L with equation 2y + x = 5, show that L is the perpendicular bisector of AB.
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Solution · 5 marks
- 1Midpoint of AB = ((3 + (−5))/2, (11 + (−5))/2) = (−1, 3)
- 2Gradient of AB = (11 − (−5))/(3 − (−5)) = 16/8 = 2
- 3L: 2y = −x + 5, so y = −1/2 x + 5/2, gradient −1/2
- 42 × (−1/2) = −1, so L is perpendicular to AB
- 5At (−1, 3): 2 × 3 + (−1) = 5, so the midpoint (−1, 3) lies on L
- 6So L is the perpendicular bisector of AB
AnswerMidpoint of AB = (−1, 3); gradient of AB = 2, so gradient of perpendicular = −1/2; 2 × 3 + −1 = 5 so (−1, 3) is on L, and gradient of L = −1/(gradient AB)
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