0606/21 May/Jun 2025, Question 4: The Perpendicular Bisector
Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 7 marks
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The question, in short
Given A(−4, 3), B(6, −9), C(15, 10), D(14, −1) and line L: y = 11x − 75, find the area of triangle CDE where E is where the perpendicular bisector of AB meets L.
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Solution · 7 marks
- 1Gradient AB = (−9 − 3)/(6 − (−4)) = −6/5, so perpendicular gradient = 5/6
- 2Midpoint of AB = ((−4 + 6)/2, (3 − 9)/2) = (1, −3)
- 3Perpendicular bisector: y + 3 = (5/6)(x − 1), i.e. y = (5/6)x − 23/6
- 411x − 75 = (5/6)x − 23/6 gives 61x = 427, x = 7, y = 2: E(7, 2)
- 5Area CDE = (1/2)|15(−1 − 2) + 14(2 − 10) + 7(10 − (−1))| = (1/2)|−45 − 112 + 77| = 40
Answer40
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