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0580/43 May/Jun 2023, Question 11: The Perpendicular Bisector

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 9 marks

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The question, in short

For points M(4, 1) and N(−2, −7), find the length and gradient of MN, then the equation of its perpendicular bisector.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 3 marks

Calculate the distance between M(4, 1) and N(−2, −7).

  1. 1Differences: 4 − (−2) = 6, 1 − (−7) = 8
  2. 2MN² = 6² + 8² = 100
  3. 3MN = 10

Answer10

Part (b) · 2 marks

Work out the gradient of the line joining M(4, 1) and N(−2, −7).

  1. 1Gradient = (1 − (−7))/(4 − (−2))
  2. 2= 8/6 = 4/3

Answer4/3

Part (c) · 4 marks

Using M(4, 1) and N(−2, −7), find the equation of the line that perpendicularly bisects MN.

  1. 1Midpoint of MN = ((4 + (−2))/2, (1 + (−7))/2) = (1, −3)
  2. 2Gradient of MN = (1 − (−7)) ÷ (4 − (−2)) = 8/6 = 4/3
  3. 3Perpendicular gradient = −3/4
  4. 4Through (1, −3): −3 = −3/4 × 1 + c, so c = −9/4
  5. 5y = −3/4 x − 9/4

Answery = −3/4 x − 9/4 or 4y + 3x + 9 = 0

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Also searched as: 0580/43/M/J/23 · 0580_s23_qp_43 · 0580 s23 43 · question 11.
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