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0580/42 Oct/Nov 2023, Question 12: The Perpendicular Bisector

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 11 marks

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The question, in short

Given O(0, 0), A(8, 1) and B(2, 5), work with vectors, the line through A and B, its perpendicular bisector, and the gap between two y-axis intercepts.

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

Part (ai) · 1 mark

Express the vector from O to B as a column vector.

  1. 1→OB = column vector (2, 5)

Answercolumn vector (2, 5)

Part (aii) · 1 mark

Express the vector from A to B as a column vector.

  1. 1→AB = (2 − 8, 5 − 1) = column vector (−6, 4)

Answercolumn vector (−6, 4)

Part (b) · 3 marks

Determine the equation of the straight line through A and B in y = mx + c form.

  1. 1Gradient = (5 − 1)/(2 − 8) = 4/−6 = −2/3
  2. 2Using (2, 5): 5 = −2/3 × 2 + c, c = 5 + 4/3 = 19/3
  3. 3y = −2/3 x + 19/3

Answery = −2/3 x + 19/3

Part (c) · 4 marks

Determine the equation of the line perpendicular to AB passing through its midpoint, in y = mx + c form.

  1. 1Midpoint of AB = ((8 + 2)/2, (1 + 5)/2) = (5, 3)
  2. 2Perpendicular gradient = −1 ÷ (−2/3) = 3/2
  3. 33 = 3/2 × 5 + c, c = 3 − 15/2 = −9/2
  4. 4y = 3/2 x − 9/2

Answery = 3/2 x − 9/2

Part (d) · 2 marks

Find the distance PQ, where P and Q are where AB and its perpendicular bisector cross the y-axis.

  1. 1Gradient of AB = (5 − 1)/(2 − 8) = −2/3
  2. 2From B(2, 5) back to x = 0: y = 5 + 2 × 2/3 = 19/3, so P = (0, 19/3)
  3. 3Midpoint of AB = (5, 3); perpendicular gradient = 3/2
  4. 4From (5, 3) back to x = 0: y = 3 − 5 × 3/2 = −9/2, so Q = (0, −9/2)
  5. 5PQ = 19/3 − (−9/2) = 38/6 + 27/6 = 65/6

Answer65/6

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Also searched as: 0580/42/O/N/23 · 0580_w23_qp_42 · 0580 w23 42 · question 12.
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