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0580/41 May/Jun 2024, Question 5: The Perpendicular Bisector

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 17 marks

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The question, in short

Given points P(1, 7) and Q(5, −5) about a circle centred at O, work out vectors, lengths, circle measures, a midpoint and a line equation, plus a ratio-based position vector.

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

Part (ai) · 2 marks

Work out the vector from P to Q using the two coordinate pairs.

  1. 1PQ = (5 − 1, −5 − 7)
  2. 2= column vector (4, −12)

Answercolumn vector (4, −12)

Part (aii) · 3 marks

Prove the distances from O to P and from O to Q are equal.

  1. 1|OP| = √(1² + 7²) = √(1 + 49) = √50
  2. 2|OQ| = √(5² + (−5)²) = √(25 + 25) = √50
  3. 3So |OP| = |OQ| = √50

AnswerBoth √50

Part (aiii) · 2 marks

Using PQ as a chord of the circle centred at O, work out its circumference.

  1. 1P is on the circle, so radius = OP.
  2. 2OP = √(1² + 7²) = √50
  3. 3Circumference = 2 × π × √50 = 44.428…
  4. 4Circumference = 44.4

Answer44.4 or 44.428… to 44.435

Part (aiv) · 2 marks

Taking PQ as a diameter of another circle centred at R, work out R's coordinates.

  1. 1R = midpoint of PQ = ((1 + 5)/2, (7 + (−5))/2)
  2. 2= (3, 1)

Answer(3, 1)

Part (av) · 4 marks

Work out the perpendicular bisector of PQ in the form y = mx + c.

  1. 1Midpoint of PQ = ((1 + 5)/2, (7 + (−5))/2) = (3, 1)
  2. 2Gradient of PQ = (−5 − 7)/(5 − 1) = −12/4 = −3
  3. 3Perpendicular gradient = 1/3, since −3 × 1/3 = −1
  4. 4Through (3, 1): 1 = 1/3 × 3 + c, so c = 0
  5. 5y = (1/3)x

Answery = (1/3)x

Part (b) · 4 marks

With position vectors a for A and b for B, and M dividing AB in the ratio 2 : 3, work out M's position vector in simplest terms.

  1. 1AB = b − a
  2. 2AM = 2/5(b − a)
  3. 3OM = a + 2/5(b − a)
  4. 4= 3/5 a + 2/5 b

Answer3/5 a + 2/5 b

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Also searched as: 0580/41/M/J/24 · 0580_s24_qp_41 · 0580 s24 41 · question 5.
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