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0580/43 May/Jun 2021, Question 4: The Perpendicular Bisector

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 13 marks

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The question, in short

Coordinate geometry and vectors: midpoint and perpendicular bisector of A(1, 5) B(3, 9), vector PQ and a scaled point R, then vectors in a triangle with a ratio.

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

Part (ai) · 2 marks

Midpoint M of A(1, 5) and B(3, 9).

  1. 1x = (1 + 3)/2 = 2
  2. 2y = (5 + 9)/2 = 7
  3. 3M = (2, 7)

Answer(2, 7)

Part (aii) · 4 marks

Line through M at right angles to AB, as y = mx + c.

  1. 1Gradient of AB = (9 − 5)/(3 − 1) = 2
  2. 2Perpendicular gradient = −1/2
  3. 37 = −1/2 × 2 + c, so c = 8
  4. 4y = −1/2 x + 8

Answery = −1/2 x + 8

Part (bi) · 2 marks

Vector PQ from position vectors (−2, 3) and (−2, 5).

  1. 1→PQ = q − p
  2. 2= (−2 − (−2), 5 − 3)
  3. 3= column vector (0, 2)

Answercolumn vector (0, 2)

Part (bii) · 2 marks

Position vector of R when PR is three times PQ.

  1. 1→PR = 3 × (0, 2) = (0, 6)
  2. 2r = p + →PR = (−2, 3) + (0, 6)
  3. 3= column vector (−2, 9)

Answercolumn vector (−2, 9)

Part (ci) · 2 marks

Y divides UT in the ratio 2 : 1; express OY using t and u.

  1. 1→UT = t − u
  2. 2→UY = 2/3(t − u)
  3. 3→OY = u + 2/3(t − u)
  4. 4= 2/3 t + 1/3 u

Answer2/3 t + 1/3 u or 1/3(2t + u)

Part (cii) · 1 mark

Z on OT with YZ parallel to UO; express OZ.

  1. 1YZ is parallel to UO, so Z divides OT in the same ratio as Y divides UT
  2. 2OZ : ZT = UY : YT = 2 : 1
  3. 3→OZ = 2/3 t

Answer2/3 t

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Also searched as: 0580/43/M/J/21 · 0580_s21_qp_43 · 0580 s21 43 · question 4.
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