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0580/22 May/Jun 2025, Question 9: Position Vectors

Cambridge IGCSE Mathematics 0580 · Paper 22 (Extended) · Non-calculator · 6 marks

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The question, in short

Given point A at (3, −1) and vector AB as column (2, −4), work out coordinates of C, a constant k in length AB = k√5, and position vector of point P on AB.

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Part (a) · 2 marks

Using the relation that vector AC equals twice vector AB, determine the coordinates of point C.

  1. 1→AC = 2 × column vector (2, −4) = column vector (4, −8)
  2. 2C = (3 + 4, −1 − 8) = (7, −9)

Answer(7, −9)

Part (b) · 2 marks

Given that the distance AB can be written as k√5, calculate the numerical value of k.

  1. 1AB = √(2² + (−4)²) = √20
  2. 2√20 = √4 × √5 = 2√5
  3. 3k = 2

Answer2

Part (c) · 2 marks

Point P lies on AB with AP to PB in ratio 1 to 3; find P's position vector.

  1. 1→AP = 1/4 →AB = column vector (0.5, −1)
  2. 2→OP = column vector (3, −1) + column vector (0.5, −1)
  3. 3= column vector (3.5, −2)

Answercolumn vector (3.5, −2)

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