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0580/41 May/Jun 2023, Question 6: Coordinate Geometry of Shapes

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 14 marks

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The question, in short

Square ABCD has A at (−2, 1) and B at (1, 5); locate C and D, then for P (−1, 3) and Q (6, 4) work out midpoint, distance, gradient and a related line.

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

Part (a) · 4 marks

Using the grid, determine where vertices C and D lie, given C has positive integer coordinates.

  1. 1A to B: across 3, up 4
  2. 2For the next side turn through 90°: across 4, down 3
  3. 3C = (1 + 4, 5 − 3) = (5, 2)
  4. 4D = (−2 + 4, 1 − 3) = (2, −2)
  5. 5C (5, 2), D (2, −2)

AnswerC (5, 2), D (2, −2)

Part (bi) · 2 marks

Work out the point halfway between P and Q.

  1. 1Midpoint = ((−1 + 6)/2, (3 + 4)/2)
  2. 2= (2.5, 3.5)

Answer(2.5, 3.5)

Part (bii) · 3 marks

Calculate how far apart P and Q are.

  1. 1PQ² = (6 − (−1))² + (4 − 3)² = 49 + 1 = 50
  2. 2PQ = √50 = 7.071… ≈ 7.07

Answer7.07 or 7.071...

Part (biii) · 2 marks

Determine the slope of the segment joining P and Q.

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

Answer1/7

Part (biv) · 3 marks

Give the equation of the line through x = 2 on the x-axis that runs parallel to PQ.

  1. 1Gradient of PQ = (4 − 3) ÷ (6 − (−1)) = 1/7
  2. 2A parallel line has the same gradient: y = (1/7)x + c
  3. 3It passes through (2, 0): 0 = 2/7 + c, so c = −2/7
  4. 4y = (1/7)x − 2/7

Answery = (1/7)x − 2/7 or 7y = x − 2

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Also searched as: 0580/41/M/J/23 · 0580_s23_qp_41 · 0580 s23 41 · question 6.
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