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0580/42 Feb/Mar 2021, Question 12: Perpendicular Lines

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 14 marks

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The question, in short

Differentiate a cubic to get the gradient at x = −2, then work with points A(7, 2) and B(−5, 8) on lengths, perpendicular lines and a parallelogram ABCD.

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

Part (a) · 3 marks

Find the gradient of y = 2x³ − 7x + 4 at x = −2.

  1. 1dy/dx = 6x² − 7
  2. 2At x = −2: 6 × 4 − 7
  3. 3= 17

Answer17

Part (bi) · 3 marks

Calculate the distance between A(7, 2) and B(−5, 8).

  1. 1AB = √((−5 − 7)² + (8 − 2)²)
  2. 2= √(144 + 36) = √180
  3. 3= 13.41… ≈ 13.4

Answer13.4 or 13.41 to 13.42

Part (bii) · 4 marks

Find the equation, in y = mx + c form, of the line perpendicular to AB through (−1, 3).

  1. 1Gradient AB = (8 − 2)/(−5 − 7) = −1/2
  2. 2Perpendicular gradient = 2
  3. 33 = 2 × (−1) + c so c = 5
  4. 4y = 2x + 5

Answery = 2x + 5

Part (biii) · 4 marks

Given BC = (−a, −b) with a, b > 0, gradient of BC is 1 and |BC| = √8, find the coordinates of D.

  1. 1Gradient of BC is 1 so a = b
  2. 2a² + b² = (√8)² = 8, so 2a² = 8, a = 2
  3. 3BC = column vector (−2, −2)
  4. 4In parallelogram ABCD, AD = BC
  5. 5D = (7 − 2, 2 − 2) = (5, 0)

Answer(5, 0)

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