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4PM1/02 Jun 2021, Question 4: Length of a Line Segment

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 16 marks

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The question, in short

Given points A(−4, −10) and B(3, 11), work with line l through them, point P dividing AB in ratio 3:4, perpendicular point Q with PQ = √10, and point R(−11, −21).

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

Part (a) · 2 marks

Find an equation of the line l passing through A and B.

  1. 1Gradient = (11 + 10)/(3 + 4) = 3
  2. 2y + 10 = 3(x + 4)
  3. 3y = 3x + 2

Answery = 3x + 2

Part (b) · 2 marks

Find the coordinates of P on l where AP : PB = 3 : 4.

  1. 1P = ((4 × (−4) + 3 × 3)/7, (4 × (−10) + 3 × 11)/7)
  2. 2P = (−1, −1)

AnswerP(−1, −1)

Part (c) · 6 marks

Find coordinates of Q, with m < 0, on the perpendicular to l through P where PQ = √10.

  1. 1Perpendicular gradient −1/3: (n + 1)/(m + 1) = −1/3, so n + 1 = −(1/3)(m + 1)
  2. 2PQ² = 10: (m + 1)² + (n + 1)² = 10
  3. 3(m + 1)² + (1/9)(m + 1)² = 10 ⇒ (m + 1)² = 9
  4. 4m = 2 or m = −4; m < 0 so m = −4
  5. 5n + 1 = 1, n = 0, so Q(−4, 0)

AnswerQ(−4, 0)

Part (d) · 4 marks

Show that AB and RQ have equal length, and that AB and RQ are parallel.

  1. 1AB = √(7² + 21²) = 7√10
  2. 2RQ = √((−4 + 11)² + (0 + 21)²) = 7√10, so AB = RQ
  3. 3Gradient of RQ = (0 − (−21))/(−4 − (−11)) = 21/7 = 3
  4. 4Gradient of AB = 3, so AB and RQ are parallel

AnswerAB = RQ = 7√10 and both gradients are 3, so AB and RQ are equal and parallel

Part (e) · 2 marks

Find the area of quadrilateral ABQR.

  1. 1ABQR is a parallelogram with base AB = 7√10
  2. 2Its height is the perpendicular distance PQ = √10
  3. 3Area = 7√10 × √10 = 70

Answer70

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