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4PM1/01R Jun 2019, Question 11: Length of a Line Segment

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 17 marks

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The question, in short

Given points A(−1, 3) and B(5, 6), find line AB, a division point P, a point C on a perpendicular through P with circle radius 5, point D from perpendicular and parallel conditions, and the trapezium area.

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

Part (a) · 2 marks

Find an equation for the line through A(−1, 3) and B(5, 6).

  1. 1Gradient = (6 − 3)/(5 − (−1)) = 1/2
  2. 2y − 6 = ½(x − 5)
  3. 32y − 12 = x − 5

Answer2y − 12 = x − 5 (x − 2y + 7 = 0)

Part (b) · 2 marks

Show that the point dividing AB in the ratio 2 : 1 has coordinates (3, 5).

  1. 1P = ((1 × (−1) + 2 × 5)/3, (1 × 3 + 2 × 6)/3)
  2. 2= (9/3, 15/3) = (3, 5)

AnswerP = (3, 5) shown

Part (c) · 6 marks

For C(m, n), m > 0, with CP perpendicular to AB and circle through A, P, C having radius 5, find m and n.

  1. 1Gradient of CP = −2: (5 − n)/(3 − m) = −2 → n = 11 − 2m
  2. 2Angle APC = 90°, so AC is a diameter: AC = 10
  3. 3(m + 1)² + (n − 3)² = 100
  4. 4Substitute: (m + 1)² + (8 − 2m)² = 100 → m² − 6m − 7 = 0
  5. 5(m − 7)(m + 1) = 0, m > 0 so m = 7, n = −3

Answerm = 7, n = −3

Part (d) · 3 marks

For D(p, q) with AD perpendicular to AB and DC parallel to AB, find p and q.

  1. 1DC through (7, −3) with gradient ½: y = (x − 13)/2
  2. 2AD through (−1, 3) with gradient −2: y = −2x + 1
  3. 3(x − 13)/2 = −2x + 1 → x = 3, y = −5
  4. 4(or AD = PC = 4i − 8j, D = (−1 + 4, 3 − 8))

Answerp = 3, q = −5

Part (e) · 4 marks

Find the area of trapezium ABCD.

  1. 1AB = √(6² + 3²) = 3√5
  2. 2CD = √(4² + 2²) = 2√5
  3. 3Height AD = √(4² + 8²) = 4√5
  4. 4Area = ½(3√5 + 2√5) × 4√5 = 50

Answer50

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