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4PM1/01R Jun 2022, Question 11: The Point Dividing a Line in a Ratio

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

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The question, in short

Line l is y − 3x = 3 with A on the y-axis and B = (10, b). C divides AB in ratio 2:3. Show k through C perpendicular to l has equation 3y + x − 49 = 0, then find D and the area of triangle ECF.

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

Part (a) · 6 marks

Show that the line k through C, perpendicular to l, has equation 3y + x − 49 = 0.

  1. 1b = 3 × 10 + 3 = 33, A = (0, 3), B = (10, 33)
  2. 2C = ((3 × 0 + 2 × 10)/5, (3 × 3 + 2 × 33)/5) = (4, 15)
  3. 3Gradient of l is 3, so gradient of k = −1/3
  4. 4y − 15 = −(1/3)(x − 4) ⇒ 3y − 45 = −x + 4
  5. 53y + x − 49 = 0

Answer3y + x − 49 = 0 shown

Part (b) · 6 marks

Given D = (p, q) with q positive, AD parallel to k and AD = 12√10, find the coordinates of D.

  1. 1A = (0, 3)
  2. 2AD parallel to k: (q − 3)/p = −1/3 ⇒ p = −3(q − 3)
  3. 3AD² = p² + (q − 3)² = 1440 ⇒ 10(q − 3)² = 1440
  4. 4(q − 3)² = 144 ⇒ q = 15 or q = −9; q > 0 so q = 15
  5. 5p = −3(12) = −36, D(−36, 15)

AnswerD(−36, 15)

Part (c) · 5 marks

E on k and F on l with DE and DF parallel to the y-axis; find the exact area of triangle ECF.

  1. 1E on k with x = −36: 3y − 36 − 49 = 0 ⇒ E(−36, 85/3)
  2. 2F on l with x = −36: y = 3(−36) + 3 ⇒ F(−36, −105)
  3. 3EF = 85/3 + 105 = 400/3 (vertical side)
  4. 4Horizontal distance from C(4, 15) to the line x = −36 is 40
  5. 5Area = (1/2) × 40 × 400/3 = 8000/3

Answer8000/3

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