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4PM1/02 Nov 2021, Question 11: Perpendicular Lines

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 15 marks

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The question, in short

Given points A(−3, −5) and B(7, 5), work through line AB, a perpendicular condition fixing p, a ratio division to find D, and lengths and 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(−3, −5) and B(7, 5).

  1. 1Gradient of AB = (5 − (−5))/(7 − (−3)) = 1
  2. 2y − 5 = 1(x − 7)
  3. 3y = x − 2

Answery = x − 2

Part (b) · 7 marks

With C = (p, 1) and p < 0, use perpendicularity of AC and BC to show p = −5.

  1. 1Gradient of BC = (5 − 1)/(7 − p) = 4/(7 − p)
  2. 2Gradient of AC = (1 − (−5))/(p − (−3)) = 6/(p + 3)
  3. 3Perpendicular: 4/(7 − p) × 6/(p + 3) = −1
  4. 424 = −(7 − p)(p + 3), so p² − 4p − 45 = 0
  5. 5(p − 9)(p + 5) = 0
  6. 6p < 0, so p = −5

Answerp = −5 shown

Part (c) · 2 marks

Given C divides straight line BD in ratio 4 : 3, find coordinates of D.

  1. 1BC : CD = 4 : 3, so BD = (7/4)BC
  2. 2C − B = (−12, −4), so D = B + (7/4)(−12, −4)
  3. 3D = (7 − 21, 5 − 7) = (−14, −2)

AnswerD = (−14, −2)

Part (d) · 4 marks

Find the exact length of AC, then use it to find the area of triangle ABD.

  1. 1(i) AC = √((1 − (−5))² + (−5 − (−3))²) = √(36 + 4)
  2. 2= √40 = 2√10
  3. 3(ii) BD = √((7 + 14)² + (5 + 2)²) = √490 = 7√10
  4. 4AC is perpendicular to the line BD, so AC is the height
  5. 5Area = (1/2) × 7√10 × 2√10 = 70

Answer(i) AC = 2√10; (ii) 70 (units²)

Stuck on a step? Read the free lesson: Perpendicular Lines

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