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WMA11/01 Oct 2020, Question 6: Perpendiculars: Feet and Areas

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 11 marks

The question, in short

Given points A(−4, 11) and B(8, 2), work with line AB, its midpoint M, a perpendicular line l through M, and point C on l so triangle ABC has area 37.5.

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

Part (a) · 2 marks

Calculate the slope of the line through A and B, expressed as a fraction in lowest terms.

  1. 1m = (2 − 11)/(8 − (−4))
  2. 2= −9/12 = −3/4

Answer−3/4

Part (b) · 4 marks

Using midpoint M, determine the equation of the line perpendicular to AB through M in the integer form px + qy + r = 0.

  1. 1M = ((−4 + 8)/2, (11 + 2)/2) = (2, 13/2)
  2. 2Perpendicular gradient = −1 ÷ (−3/4) = 4/3
  3. 3y − 13/2 = (4/3)(x − 2)
  4. 46y − 39 = 8x − 16, so 8x − 6y + 23 = 0

Answer8x − 6y + 23 = 0

Part (c) · 5 marks

Locate both positions of C on line l that make triangle ABC have area 37.5 square units, giving coordinate pairs.

  1. 1AB = √((8 + 4)² + (2 − 11)²) = √225 = 15
  2. 2Area = (1/2) × MC × AB = 37.5, so MC = 5
  3. 3l has gradient 4/3, so a step of 5 along l is 3 across and 4 up
  4. 4C = (2 − 3, 13/2 − 4) or (2 + 3, 13/2 + 4)
  5. 5C = (−1, 2.5) or (5, 10.5)

AnswerC = (−1, 2.5) and C = (5, 10.5)

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Also searched as: WMA11 October 2020 · WMA11/01 · wma11 oct20 · IAS Maths Pure Mathematics 1 October 2020 · question 6.
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