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0580/43 May/Jun 2022, Question 8: Simultaneous Equations with a Curve

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 18 marks

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The question, in short

Given points A(−2, 7), B(1, −5), C(5, 4), work out midpoint, vector, length, and two line equations; then find intersections of y + 5x = 8 with y = 2x² + 6x − 13.

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

Part (ai) · 2 marks

Find the midpoint coordinates of the line segment AB.

  1. 1Midpoint = ((−2 + 1)/2, (7 + (−5))/2)
  2. 2= (−0.5, 1)

Answer(−0.5, 1)

Part (aii) · 2 marks

Find the vector AC.

  1. 1→AC = (5 − (−2), 4 − 7)
  2. 2= column vector (7, −3)

Answercolumn vector (7, −3)

Part (aiii) · 2 marks

Find the magnitude of vector AC.

  1. 1|→AC| = √(7² + (−3)²) = √58
  2. 2= 7.615… ≈ 7.62

Answer7.62 or 7.615 to 7.616

Part (aiv) · 3 marks

Find the equation of line AB in the form y = mx + c.

  1. 1Gradient = (−5 − 7)/(1 − (−2)) = −12/3 = −4
  2. 27 = −4 × (−2) + c, so c = −1
  3. 3y = −4x − 1

Answery = −4x − 1

Part (av) · 3 marks

Find the equation, in y = mx + c form, of the line through C perpendicular to AB.

  1. 1Gradient of AB = (−5 − 7)/(1 − (−2)) = −12/3 = −4
  2. 2Perpendicular gradient = −1 ÷ (−4) = 1/4
  3. 3Substitute C(5, 4): 4 = 1/4 × 5 + c, so c = 11/4
  4. 4y = 1/4 x + 11/4

Answer1/4 x + 11/4

Part (b) · 6 marks

Find the coordinates of both intersection points P and Q of the line and the curve, showing all working.

  1. 1y = 8 − 5x, so 8 − 5x = 2x² + 6x − 13
  2. 22x² + 11x − 21 = 0
  3. 3(2x − 3)(x + 7) = 0, so x = 3/2 or x = −7
  4. 4x = 3/2: y = 8 − 7.5 = 1/2; x = −7: y = 8 + 35 = 43
  5. 5P(3/2, 1/2) and Q(−7, 43)

Answer(3/2, 1/2) and (−7, 43)

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Also searched as: 0580/43/M/J/22 · 0580_s22_qp_43 · 0580 s22 43 · question 8.
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