0580/13 May/Jun 2021, Question 21: The Equation y = mx + c
Cambridge IGCSE Mathematics 0580 · Paper 13 (Core) · Calculator · 3 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
Use the given graph of straight line L to determine its equation in the form y = mx + c, then find a second line's equation from given conditions.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 2 marks
From the graph of line L, find its gradient and intercept to write the equation as y = mx + c.
- 1Line L crosses the y-axis at (0, 2) and the x-axis at (4, 0)
- 2Gradient = (0 − 2)/(4 − 0) = −1/2
- 3y = −1/2 x + 2
Answer−0.5x + 2 or −1/2 x + 2
Part (b) · 1 mark
Find the equation of the line parallel to y = 3x − 5 that passes through (0, 17).
- 1Parallel lines have the same gradient, m = 3
- 2The line passes through (0, 17), so c = 17
- 3y = 3x + 17
Answery = 3x + 17
Stuck on a step? Read the free lesson: The Equation y = mx + c
Practise more the equation y = mx + c questions
Real 0580 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on the equation y = mx + c
Also searched as: 0580/13/M/J/21 · 0580_s21_qp_13 · 0580 s21 13 · question 21.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.