EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
Exam Practice: More Linear Graphs
More linear graphs · Graphs · Lesson 2 of 8 · 18 marks · 25 minutes
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Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Show your working.
Question 1 NON-CALCULATOR (3 marks)
Find the equation of the straight line that passes through the points (1, 3) and (4, 12).
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(Total for Question 1 = 3 marks)
Question 2 NON-CALCULATOR (3 marks)
Find the equation of the line that is parallel to y = 3x − 2 and passes through the point (2, 9).
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(Total for Question 2 = 3 marks)
Question 3 NON-CALCULATOR (2 marks)
Are the lines y = 4x + 1 and 2y = 8x − 6 parallel? Explain your answer.
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(Total for Question 3 = 2 marks)
Question 4 NON-CALCULATOR (3 marks)
Find the coordinates of the point where the lines y = x + 1 and y = −x + 5 cross.
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(Total for Question 4 = 3 marks)
Question 5 NON-CALCULATOR (4 marks)
Find the equation of the line that is perpendicular to y = 2x + 1 and passes through the point (4, 3).
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(Total for Question 5 = 4 marks)
Question 6 NON-CALCULATOR (3 marks)
The line L₁ has equation 3y = 6x − 2. The line L₂ is perpendicular to L₁. Work out the gradient of L₂.
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(Total for Question 6 = 3 marks)
TOTAL FOR PAPER = 18 MARKS
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Answers and mark scheme
Check your answer only once you have written one.
Question 1 (3 marks)
Gradient = (12 − 3)/(4 − 1) = 3. Substituting (1, 3): 3 = 3 + c, so c = 0. The equation is y = 3x.
• Gradient of 3 M1
• Substituting a point M1
• y = 3x A1
Question 2 (3 marks)
The gradient is 3, so y = 3x + c. 9 = 6 + c, so c = 3. The equation is y = 3x + 3.
• Gradient 3 M1
• 9 = 3 × 2 + c M1
• y = 3x + 3 A1
Question 3 (2 marks)
2y = 8x − 6 rearranges to y = 4x − 3. Both lines have gradient 4, so they are parallel.
• Rearranging to y = 4x − 3 M1
• Same gradient, so parallel C1
Question 4 (3 marks)
x + 1 = −x + 5, so 2x = 4 and x = 2. Then y = 3. The point is (2, 3).
• Setting the expressions equal M1
• x = 2 A1
• (2, 3) A1
Question 5 (4 marks)
The perpendicular gradient is −½. 3 = −½ × 4 + c, so c = 5. The equation is y = −½x + 5.
• Perpendicular gradient −½ M1
• 3 = −½ × 4 + c M1
• c = 5 A1
• y = −½x + 5 A1
Question 6 (3 marks)
y = 2x − ⅔, so the gradient of L₁ is 2. The gradient of L₂ is −½.
• Rearranging to y = … M1
• Gradient of L₁ is 2 A1
• −½ A1