AQA GCSE Mathematics Paper 3 (Higher), June 2025: Question 11

Straight Line Graphs: Parallel Lines & Equations

📋 What This Question Tests

This question assesses your ability to work with linear graphs in the form y = mx + c. Specifically, you need to understand that parallel lines share the exact same gradient, rearrange equations into gradient-intercept form, and find the equation of a line given its gradient and a coordinate point on the line.

Question 11 (a)

Writing an equation of a parallel line [1 mark]

✅ Acceptable Answers

Any line with gradient m = 2 and a y-intercept other than 9, for example:

y = 2x + 1
y = 2x
y = 2x - 5
y - 2x = 3

💡 Key Knowledge

Rearrange into standard form y = mx + c :
Add 2x to both sides of y - 2x = 9 gives y = 2x + 9 .
The gradient is the coefficient of x, so gradient (m) = 2.
Parallel lines have identical gradients.
A distinct parallel line must have a different y-intercept ( c ≠ 9 ).

🧠 Exam Technique

The question says "Write down the equation...", which means you need to supply a specific numeric line equation. Choose the simplest possible constant, such as y = 2x or y = 2x + 1 .

❌ Common Errors & Examiner Traps

Writing the letter 'c': Writing y = 2x + c scores 0 marks. You must state an actual number for the constant!
Writing the identical line: Leaving the constant as 9 gives the same line, not a parallel one.
Rearranging with y - 2x + 1 = 10 : This simplifies back to y - 2x = 9 (the exact same line), scoring 0.

Mark Scheme: [B1] for y = 2x + c or y - 2x = c (or equivalent) where c is an actual numerical value and c ≠ 9.

Question 11 (b)

Finding the equation from a gradient and point [1 mark]

📐 Step-by-Step Calculation

Identify gradient: m = 5 , so the equation begins as y = 5x + c .
Substitute the point (3, 7): Replace x = 3 and y = 7 :
7 = 5(3) + c
7 = 15 + c
Solve for c:
c = 7 - 15 = -8
Form complete equation:
y = 5x - 8

✅ Correct Answer

The correct option to circle is:

y = 5x - 8

This matches our calculated gradient of 5 and y-intercept of -8.

🧠 Quick Elimination Method

Since gradient = 5, the term in front of x must be 5. Instantly eliminate y = 3x - 2 and y = 3x + 7 .
Now test (3, 7) in the remaining two options:
• If y = 5x , when x = 3 , y = 15 ≠ 7 ❌
• If y = 5x - 8 , when x = 3 , y = 15 - 8 = 7 ✔️

❌ Common Errors

Swapping gradient and coordinate: Picking an option starting with 3x because the x-coordinate was 3.
Sign mistake in finding c: Calculating c = 15 - 7 = 8 instead of 7 - 15 = -8 .
Selecting without checking: Circling multiple options cancels out the mark!

Mark Scheme: [B1] for correctly circling y = 5x - 8 .