Parallel lines and their equations
Use a shared gradient and a new point to place a distinct parallel line.
Core and Extended. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Read the gradient in y = mx + c.
- Find c by substituting a point.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the gradient of y = −3x + 5?
What you will learn
- Recognise distinct parallel non-vertical lines by their gradients.
- Find a parallel line through a given point.
- Distinguish parallel, intersecting and coincident lines.
Same direction, different position
Distinct non-vertical straight lines are parallel when their gradients are equal. They rise or fall at the same rate and never meet. The lines y = 2x + 1 and y = 2x − 4 both have gradient two, but their different intercepts place them at different positions.
If both m and c are equal, the equations describe the same line, rather than two distinct parallel lines. Different gradients make non-vertical straight lines intersect somewhere, even if the intersection lies beyond the visible grid. Compare the exact gradients, not the apparent angles in a stretched diagram.
Keep m; recalculate c
To find a line parallel to y = −3x + 5 through (2, 1), keep m = −3 and write y = −3x + c. Substitute the new point: 1 = −6 + c, so c = 7. The answer y = −3x + 7 shares the original gradient and passes through the required point.
Do not copy the original intercept unless the supplied point lies on the original line. Check both requirements separately: the gradient must match, and substituting the point must make the equation true. These checks catch different mistakes and neither replaces the other.
Special cases and equivalent descriptions
Distinct vertical lines x = a and x = b, with a ≠ b, are parallel even though their gradients are undefined. Horizontal lines y = a and y = b are parallel when their fixed heights differ. A line parallel to x = 2 through (−3, 5) is x = −3.
For Extended work, an original equation may need rearranging before its gradient is apparent. For 2x + 3y = 9, m = −2/3. Keeping this gradient is enough to find a parallel line through a point. Equivalent equations can conceal that two apparent routes are actually the same line; simplify before concluding they are distinct.
Pause and explain
Which line is distinct and parallel to y = 2x + 1?
Worked example
Find the line parallel to y = 3x + 2 passing through (2, −1).
Show the worked solution
- Keep the gradient three and write y = 3x + c.
- Substitute the point: −1 = 6 + c, so c = −7.
- The required line is y = 3x − 7; it differs from the original in intercept.
Answer y = 3x − 7
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find a line parallel to y = 4x − 1 through (1, 5).
Give me a hint
The new point determines the intercept after retaining m = 4.
Compare my reasoning
- Write y = 4x + c.
- 5 = 4 + c gives c = 1.
- The required line is y = 4x + 1; the gradient matches and the intercept differs.
y = 4x + 1
Look for these in your work
- Retained the original gradient.
- Verified the new point and distinct intercept.
Find the line parallel to y = −2x + 6 passing through (−1, 5).
Give me a hint
Keep the negative gradient and bracket the negative input.
Compare my reasoning
- Write y = −2x + c.
- 5 = −2(−1) + c = 2 + c, so c = 3.
- The line y = −2x + 3 has the same gradient and a different intercept.
y = −2x + 3
Look for these in your work
- Matched the signed gradient.
- Verified the new point and distinct intercept.
A map boundary is x = 2. Give the distinct parallel boundary passing through (−3, 5) and its horizontal separation from the original.
Give me a hint
A vertical parallel line keeps one horizontal coordinate fixed.
Compare my reasoning
- The new line has x = −3.
- The original has x = 2, so separation is |2 − (−3)| = 5 units.
- Both are vertical; comparing finite gradients is unnecessary.
x = −3; separation 5 units
Look for these in your work
- Used a vertical-line equation.
- Reported a positive separation.
Common mistakes
- Copying both gradient and intercept without checking the new point.
- Comparing intercepts instead of gradients.
- Calling coincident equations distinct parallel lines.
What can you explain now?
Are y = 3x + 2 and y = 3x − 7 distinct parallel lines?
Compare with the explanation
Yes.
Their gradients are equal and their intercepts differ, so they never intersect.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, find two lines parallel to a chosen line and one coincident description. Explain how to tell them apart.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.