Use parallel and perpendicular gradients
Compare line gradients to identify parallel and perpendicular lines, including horizontal and vertical cases.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the gradient of a non-vertical line parallel to y = −2x + 5?
What you will learn
- Recognise parallel lines from equal gradients
- Use negative reciprocal gradients for perpendicular non-vertical lines
- Find an unknown line equation from a gradient relationship and point
Parallel lines keep the same direction
Two distinct non-vertical parallel lines have equal gradients. They can have different intercepts; equal gradients alone do not mean they are the same line.
Two vertical lines are parallel even though their gradients are undefined. A vertical and a horizontal line are perpendicular.
Perpendicular gradients multiply to −1
For two non-vertical, non-horizontal perpendicular lines, m₁m₂ = −1. Thus a line perpendicular to gradient m has gradient −1/m.
A horizontal line has gradient 0, and its perpendicular is vertical. The product rule cannot represent this special case because a vertical gradient is undefined.
First rearrange a given equation into y = mx + c to read its gradient. Then use the parallel or perpendicular condition before applying point-gradient form.
Pause and explain
Which gradient is perpendicular to a line of gradient 3?
Worked example
Find the gradient of a line perpendicular to 3x + 2y = 8.
Show the worked solution
- Rearrange: 2y = −3x + 8, so the original gradient is −3/2.
- For perpendicular lines, multiply the gradients to get −1.
- The perpendicular gradient is 2/3 because (−3/2)(2/3) = −1.
Answer 2/3.
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 the equation through (2, 1) parallel to y = −4x + 2.
Give me a hint
Keep the same gradient and substitute the given point.
Compare my reasoning
- The parallel gradient is −4.
- Use y − 1 = −4(x − 2).
- Expand to y = −4x + 9.
y = −4x + 9.
Look for these in your work
- Parallel gradient retained
- Point used correctly
- Intercept checked
Find the equation through (3, 1) perpendicular to y = 2x + 5.
Give me a hint
Take the negative reciprocal of 2, then use point-gradient form.
Compare my reasoning
- The perpendicular gradient is −1/2.
- Use y − 1 = −(1/2)(x − 3).
- Simplify to y = −(1/2)x + 5/2.
y = −(1/2)x + 5/2.
Look for these in your work
- Gradient relationship correct
- Point substituted
- Equation simplified
A ramp edge has gradient 3/4. A safety rail must be perpendicular and pass through (4, 6). Find the rail equation.
Give me a hint
A perpendicular gradient is the negative reciprocal of 3/4.
Compare my reasoning
- The rail gradient is −4/3.
- Use y − 6 = −(4/3)(x − 4).
- The point (4, 6) satisfies the equation by construction.
y − 6 = −(4/3)(x − 4).
Look for these in your work
- Negative reciprocal used
- Required point included
- Relationship justified
Common mistakes
- Assuming perpendicular gradients are equal
- Using the reciprocal without changing its sign
- Applying m₁m₂ = −1 to a horizontal/vertical pair
What can you explain now?
What gradient must a line parallel to y = 4x − 1 have?
Compare with the explanation
4
Distinct parallel lines have equal gradients; their intercepts may differ.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State the gradient conditions for parallel and perpendicular non-vertical lines, and name the horizontal/vertical exception.
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.