Find perpendicular lines and bisectors
Use negative reciprocal gradients to write perpendicular lines and construct a perpendicular bisector from endpoints.
Use this pathway for Cambridge IGCSE International Mathematics 0607, 2028–2030. Core and Extended objectives are labelled in each lesson; Extended adds requirements beyond Core. Paper 1 and Paper 2 are non-calculator. Papers 3–6 require a graphic display calculator (GDC). These original lessons use self-checked practice; they are not a Cambridge grade, full timed paper or teacher review.
Before you begin
- Check whether your school has entered you for Core or Extended. Core and Extended share most Number ideas, but some notation and methods are Extended only.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the negative reciprocal of 3/5?
What you will learn
- Find the gradient of a line perpendicular to a given line
- Write the perpendicular line through a point
- Construct a perpendicular bisector using midpoint and gradient
Perpendicular gradients multiply to −1
For non-vertical, non-horizontal lines with gradients m₁ and m₂, perpendicularity means m₁m₂ = −1. So the perpendicular gradient is −1/m.
A horizontal line is perpendicular to a vertical line; these special cases do not have two finite reciprocal gradients.
A perpendicular bisector combines two skills
Find the segment midpoint, then use the negative reciprocal of the segment gradient. The bisector must pass through that midpoint.
C4.6 is Extended content only in the Core list. In the final equation, check both the midpoint and the right-angle gradient relationship.
Pause and explain
What condition do two non-vertical, non-horizontal perpendicular gradients satisfy?
Worked example
Find the gradient of a line perpendicular to 2y = 3x + 1.
Show the worked solution
- Rearrange to y = (3/2)x + 1/2, so its gradient is 3/2.
- The perpendicular gradient is the negative reciprocal −2/3.
- Their product (3/2)(−2/3) = −1.
Answer Perpendicular gradient −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 of a line perpendicular to y = 2x + 1 through (2, 3).
Give me a hint
The original gradient is 2; use its negative reciprocal and substitute the point.
Compare my reasoning
- The perpendicular gradient is −1/2.
- Use y = −(1/2)x + c.
- At (2, 3), c = 4, so y = −(1/2)x + 4.
y = −(1/2)x + 4.
Look for these in your work
- Negative reciprocal used
- Point substituted
- Equation verified
Find the equation of the perpendicular bisector of A(−3, 8) and B(9, −2).
Give me a hint
Find the midpoint and segment gradient; then use the perpendicular gradient through that midpoint.
Compare my reasoning
- Midpoint is (3, 3).
- Segment gradient = (−2 − 8)/(9 − (−3)) = −10/12 = −5/6.
- Perpendicular gradient = 6/5; y − 3 = (6/5)(x − 3).
y − 3 = (6/5)(x − 3).
Look for these in your work
- Midpoint correct
- Original gradient correct
- Bisector uses reciprocal through midpoint
Two sensors are located at (−3, 8) and (9, −2). A cable route must pass through the exact halfway point and meet their joining line at 90°. Give its equation and one check.
Give me a hint
The required route is the perpendicular bisector; use the midpoint and perpendicular gradient.
Compare my reasoning
- The midpoint is (3, 3).
- The joining line has gradient −5/6, so the route gradient is 6/5.
- The route is y − 3 = (6/5)(x − 3); its gradient product with −5/6 is −1.
y − 3 = (6/5)(x − 3); it passes through the midpoint and is perpendicular.
Look for these in your work
- Halfway point identified
- Perpendicular direction found
- Both defining properties checked
Common mistakes
- Changing only the sign instead of taking the reciprocal
- Using the original gradient for a perpendicular line
- Forgetting the midpoint in a perpendicular bisector
- Applying the product rule to horizontal/vertical special cases without noting them
What can you explain now?
What is perpendicular to a line with gradient 4?
Compare with the explanation
−1/4
Take the negative reciprocal.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How does a perpendicular bisector use both a midpoint and a negative reciprocal gradient?
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.