Use distance, midpoint and perpendicular gradients together
Connect coordinate measures to parallel lines, perpendicular bisectors and equidistant points.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use algebra and elementary geometry; implicit differentiation is outside this section. Written practice is self-checked.
Before you begin
- Build a line from one point and a gradient.
- Use Pythagoras and simplify a square root.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is the midpoint of (0, 0) and (6, 8)?
What you will learn
- Find exact distances, gradients and midpoints.
- Build parallel lines and perpendicular bisectors, including horizontal and vertical cases.
A midpoint averages; a distance uses differences
The midpoint of A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). The distance is √((x₂ − x₁)² + (y₂ − y₁)²), found from a right triangle of coordinate differences. Keep an exact surd when exact form is requested. Squaring removes the signs of the two differences, but not their magnitudes.
For A(−1, 2) and B(5, 6), the midpoint is (2, 4), the gradient is 4/6 = 2/3 and the distance is √52 = 2√13. These are three different quantities: do not use half the coordinate differences as the midpoint or divide distance by two to find an average coordinate.
Perpendicular and parallel have different conditions
Parallel non-vertical lines have equal gradients. Perpendicular non-horizontal, non-vertical lines have gradients m₁m₂ = −1; the required gradient is the negative reciprocal. The perpendicular bisector passes through the midpoint and is perpendicular to the segment, so it is the locus of points equidistant from its two ends.
For the segment above, use gradient −3/2 through (2, 4): y − 4 = −3(x − 2)/2, or 3x + 2y = 14. A horizontal segment has a vertical perpendicular bisector; a vertical segment has a horizontal one. Handle these geometrically instead of trying to take a reciprocal of zero or an undefined gradient.
Pause and explain
Which finite gradient is perpendicular to 2/3?
Worked example
For A(−1, 2) and B(5, 6), find AB, its midpoint and its perpendicular bisector.
Show the worked solution
- The differences are 6 and 4, so AB = √(36 + 16) = 2√13 and the midpoint is (2, 4).
- The segment gradient is 2/3, hence the perpendicular gradient is −3/2.
- Use the midpoint: y − 4 = −3(x − 2)/2, simplifying to 3x + 2y = 14.
Answer AB = 2√13; midpoint (2, 4); perpendicular bisector 3x + 2y = 14.
Try a topic test
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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 distance and midpoint between (0, 0) and (6, 8).
Give me a hint
Use a right triangle for distance and coordinate averages for midpoint.
Compare my reasoning
- The horizontal and vertical differences are 6 and 8.
- The distance is √(36 + 64) = 10.
- The midpoint averages the coordinates to give (3, 4).
Distance 10; midpoint (3, 4).
Look for these in your work
- Use the square root of the sum of squared differences.
- Compute both midpoint coordinates independently.
Find the locus of points equidistant from A(2, 1) and B(6, 5).
Give me a hint
The locus is the perpendicular bisector, not the segment itself.
Compare my reasoning
- The midpoint is (4, 3), and AB has gradient 1.
- The perpendicular bisector has gradient −1 through (4, 3).
- Use y − 3 = −(x − 4), giving y = −x + 7.
y = −x + 7.
Look for these in your work
- Identify the correct locus and its midpoint.
- Use a perpendicular gradient to build its equation.
For A(−3, 4) and B(5, 4), find the perpendicular bisector and explain why y = mx + c is unsuitable for it.
Give me a hint
The original segment is horizontal, so its bisector is vertical.
Compare my reasoning
- The midpoint is (1, 4).
- A line perpendicular to this horizontal segment is vertical.
- The vertical line through the midpoint is x = 1; it has no finite gradient and cannot be written as y = mx + c.
x = 1; the perpendicular bisector is vertical.
Look for these in your work
- Use the midpoint to position the vertical line.
- Explain the undefined-gradient exception.
Common mistakes
- Average coordinates for a midpoint; subtract coordinates for a displacement.
- Perpendicular gradients are negative reciprocals, not merely negatives.
- Treat horizontal and vertical cases separately.
What can you explain now?
Give equations of the parallel and perpendicular lines through (2, 1) to y = 3x − 4.
Compare with the explanation
Parallel: y = 3x − 5. Perpendicular: y = −x/3 + 5/3.
Use gradients 3 and −1/3 respectively in the point-gradient form through (2, 1).
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Make it stick
Reconstruct a perpendicular bisector from two points. Explain why averaging, taking differences and taking a negative reciprocal serve different purposes.
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.