Find distance and midpoint from coordinates
Calculate segment lengths with horizontal and vertical changes and locate midpoints by averaging coordinates.
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 operation combines horizontal and vertical changes into a distance?
What you will learn
- Find the length between two coordinate points
- Find a segment's midpoint from its endpoints
- Keep exact roots or appropriate units in context
Distance combines perpendicular changes
Between (x₁, y₁) and (x₂, y₂), the horizontal change is x₂ − x₁ and the vertical change is y₂ − y₁. Pythagoras gives length √((x₂ − x₁)² + (y₂ − y₁)²).
Square both signed differences, add, then take the positive root for distance. A horizontal or vertical segment is a useful check because one difference is zero.
A midpoint averages each coordinate
The midpoint is ((x₁ + x₂)/2, (y₁ + y₂)/2). Average x-values together and y-values together; do not average an x-coordinate with a y-coordinate.
Keep a surd exact unless a question requests rounding. Include the unit of length when coordinates describe a physical map or plan.
Pause and explain
What is the midpoint's x-coordinate for endpoints x = −4 and x = 6?
Worked example
Find the distance and midpoint between A(−2, 1) and B(4, 10).
Show the worked solution
- The changes are 6 horizontally and 9 vertically.
- Distance = √(6² + 9²) = √117 = 3√13.
- Midpoint = ((−2 + 4)/2, (1 + 10)/2) = (1, 11/2).
Answer Distance 3√13 units; midpoint (1, 11/2).
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 distance and midpoint between (1, 2) and (7, 10).
Give me a hint
Calculate the two coordinate changes, then apply Pythagoras and coordinate averages.
Compare my reasoning
- Changes are 6 and 8.
- Distance = √(36 + 64) = 10.
- Midpoint = (4, 6).
Distance 10 units; midpoint (4, 6).
Look for these in your work
- Differences correct
- Distance positive
- Coordinates averaged separately
Find the length and midpoint between (−3, 5) and (9, −2).
Give me a hint
Use the signed changes for the length and average corresponding coordinates.
Compare my reasoning
- Changes are 12 and −7.
- Length = √(12² + (−7)²) = √193.
- Midpoint = (3, 3/2).
Length √193 units; midpoint (3, 3/2).
Look for these in your work
- Signs handled before squaring
- Exact length retained
- Midpoint coordinates paired correctly
A rectangular park map marks two gates at (1, 2) and (7, 10), in equal grid units. Find the straight-line route and its halfway point; explain what changes if the map's grid is not square.
Give me a hint
Use the coordinate distance and midpoint; equal grid units are needed for the direct Pythagoras interpretation.
Compare my reasoning
- Coordinate changes are 6 and 8.
- Route length is √(6² + 8²) = 10 grid units.
- The halfway point is (4, 6); unequal axis scales require converting to common physical units first.
Route 10 grid units; halfway point (4, 6).
Look for these in your work
- Distance calculated
- Midpoint calculated
- Scale assumption identified
Common mistakes
- Using a negative distance
- Forgetting to square one coordinate difference
- Averaging coordinates in the wrong pairs
- Rounding a surd when exact form is requested
What can you explain now?
Find the midpoint of (2, 3) and (8, 7).
Compare with the explanation
(5, 5)
Average x-values and y-values separately.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why does distance use Pythagoras while midpoint averages the two x-values and the two y-values separately?
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.