The midpoint of a line segment is the point exactly halfway between its two endpoints, found by averaging the x-coordinates and averaging the y-coordinates. The length of the segment comes from Pythagoras: treat the horizontal and vertical distances as the two shorter sides of a right-angled triangle. Both skills appear together in GCSE coordinate geometry questions.
What is the midpoint of a line segment?
The midpoint M of the segment joining A(x₁, y₁) and B(x₂, y₂) is the average of each coordinate:
M = ((x₁ + x₂)/2 , (y₁ + y₂)/2)
Think of it this way: to travel halfway along the segment you move half the horizontal distance and half the vertical distance. Adding half of each difference to the starting coordinate gives the midpoint.
How do you calculate the midpoint step by step?
Worked example: Find the midpoint of A(2, 5) and B(8, 1).
- Add the x-coordinates: 2 + 8 = 10. Halve: 10 ÷ 2 = 5.
- Add the y-coordinates: 5 + 1 = 6. Halve: 6 ÷ 2 = 3.
- Midpoint M = (5, 3).
Check by sketching: A is at (2, 5), B at (8, 1). M should sit between them — (5, 3) is three units right of A and three units left of B, and two units below A and two units above B. ✓
Second example — with negative coordinates: Find the midpoint of P(−4, 3) and Q(6, −7).
- x: (−4 + 6)/2 = 2/2 = 1.
- y: (3 + (−7))/2 = −4/2 = −2.
- Midpoint = (1, −2).
How do you find the distance between two points?
The distance d between A(x₁, y₁) and B(x₂, y₂) follows from Pythagoras' theorem. The horizontal separation is |x₂ − x₁| and the vertical separation is |y₂ − y₁|, which form the two legs of a right-angled triangle. The hypotenuse is the straight-line distance:
d = √((x₂ − x₁)² + (y₂ − y₁)²)
How do you calculate the distance step by step?
Worked example: Find the exact length of the segment from A(1, 3) to B(7, 11).
- Find the horizontal difference: 7 − 1 = 6.
- Find the vertical difference: 11 − 3 = 8.
- Square both: 6² = 36, 8² = 64.
- Add: 36 + 64 = 100.
- Take the square root: √100 = 10.
The length of AB is 10 units.
Note: you could also recognise 6, 8, 10 as a Pythagorean triple (a multiple of the 3-4-5 triple), which lets you skip the squaring step once you are confident.
How do the two formulas compare?
| Task | Formula | What you calculate |
|---|---|---|
| Midpoint | ((x₁+x₂)/2, (y₁+y₂)/2) | Average of each coordinate |
| Distance | √((x₂−x₁)² + (y₂−y₁)²) | Hypotenuse via Pythagoras |
Both use the differences between the x-coordinates and between the y-coordinates. For the midpoint you add and halve; for the distance you square the differences, add, and root.
How do you find a missing endpoint given the midpoint?
This is a common exam twist: the midpoint and one endpoint are given, and you must find the other endpoint.
Worked example: The midpoint of PQ is M(4, −1). P = (1, 3). Find Q.
Let Q = (a, b).
- Set up the midpoint equation: (1 + a)/2 = 4 and (3 + b)/2 = −1.
- Solve for a: 1 + a = 8, so a = 7.
- Solve for b: 3 + b = −2, so b = −5.
- Q = (7, −5).
- Check: midpoint of (1, 3) and (7, −5) = ((1+7)/2, (3+(−5))/2) = (4, −1). ✓
How do these formulas appear in GCSE questions?
Midpoint and distance calculations rarely stand alone on exam papers. They most commonly appear:
- As part of a proof, e.g. showing that the diagonals of a quadrilateral bisect each other by proving the midpoints coincide.
- Combined with the equation of a line, e.g. finding the midpoint of a chord then using it to find the perpendicular bisector.
- In circle problems, where you must confirm that a given point lies on a circle by checking its distance from the centre equals the radius.
Frequently asked questions
Does it matter which point I call (x₁, y₁) and which I call (x₂, y₂)?
No. Addition is commutative, so (x₁ + x₂)/2 = (x₂ + x₁)/2, giving the same midpoint either way. For the distance formula, each difference is squared before being added, so the sign of the difference does not affect the result: (x₂ − x₁)² = (x₁ − x₂)². Label points whichever way feels natural.
What if the answer involves a surd?
Leave it in surd form unless the question asks for a decimal. For example, the distance between (0, 0) and (3, 4) is √(9 + 16) = √25 = 5. But between (0, 0) and (2, 3) it is √(4 + 9) = √13, which is already in simplest form and should not be converted to 3.606 unless the question specifically requests a decimal or a given number of decimal places.
Can these formulas apply to 3D coordinates?
Yes, with an extra term. In three dimensions with points A(x₁, y₁, z₁) and B(x₂, y₂, z₂), the distance is √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²). This extends Pythagoras into 3D and appears in some GCSE Higher questions on 3D coordinates.
Why does the distance formula use squares and a square root?
Because it is Pythagoras' theorem applied to a coordinate grid. The horizontal separation (x₂ − x₁) and vertical separation (y₂ − y₁) form the two legs of a right-angled triangle; the straight-line distance is the hypotenuse, which Pythagoras gives as √(leg₁² + leg₂²). The coordinate formula is simply Pythagoras restated in algebraic notation.
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