The distance between two points (x₁, y₁) and (x₂, y₂) is √((x₂ − x₁)² + (y₂ − y₁)²), which is Pythagoras on a coordinate grid. The midpoint is the average of the coordinates, ((x₁ + x₂)/2, (y₁ + y₂)/2). Together they let you prove a triangle is right-angled, find missing coordinates, and get the centre and radius of a circle from a diameter.
Distance and midpoint formulae
- Distance: AB = √((x₂ − x₁)² + (y₂ − y₁)²)
- Midpoint: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
- Working backwards from a midpoint: x_B = 2x_M − x_A and y_B = 2y_M − y_A
Leave distances as simplified surds unless told otherwise, for example √45 = 3√5.
Where does the distance formula come from?
Join the two points and draw a right-angled triangle underneath. Its horizontal side is x₂ − x₁, its vertical side is y₂ − y₁, and the distance between the points is the hypotenuse. Pythagoras gives the formula directly.
Because both differences are squared, the order of the points does not matter, and a negative difference squares to the same value.
The midpoint, forwards and backwards
The midpoint is simply the average of the two x-coordinates and the average of the two y-coordinates. It also works backwards: if you know one end, A, and the midpoint, M, the other end is as far beyond M as A is before it.
A fact you will use often: the diagonals of a parallelogram bisect each other, so both diagonals share the same midpoint. That is the quickest way to find a missing fourth vertex.
How do you prove a triangle is right-angled?
Find the squares of all three side lengths and use the converse of Pythagoras: if the two smaller squares add up to the largest, the angle opposite the longest side is 90°.
You never need the square roots for this test. Working with AB², BC² and AC² is faster and avoids surds. The same comparisons prove a triangle is isosceles (two sides equal) or a quadrilateral is a rhombus (all sides equal).
Finding unknown coordinates
Turn the condition in the question into an equation. “The distance is d” means squaring both sides, which usually gives a quadratic and often two answers. “Equidistant from A and B” means setting PA² = PB²; the x² and y² terms cancel, leaving a linear equation.
If a point lies on the x-axis its y-coordinate is 0, and on the y-axis its x-coordinate is 0. That removes one unknown before you start.
Using both formulae together: circles
If AB is a diameter of a circle, the centre is the midpoint of AB and the radius is half the length of AB. That is all you need to write down the circle's equation, which is why these two formulae lead straight into the circle courses.
Areas of triangles are easiest when one side is horizontal or vertical: use it as the base, and the height is the difference in the other coordinate.
Worked examples
Example 1. Find the length AB, where A is (1, 2) and B is (7, 10).
- Differences: 7 − 1 = 6 and 10 − 2 = 8.
- AB = √(6² + 8²) = √(36 + 64) = √100.
- AB = 10.
Answer: AB = 10.
Marks: M1 for the formula, A1 for the value.
Example 2. Show that the triangle with vertices A(1, 2), B(7, 5) and C(0, 4) is right-angled, and state where the right angle is.
- AB² = (7 − 1)² + (5 − 2)² = 36 + 9 = 45.
- AC² = (0 − 1)² + (4 − 2)² = 1 + 4 = 5.
- BC² = (0 − 7)² + (4 − 5)² = 49 + 1 = 50.
- BC² is largest, and AB² + AC² = 45 + 5 = 50 = BC².
- By the converse of Pythagoras, the right angle is opposite BC, at A.
Answer: Right-angled at A.
Marks: M1 for all three squares, M1 for testing Pythagoras, A1 for stating where.
Example 3. The point P lies on the x-axis and is equidistant from A(1, 3) and B(5, 1). Find P.
- P is on the x-axis, so P = (x, 0).
- PA² = (x − 1)² + 9 and PB² = (x − 5)² + 1.
- Set them equal and expand: x² − 2x + 10 = x² − 10x + 26.
- The x² terms cancel: 8x = 16, so x = 2.
- Check: PA² = 1 + 9 = 10 and PB² = 9 + 1 = 10.
Answer: P(2, 0).
Marks: M1 for using the axis, M1 for PA² = PB², A1 for solving.
Example 4. A(1, −2) and B(7, 6) are the ends of a diameter of a circle. Find the centre and the radius.
- Centre = midpoint of AB = ((1 + 7)/2, (−2 + 6)/2) = (4, 2).
- AB = √(6² + 8²) = √100 = 10.
- Radius = 10 / 2 = 5. Check: the distance from (4, 2) to A is √(9 + 16) = 5.
Answer: Centre (4, 2), radius 5.
Marks: M1 for the midpoint, M1 for the length, A1 for halving it.
Common mistakes
- Adding instead of subtracting. The distance formula uses differences, x₂ − x₁, not sums. The midpoint formula uses sums.
- Forgetting to square root. (x₂ − x₁)² + (y₂ − y₁)² is the distance squared.
- Not simplifying surds. Write √45 as 3√5 when an exact answer is asked for.
- Square-rooting before testing for a right angle. Compare the squared lengths directly; it is quicker and avoids rounding.
- Missing the second answer. A condition like “the distance is 5” usually gives a quadratic with two possible coordinates.
How distance and midpoint questions are examined
In Edexcel International A-Level these formulae sit in Pure 1 (paper WMA11), in the chapter on straight line graphs, and they come back in Pure 2 circle questions. In Edexcel A-Level (9MA0) they are first-year Pure content. Typical tasks:
- find a length as an exact surd, and a midpoint
- show that a triangle is right-angled or isosceles
- find an unknown coordinate from a distance or an equidistance condition
- find the centre and radius of a circle from the ends of a diameter
The Pure 1 straight line graphs exam questions include these skills with mark schemes. The straight line graphs worksheet and the A-Level coordinate geometry revision engine give more practice.
Next: straight lines, then the equation of a circle.
Frequently asked questions
Is the distance formula the same as Pythagoras?
Yes. It is Pythagoras applied to the right-angled triangle formed by the horizontal and vertical differences between the two points.
Does the order of the points matter?
No. The differences are squared, so (x₂ − x₁)² gives the same value as (x₁ − x₂)².
How do I find the other end of a line segment from its midpoint?
Double the midpoint and subtract the known end: x_B = 2x_M − x_A and y_B = 2y_M − y_A.