Distance Between Two Points Using Pythagoras Revision Notes | GCSE AQA Higher Further Maths

1. Why This Matters

Imagine you need to find the exact distance between two locations on a map grid. You can count across and up, but the straight-line diagonal distance? That needs a different approach — and Pythagoras' theorem is the tool.

This technique appears across GCSE papers in coordinate geometry questions, often as part of a longer problem involving midpoints, gradients, or circle equations. Mastering it here pays dividends throughout the course.

2. The Core Idea: Building a Right-Angled Triangle

When you have two points on a coordinate grid, you can always draw a right-angled triangle connecting them. The straight-line distance between the points becomes the hypotenuse of that triangle.

The two shorter sides are the horizontal gap (difference in $x$-values) and the vertical gap (difference in $y$-values). Once you have those two lengths, Pythagoras' theorem gives you the hypotenuse — which is the distance you want.

3. Worked Examples

Theory clicks when you see it in action. Work through the following examples step by step — try each stage yourself before revealing the answer.

Worked example: Finding the Distance Between Two Points

Calculate the distance between $A(1, 2)$ and $B(5, 5)$. Give your answer to 1 decimal place.

1. $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ — State Pythagoras' theorem in coordinate form. This earns a method mark [M1] even if arithmetic goes wrong later.
2. $x_2 - x_1 = 5 - 1 = 4$ $y_2 - y_1 = 5 - 2 = 3$ — Subtract the coordinates carefully. The order matters for squaring — but since you square both values, a negative difference gives the same result.
3. $4^2 = 4 \times 4 = 16$ $3^2 = 3 \times 3 = 9$ — Expand each square longhand to avoid arithmetic slips. [M1] for correct substitution and squaring.
4. $16 + 9 = 25$ — This gives the square of the hypotenuse — you still need to square root.
5. $d = \sqrt{25} = 5$ — The distance is exactly 5 units. [A1] for the correct final answer. This is a 3-4-5 Pythagorean triple, so no rounding needed here.

Answer: $AB = 5$ units

That was a clean example using a Pythagorean triple. Most exam questions will not give such neat numbers, so the next example shows what to do when the answer is a surd or a decimal.

Worked example: Distance Between Two Points (Non-Integer Answer)

Find the distance between $P(-2, 3)$ and $Q(4, -1)$. Give your answer to 2 decimal places.

1. $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ — Always start by writing the formula. It signals to the examiner that you know the method [M1].
2. $x_2 - x_1 = 4 - (-2) = 4 + 2 = 6$ $y_2 - y_1 = -1 - 3 = -4$ — Take care subtracting a negative: $4 - (-2) = 6$. The vertical difference is negative, but squaring removes the sign.
3. $6^2 = 6 \times 6 = 36$ $(-4)^2 = (-4) \times (-4) = 16$ — A negative squared gives a positive result. [M1] for correct squaring of both values.
4. $36 + 16 = 52$ — The sum under the square root is 52.
5. $d = \sqrt{52} \approx 7.2111...$ $d \approx 7.21$ (2 d.p.) — Write the unrounded value first, then round to the precision stated in the question. [A1] for 7.21.

Answer: $PQ \approx 7.21$ units (2 d.p.)

4. Guided Practice

Now it is your turn. These problems follow the same method — write the formula, find the separations, square, add, and square root. Check your answers once you have finished.

5. Exam Technique and Common Mistakes

You now have the method. The final step is learning where marks are lost in exam conditions — and how to make sure you keep every one of them.

Applying Pythagoras' theorem to find the distance between two coordinate points is the single core skill for this topic — write the formula, find the separations, square, add, and square root, showing every step for full marks.