Distance between two points

Edexcel International GCSE Further Pure Mathematics· Specification section 8A· 15 min read

1. Deriving & Recalling the Distance Formula★☆☆☆☆⏱ 5 min

The distance between two points on a Cartesian plane is the length of the straight line segment connecting them. This formula is derived directly from Pythagoras’ theorem, where the horizontal and vertical differences between points form the two shorter sides of a right-angled triangle.

📘 Definition

Distance between two points formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

Where $d$ is the distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . You may also use the squared form $d^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}$ to avoid surds in intermediate steps.

Example:

For points (1,2) and (4,6): $d^{2} = (4 - 1)^{2} + (6 - 2)^{2} = 9 + 16 = 25$ , so $d = 5$ .

📐 Worked Example

Derive the distance formula using points A $(x_{1}, y_{1})$ and B $(x_{2}, y_{2})$ .

1
1. Calculate the horizontal difference between the points: $Δ x = x_{1} - x_{2}$
2
1. Calculate the vertical difference between the points: $Δ y = y_{1} - y_{2}$
3
1. The straight line AB forms the hypotenuse of a right triangle with legs $Δ x$ and $Δ y$ . By Pythagoras' theorem:
4
$AB^2 = (\Delta x)^2 + (\Delta y)^2$
5
1. Take the positive square root (distance is always positive) to get the final formula:
6
$AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Exam tip:

Always write the distance formula first in your exam response to earn method marks even if you make an arithmetic error.

2. Calculating Exact Distances Between Two Points★★☆☆☆⏱ 6 min

✓ Calculator OK

You will be expected to calculate distances between points with integer, negative and fractional coordinates, and leave answers as simplified surds unless told otherwise. You can use a calculator to check your arithmetic, but show all working for full marks.

📐 Worked Example

Calculate the exact distance between the points P $(- 2, 5)$ and Q $(3, - 1)$ . Give your answer in simplest surd form.

1
1. Label the coordinates: let $x_{1} = - 2, y_{1} = 5$ and $x_{2} = 3, y_{2} = - 1$ .
2
1. Substitute into the distance formula:
3
$d = \sqrt{(-2 - 3)^2 + (5 - (-1))^2}$
4
1. Simplify the terms inside the square root:
5
$d = \sqrt{(-5)^2 + (6)^2} = \sqrt{25 + 36} = \sqrt{61}$
6
1. Check if the surd simplifies: 61 is a prime number, so $61$ is the simplest exact form.

✓ Quick check

What is the distance between (0,0) and (3,4)?
- 5
- 7
- 25
- \sqrt{7}
Reveal answer
5 —
Correct: $d = 3^{2} + 4^{2} = 25 = 5$ .
What is the exact distance between (1, 2) and (5, 5)?
- 7
- 5
- \sqrt{7}
- 25
Reveal answer
5 —
Correct: $d = (5 - 1^{2} + (5 - 2)^{2}) = 16 + 9 = 25 = 5$ .

Exam tip:

If the question asks for an exact distance, do not give a decimal approximation: you will lose marks unless decimal form is explicitly requested.

3. Using Distance to Solve Straight Line Geometry Problems★★★☆☆⏱ 7 min

✓ Calculator OK

Distance calculations are often combined with other straight line coordinate geometry concepts in exam questions, for example to find the length of a side of a triangle or quadrilateral on a coordinate grid.

📐 Worked Example

Three vertices of a rhombus are A $(1, 1)$ , B $(4, 5)$ , and C $(9, 5)$ . Find the length of side AD.

1
1. Recall all sides of a rhombus are equal length, so length AD = length AB.
2
1. Calculate distance AB using coordinates of A and B:
3
$AB = \sqrt{(4 - 1)^2 + (5 -1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} =5$
4
1. Therefore length AD = 5 units.

4. Common Pitfalls

Wrong move:

Forgetting to square the differences between x and y coordinates.

Why:

This arithmetic error leads to an incorrect, too-small distance value, losing accuracy marks.

Correct move:

Write the formula first, then substitute values and explicitly show the squaring step in your working.

Wrong move:

Using the negative square root as your final distance answer.

Why:

Distance is a scalar quantity that is always positive, so negative values are invalid.

Correct move:

Always take the positive square root when calculating d from d².

Wrong move:

Leaving surd answers unsimplified, e.g. writing \sqrt{18} instead of 3\sqrt{2}.

Why:

Exam questions requiring exact answers expect surds to be in their simplest form for full marks.

Correct move:

Factor the number under the square root to pull out any perfect square factors before writing your final answer.

Wrong move:

Mixing up x and y coordinates when substituting into the formula, e.g. using (x1 - y1) instead of (x1 -x2).

Why:

This leads to a completely incorrect calculation, losing both method and accuracy marks.

Correct move:

Label your coordinates x1, y1 and x2, y2 clearly before substituting into the formula, and double check your substitutions.

5. Quick Reference Cheatsheet

Formula	Notes	Exam Reminder
$d = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}$	Order of x1/x2 and y1/y2 does not matter	Recall formula, not given on formula sheet
$d^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}$	Use squared form for intermediate steps to avoid surds early	Give final distance as positive value
Simplified surd answer	Factor out perfect squares from under root	Exact answers required unless decimal is requested

6. Frequently Asked

Is the distance formula provided on the Edexcel 4PM1 formula sheet?

No. All coordinate geometry formulae, including the distance between two points, must be recalled for your exam.

Do I need to simplify distance answers for the exam?

Yes. If your distance is a surd, you must give it in its simplest exact form unless the question explicitly asks for a decimal approximation.

Going deeper

formula_sheetEdexcel 4PM1 Official Formula SheetLists all formulae provided in 4PM1 exams; distance formula is not included

What's Next

Now that you can calculate the distance between two Cartesian points, you are ready to progress to the next topics in the Edexcel 4PM1 rectangular Cartesian coordinates unit. Next, you will learn to find the coordinates of a point that divides a line segment in a given ratio, followed by calculating the gradient of a straight line between two points. These foundational skills build up to writing equations of straight lines, and solving problems involving parallel and perpendicular lines, which are all frequent questions in the 4PM1 exam. Mastering the distance formula first will make these subsequent topics much easier to grasp, as they all rely on manipulating differences between coordinates of two points on a plane.