# Distance between two points

> Edexcel International GCSE Further Pure Mathematics · 4PM1 Higher
> 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s8-distance-between-two-points/

This guide covers calculating the distance between two points in rectangular Cartesian coordinates, aligned to Edexcel IGCSE Further Pure Math 4PM1 specification 8A, including formula recall, worked examples and exam pitfalls.

**先修:** [Simplifying surds](https://www.owlsprep.com/zh/study/edexcel-igcse-maths-simplifying-surds/); [Cartesian coordinate system basics](https://www.owlsprep.com/zh/study/edexcel-igcse-maths-cartesian-coordinates-basics/); [Pythagoras' theorem](https://www.owlsprep.com/zh/study/edexcel-igcse-maths-pythagoras-theorem/)

## 学习目标

- Recall the distance formula for two Cartesian points from memory
- Apply the formula to calculate exact distances between points with integer, negative and fractional coordinates
- Avoid common exam errors when simplifying surd distance answers
- Use distance calculations to solve basic straight line geometry problems

## Deriving & Recalling the Distance Formula

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.

**Distance between two points formula** — 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.

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

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

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

> **tip**
>
> The order of $x_1$ and $x_2$, or $y_1$ and $y_2$, does not matter because the values are squared: $(a - b)^2 = (b - a)^2$.

> **考试提示:** Always write the distance formula first in your exam response to earn method marks even if you make an arithmetic error.

## Calculating Exact Distances Between Two Points

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.

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

**概念自测**

1. What is the distance between (0,0) and (3,4)?

   - 5
   - 7
   - 25
   - \sqrt{7}

   *解析:* Correct: $d = \sqrt{3² + 4²} = \sqrt{25} = 5$.

2. What is the exact distance between (1, 2) and (5, 5)?

   - 7
   - 5
   - \sqrt{7}
   - 25

   *解析:* Correct: $d = \sqrt{(5-1}² + (5-2)²) = \sqrt{16 +9} = \sqrt{25} =5$.

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

*计算器:* allowed

## Using Distance to Solve Straight Line Geometry Problems

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.

**例题:** 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. 2. 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. 3. Therefore length AD = 5 units.

**考试命令词**

- **Find the exact distance** — Calculate the distance and leave your answer as a simplified surd, not a decimal. *(Find exact distance between (2, 3) and (5, 7): answer is 5, not 5.0.)*

- **Show that the distance is X** — Use the distance formula to prove the given value, showing every step of your calculation.

*计算器:* allowed

## 常见错误

- **错误做法:** Forgetting to square the differences between x and y coordinates.
  - 原因: This arithmetic error leads to an incorrect, too-small distance value, losing accuracy marks.
  - 正确做法: Write the formula first, then substitute values and explicitly show the squaring step in your working.
- **错误做法:** Using the negative square root as your final distance answer.
  - 原因: Distance is a scalar quantity that is always positive, so negative values are invalid.
  - 正确做法: Always take the positive square root when calculating d from d².
- **错误做法:** Leaving surd answers unsimplified, e.g. writing \sqrt{18} instead of 3\sqrt{2}.
  - 原因: Exam questions requiring exact answers expect surds to be in their simplest form for full marks.
  - 正确做法: Factor the number under the square root to pull out any perfect square factors before writing your final answer.
- **错误做法:** Mixing up x and y coordinates when substituting into the formula, e.g. using (x1 - y1) instead of (x1 -x2).
  - 原因: This leads to a completely incorrect calculation, losing both method and accuracy marks.
  - 正确做法: Label your coordinates x1, y1 and x2, y2 clearly before substituting into the formula, and double check your substitutions.

## 速查表

| Formula | Notes | Exam Reminder |
| --- | --- | --- |
| $d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ | 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 |

## 下一步

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.

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