A-Level Mathematics: Midpoint Coordinates & Perpendicular Bisector Equations | A-Level 数学：中点坐标与垂直平分线方程

📚 A-Level Mathematics: Midpoint Coordinates & Perpendicular Bisector Equations | A-Level 数学：中点坐标与垂直平分线方程

Coordinate geometry is a core pillar of A-Level Mathematics, and two of the most frequently tested skills are finding the midpoint of a line segment and writing the equation of its perpendicular bisector. These tools appear in straight-line questions, circle geometry, and vector problems, so mastering them saves time and earns reliable marks.

坐标几何是 A-Level 数学的核心板块，而线段中点坐标的求解与垂直平分线方程的书写，是其中两项最高频的考点。它们出现在直线方程、圆的几何以及向量问题中，掌握好这两项技能，既能节省时间，也能稳定得分。

1. The Midpoint Formula | 中点公式

Given two points A(x₁, y₁) and B(x₂, y₂), the midpoint M of the segment AB is the point halfway between them. Its coordinates are the averages of the x-coordinates and the y-coordinates:

已知两点 A(x₁, y₁) 和 B(x₂, y₂)，线段 AB 的中点 M 是位于它们正中间的点。中点坐标等于两个 x 坐标和两个 y 坐标分别取平均：

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

For example, if A = (2, 3) and B = (6, 7), then M = ((2 + 6)/2, (3 + 7)/2) = (4, 5). Notice that the midpoint always lies on the line segment AB, regardless of the quadrant or the order of the points.

例如，若 A = (2, 3)，B = (6, 7)，则 M = ((2 + 6)/2, (3 + 7)/2) = (4, 5)。注意，中点始终位于线段 AB 上，无论两点在哪个象限，也无论两点的先后顺序如何。

2. Deriving the Midpoint Formula | 中点公式的推导

Imagine moving from A to B horizontally. The total horizontal distance is x₂ − x₁. Halfway across this distance gives x₁ + (x₂ − x₁)/2 = (x₁ + x₂)/2. The same logic applies vertically, producing (y₁ + y₂)/2.

可以这样理解：从 A 水平移动到 B，总水平距离为 x₂ − x₁。走完这段距离的一半，得到 x₁ + (x₂ − x₁)/2 = (x₁ + x₂)/2。垂直方向同理，得到 (y₁ + y₂)/2。

This derivation works for negative coordinates as well. If A = (−4, 1) and B = (2, −5), then M = ((−4 + 2)/2, (1 + (−5))/2) = (−1, −2). The midpoint formula does not require any diagram, but sketching always helps to check the answer.

该推导对负坐标同样成立。若 A = (−4, 1)，B = (2, −5)，则 M = ((−4 + 2)/2, (1 + (−5))/2) = (−1, −2)。中点公式不依赖图形，但画草图总有助于检验答案。

3. Worked Examples for the Midpoint | 中点坐标计算例题

Let us test the formula with a typical exam-style question. Find the midpoint of the line segment joining P(−3, 8) and Q(5, −2).

我们用一道典型考题来测试公式。求连接 P(−3, 8) 和 Q(5, −2) 的线段的中点。

Using the formula:

套用公式：

M = ((−3 + 5)/2, (8 + (−2))/2) = (2/2, 6/2) = (1, 3)

So the midpoint is (1, 3). A quick check: the x-coordinate 1 is exactly halfway between −3 and 5, and the y-coordinate 3 is exactly halfway between 8 and −2.

因此中点为 (1, 3)。快速检验：x 坐标 1 恰好是 −3 和 5 的中点，y 坐标 3 恰好是 8 和 −2 的中点。

4. Gradient and the Perpendicular Condition | 斜率与垂直条件

To write the equation of a perpendicular bisector, we first need the gradient of the original segment. The gradient m of the line through A(x₁, y₁) and B(x₂, y₂) is:

要写出垂直平分线的方程，首先需要求出原线段的斜率。经过 A(x₁, y₁) 和 B(x₂, y₂) 的直线的斜率 m 为：

m = (y₂ − y₁)/(x₂ − x₁)

Two lines are perpendicular if the product of their gradients is −1. Therefore, if the original segment has gradient m, the perpendicular gradient is:

两条直线垂直的条件是它们的斜率乘积为 −1。因此，若原线段斜率为 m，则垂直斜率为：

m⊥ = −1/m = −(x₂ − x₁)/(y₂ − y₁)

For example, if m = 3/4, then m⊥ = −4/3. If m = −2, then m⊥ = 1/2. Remember to invert the fraction and change the sign.

例如，若 m = 3/4，则 m⊥ = −4/3；若 m = −2，则 m⊥ = 1/2。记住要取倒数并改变符号。

5. What Is a Perpendicular Bisector? | 什么是垂直平分线？

A perpendicular bisector of a segment AB is a straight line that meets AB at its midpoint and is perpendicular to AB. Every point on this line is equidistant from A and B. This property is the geometric definition that connects bisector and perpendicular concepts.

线段 AB 的垂直平分线是一条经过 AB 中点且与 AB 垂直的直线。该直线上的每一点到 A 与到 B 的距离都相等。这一性质正是“垂直”与“平分”两个概念相结合的几何定义。

In coordinate geometry, we find the perpendicular bisector by combining two pieces of information: the midpoint, which gives a point on the line, and the negative reciprocal of the original gradient, which gives the slope of the line.

在坐标几何中，求垂直平分线需要结合两个信息：中点提供了直线上的一个点，而原斜率的负倒数提供了该直线的斜率。

6. Step-by-Step Method | 分步求解方法

To find the equation of the perpendicular bisector of the segment joining A(x₁, y₁) and B(x₂, y₂), follow these steps:

要求连接 A(x₁, y₁) 和 B(x₂, y₂) 的线段的垂直平分线方程，请按以下步骤进行：

Find the midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2).
Find the gradient of AB: m = (y₂ − y₁)/(x₂ − x₁).
Write the perpendicular gradient: m⊥ = −1/m.
Substitute M and m⊥ into y − y₁ = m(x − x₁), using the midpoint as the known point.
Rearrange into the required form, usually ax + by + c = 0 or y = mx + c.

求出中点 M = ((x₁ + x₂)/2, (y₁ + y₂)/2)。
求出 AB 的斜率：m = (y₂ − y₁)/(x₂ − x₁)。
写出垂直斜率：m⊥ = −1/m。
将 M 和 m⊥ 代入 y − y₁ = m(x − x₁)，以中点作为已知点。
整理为所需形式，通常是 ax + by + c = 0 或 y = mx + c。

The straight-line equation y − y₁ = m(x − x₁) is the most reliable form to use because it avoids confusing the midpoint with the original endpoints.

直线方程 y − y₁ = m(x − x₁) 是最可靠的形式，因为它可以避免将中点与原端点混淆。

7. Worked Example: Perpendicular Bisector | 垂直平分线完整例题

Find the equation of the perpendicular bisector of the segment joining A(2, 5) and B(8, 1).

求连接 A(2, 5) 和 B(8, 1) 的线段的垂直平分线方程。

Step 1: Midpoint:

第一步：中点：

M = ((2 + 8)/2, (5 + 1)/2) = (5, 3)

Step 2: Gradient of AB:

第二步：AB 的斜率：

m = (1 − 5)/(8 − 2) = −4/6 = −2/3

Step 3: Perpendicular gradient:

第三步：垂直斜率：

m⊥ = 3/2

Step 4: Substitute into y − y₁ = m(x − x₁):

第四步：代入 y − y₁ = m(x − x₁)：

y − 3 = (3/2)(x − 5)

Step 5: Rearrange:

第五步：整理：

2y − 6 = 3x − 15 → 3x − 2y − 9 = 0

Thus the perpendicular bisector is 3x − 2y − 9 = 0. Check that (5, 3) satisfies the equation: 15 − 6 − 9 = 0.

因此垂直平分线为 3x − 2y − 9 = 0。检验 (5, 3) 满足方程：15 − 6 − 9 = 0。

8. Handling Fractions and Surds | 分数与根号的处理

Exam questions often involve fractional coordinates or gradients. Suppose A(1, 2) and B(4, 6). The midpoint is (2.5, 4), and the gradient of AB is (6 − 2)/(4 − 1) = 4/3. The perpendicular gradient is −3/4, so the bisector is:

考题中经常出现分数坐标或分数斜率。设 A(1, 2)，B(4, 6)。中点为 (2.5, 4)，AB 的斜率为 (6 − 2)/(4 − 1) = 4/3。垂直斜率为 −3/4，因此垂直平分线为：

y − 4 = (−3/4)(x − 2.5)

Multiplying through by 4 gives 4y − 16 = −3x + 7.5, so 3x + 4y − 23.5 = 0. To avoid decimals, multiply by 2: 6x + 8y − 47 = 0.

两边乘以 4 得 4y − 16 = −3x + 7.5，即 3x + 4y − 23.5 = 0。为了避免小数，两边再乘以 2：6x + 8y − 47 = 0。

If the original gradient is an integer such as 5, the perpendicular gradient is −1/5. If the gradient is a fraction such as 2/7, the perpendicular gradient is −7/2. If coordinates involve surds, treat them as exact values and do not round until the final answer.

若原斜率为整数如 5，则垂直斜率为 −1/5；若原斜率为分数如 2/7，则垂直斜率为 −7/2。若坐标涉及根号，应保留精确值，直到最后答案才考虑近似。

9. Special Cases: Horizontal and Vertical Segments | 特殊情况：水平与垂直线段

If A and B have the same y-coordinate, then AB is horizontal. Its gradient is 0, and the perpendicular bisector is a vertical line. For example, if A(2, 3) and B(8, 3), the midpoint is (5, 3), and the perpendicular bisector is simply x = 5.

若 A 与 B 的 y 坐标相同，则 AB 为水平线段。其斜率为 0，垂直平分线为竖直线。例如，若 A(2, 3)，B(8, 3)，则中点为 (5, 3)，垂直平分线就是 x = 5。

If A and B have the same x-coordinate, then AB is vertical. The gradient is undefined, and the perpendicular bisector is a horizontal line. For example, if A(4, 1) and B(4, 9), the midpoint is (4, 5), and the perpendicular bisector is y = 5.

若 A 与 B 的 x 坐标相同，则 AB 为竖直线段。其斜率不存在，垂直平分线为水平线。例如，若 A(4, 1)，B(4, 9)，则中点为 (4, 5)，垂直平分线就是 y = 5。

In these cases, do not try to calculate −1/m. Instead, recognise the horizontal or vertical nature immediately and state the constant line equation.

遇到这些情况，不要强行计算 −1/m，而应立即识别水平或垂直性质，直接写出常值直线方程。

10. Connecting to Circle Geometry | 与圆几何的联系

The perpendicular bisector of any chord of a circle passes through the centre of the circle. This is a powerful theorem used in A-Level questions to find the centre when two chords are given.

圆的任意弦的垂直平分线都经过圆心。这是 A-Level 中非常有用的定理，当已知两条弦时，可以用它来求圆心。

For instance, given two endpoints of a diameter, the midpoint is the centre of the circle. Given a chord and its perpendicular bisector, the centre must lie on that bisector. Combining two such bisectors locates the centre exactly.

例如，已知直径的两个端点，其中点就是圆心。已知一条弦及其垂直平分线，圆心必在该平分线上。两条这样的平分线相交即可精确确定圆心位置。

This connection means that mastering midpoint and perpendicular bisector equations directly supports questions on circle equations, which are common in both pure mathematics and coordinate geometry papers.

这一联系意味着，掌握中点与垂直平分线方程可以直接支持圆的方程类题目，而这些题目在纯数学和坐标几何试卷中都很常见。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

The most common mistake is forgetting to negate the reciprocal of the gradient. Another is using the original point instead of the midpoint when substituting into the line equation. Always label your midpoint clearly.

最常见的错误是忘记取斜率的负倒数。另一个常见错误是在代入直线方程时使用原端点而非中点。始终清晰标注中点。

Also, when the gradient is negative, the perpendicular gradient is positive, and vice versa. A simple check: multiplying the two gradients together should give −1. If it does not, you have made an error.

另外，当斜率为负时，垂直斜率为正，反之亦然。一个简单的检验方法：两个斜率相乘应等于 −1。若不等于 −1，说明出现了错误。

In exams, show every step. A mark is often awarded for the midpoint, another for the gradient, another for the perpendicular gradient, and the final mark for the equation in the required form. Do not skip the rearrangement step.

考试中要写出每一步。通常中点、斜率、垂直斜率各占一个得分点，最后一步方程整理为所需形式再得一分。不要跳过整理步骤。

12. Practice Questions and Conclusion | 练习题与总结

Try these two questions to consolidate your understanding:

请尝试以下两道题来巩固理解：

Find the perpendicular bisector of the segment joining (3, −1) and (7, 5).
The midpoint of AB is (2, 4) and A = (−1, 6). Find B, and then find the equation of the perpendicular bisector of AB.

求连接 (3, −1) 和 (7, 5) 的线段的垂直平分线。
已知 AB 的中点为 (2, 4)，且 A = (−1, 6)。求 B 的坐标，并求 AB 的垂直平分线方程。

For the first question, the midpoint is (5, 2), the gradient of the segment is (5 − (−1))/(7 − 3) = 6/4 = 3/2, so the perpendicular gradient is −2/3. The equation is y − 2 = (−2/3)(x − 5), which simplifies to 2x + 3y − 16 = 0.

第一题中，中点为 (5, 2)，线段斜率为 (5 − (−1))/(7 − 3) = 6/4 = 3/2，因此垂直斜率为 −2/3。方程为 y − 2 = (−2/3)(x − 5)，整理得到 2x + 3y − 16 = 0。

For the second question, use (A + B)/2 = (2, 4) to find B = (5, 2). The gradient of AB is (2 − 6)/(5 − (−1)) = −4/6 = −2/3, so the perpendicular gradient is 3/2. Using the midpoint (2, 4), the equation is y − 4 = (3/2)(x − 2), which gives 3x − 2y + 2 = 0.

第二题中，利用 (A + B)/2 = (2, 4) 可得 B = (5, 2)。AB 的斜率为 (2 − 6)/(5 − (−1)) = −4/6 = −2/3，因此垂直斜率为 3/2。以中点 (2, 4) 代入，方程为 y − 4 = (3/2)(x − 2)，整理得到 3x − 2y + 2 = 0。

Midpoint coordinates and perpendicular bisector equations are straightforward once you organise your work. Always find the midpoint first, then the gradient, then the negative reciprocal, and finally write the equation in the form requested by the question. With regular practice, these questions become quick marks in the exam.

只要步骤清晰，中点坐标与垂直平分线方程并不困难。始终先求中点，再求斜率，然后取负倒数，最后按题目要求写出方程形式。通过有规律的练习，这类题目在考试中将成为快速拿分点。

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A-Level Mathematics: Midpoint Coordinates & Perpendicular Bisector Equations | A-Level 数学：中点坐标与垂直平分线方程