📚 Example 5.2.1: Perpendicular Bisector of a Line Segment | 例题 5.2.1：线段的垂直平分线
In this worked example we use midpoints, gradients and the negative reciprocal rule to find the equation of a perpendicular bisector. We then test whether a given point lies on that line, and we finish with a second method that confirms the answer by equating squared distances. Both skills - producing the equation and testing a point - appear constantly in AQA AS and A-level pure mathematics, and they also support later work on circles and loci.
在本例题中，我们将运用中点、斜率以及负倒数法则，求出一条线段的垂直平分线的方程。随后我们会检验某个给定点是否在这条直线上，并最后用一种通过"平方距离相等"来验证答案的第二种方法。求方程与检验点这两项技能，在 AQA AS 与 A-level 纯数学中反复出现，并且为后续的圆与轨迹内容打下基础。
1. The Problem Statement | 题目陈述
Consider two points A(2, 5) and B(−4, −1). Find the equation of the perpendicular bisector of the line segment AB. Then determine whether the point C(1, 2) lies on this perpendicular bisector.
已知两点 A(2, 5) 和 B(−4, −1)。求线段 AB 的垂直平分线方程。然后判断点 C(1, 2) 是否在这条垂直平分线上。
Read the wording carefully. The word "bisector" tells us the line must pass through the midpoint of AB. The word "perpendicular" tells us the line must meet AB at a right angle, so its gradient is linked to the gradient of AB. A perpendicular bisector therefore carries two pieces of information at once, and both of them are needed before we can write down an equation.
请仔细读题。"平分线"（bisector）告诉我们这条直线必须经过 AB 的中点；"垂直"（perpendicular）告诉我们这条直线与 AB 相交成直角，因此它的斜率与 AB 的斜率之间存在固定关系。因此，一条垂直平分线同时携带两条信息，缺一不可，必须两者都用上才能写出方程。
2. Key Concepts and Formulae | 核心概念与公式
s paired with the average of the two y-values. Second, the gradient of that segment is the change in y divided by the change in x. Third, if two non-vertical straight lines are perpendicular, the product of their gradients is −1. Fourth, the equation of the straight line with gradient m passing through the point (x₁, y₁) is y − y₁ = m(x − x₁), the point-gradient form.
在动手计算之前，先回顾四条标准结论。第一，连接 (x₁, y₁) 与 (x₂, y₂) 的线段中点，就是两个 x 坐标的平均值与两个 y 坐标的平均值所组成的点。第二，该线段的斜率等于 y 的变化量除以 x 的变化量。第三，如果两条非竖直直线互相垂直，那么它们斜率的乘积等于 −1。第四，斜率为 m 且经过点 (x₁, y₁) 的直线方程为 y − y₁ = m(x − x₁)，即点斜式。
- Midpoint formula: M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2) — 中点公式：M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
- Gradient formula: m = Δy ÷ Δx = (y₂ − y₁) ÷ (x₂ − x₁) — 斜率公式：m = Δy ÷ Δx = (y₂ − y₁) ÷ (x₂ − x₁)
- Perpendicular condition for non-vertical lines: m₁ × m₂ = −1 — 非竖直直线的垂直条件：m₁ × m₂ = −1
- Point-gradient form: y − y₁ = m(x − x₁) — 点斜式：y − y₁ = m(x − x₁)
These four results are all you need for the whole question, and they are all given in the AQA formula booklet or expected as standard knowledge. Notice that the perpendicular condition is a statement about the product of the gradients, not about the gradients themselves, which is where many students lose marks.
整道题只需要这四条结论，它们要么出现在 AQA 的公式册中，要么属于必须掌握的标准知识。请注意，垂直条件说的是两个斜率之积，而不是斜率本身，这正是许多同学失分的地方。
3. Step 1: Find the Midpoint of AB | 第 1 步：求 AB 的中点
= −1. The y-coordinate is (5 + (−1)) ÷ 2 = 4 ÷ 2 = 2. Therefore the midpoint is M(−1, 2).
设 A = (2, 5)，B = (−4, −1)。代入中点公式。x 坐标为 (2 + (−4)) ÷ 2 = (−2) ÷ 2 = −1。y 坐标为 (5 + (−1)) ÷ 2 = 4 ÷ 2 = 2。因此中点 M 为 (−1, 2)。
M = ((2 + (−4)) ÷ 2, (5 + (−1)) ÷ 2) = (−1, 2)
The most frequent slip here is to add the coordinates without dividing by 2, which gives (−2, 4) instead of (−1, 2). Always finish the midpoint calculation by halving both sums, and keep the negative signs attached to their numbers as you work. Every point on the perpendicular bisector is equidistant from A and B, and M itself is the one such point that also lies on AB.
这里最常见的失误是把坐标相加后忘记除以 2，从而得到 (−2, 4) 而不是 (−1, 2)。务必在最后对两个和都取一半，并在运算过程中把负号牢牢绑定在对应的数字上。垂直平分线上的每一点到 A 与 B 的距离都相等，而 M 正是其中唯一同时落在 AB 上的点。
4. Step 2: Find the Gradient of AB | 第 2 步：求 AB 的斜率
−6 and the change in x is (−4) − 2 = −6. Therefore the gradient of AB is (−6) ÷ (−6) = 1.
对 A(2, 5) 与 B(−4, −1) 使用斜率公式。y 的变化量是 (−1) − 5 = −6，x 的变化量是 (−4) − 2 = −6。因此 AB 的斜率为 (−6) ÷ (−6) = 1。
mAB = ((−1) − 5) ÷ ((−4) − 2) = (−6) ÷ (−6) = 1
om: ((5 − (−1)) ÷ (2 − (−4))) = 6 ÷ 6 = 1 gives exactly the same answer. Second, dividing a negative by a negative gives a positive, so the line rises from left to right at 45° to the horizontal, which matches the visual picture of moving 6 right and 6 up from B to A.
这一步有两点值得记住。第一，只要分子与分母的相减顺序保持一致，谁减谁并不影响结果：((5 − (−1)) ÷ (2 − (−4))) = 6 ÷ 6 = 1，答案完全相同。第二，负数除以负数得正数，所以这条直线从左下向右上以与水平方向成 45° 的角度上升，这与从 B 到 A"向右 6、向上 6"的图像完全吻合。
5. Step 3: Apply the Negative Reciprocal Rule | 第 3 步：应用负倒数法则
The perpendicular bisector is perpendicular to AB, so its gradient m must satisfy mAB × m = −1. Substituting mAB he gradient of the perpendicular is −1 divided by the gradient of AB, that is −1 ÷ 1 = −1.
垂直平分线与 AB 垂直，因此其斜率 m 必须满足 mAB × m = −1。代入 mAB = 1，得 1 × m = −1，所以 m = −1。等价地，垂直直线的斜率为 −1 除以 AB 的斜率，即 −1 ÷ 1 = −1。
mAB × m = 1 × (−1) = −1 ⇒ m = −1
h an undefined gradient; if AB were vertical, the perpendicular bisector would be horizontal with gradient 0. The rule m₁ × m₂ = −1 therefore applies only to non-vertical lines, and a vertical line has no finite gradient at all.
考试中需要留意特例。如果 AB 是水平的（斜率为 0），那么垂直平分线就是竖直的，方程为 x = k，斜率不存在；如果 AB 是竖直的，那么垂直平分线就是水平的，斜率为 0。因此 m₁ × m₂ = −1 这一法则只适用于非竖直直线，而竖直直线根本没有有限的斜率。
6. Step 4: Use the Point-Gradient Form | 第 4 步：使用点斜式
to y − y₁ = m(x − x₁) gives y − 2 = −1 × (x − (−1)), which simplifies to y − 2 = −(x + 1).
垂直平分线经过中点 M(−1, 2)，斜率为 −1。代入 y − y₁ = m(x − x₁)，得 y − 2 = −1 × (x − (−1))，化简为 y − 2 = −(x + 1)。
y − 2 = −1 × (x − (−1)) = −(x + 1)
− 1), which produces the wrong line entirely. It is also worth remembering that the point you substitute here must be the midpoint M, not A or B: A and B lie on AB, and only their midpoint lies on the bisector.
关键细节在于 x − (−1) 中的双负号，它化成了 x + 1。非常常见的错误是写成 y − 2 = −(x − 1)，那样得到的就是一条完全不同的直线。同时要记住，此处代入的点必须是中点 M，而不是 A 或 B：A 与 B 位于 AB 上，只有它们的中点才落在垂直平分线上。
7. Step 5: Rearranging into Standard Forms | 第 5 步：整理成标准形式
se equivalent forms is acceptable in an examination as long as it is clearly an equation of a straight line.
展开并整理。由 y − 2 = −(x + 1) 得 y − 2 = −x − 1，因此 y = −x + 1。两边同时加 x 得 x + y = 1，再减
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