Example 5.2.1: Perpendicular Bisector of a Line Segment | 例题 5.2.1：线段的垂直平分线

📚 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. You will also see how to test whether a point lies on a given line, which is a very common AQA AS and A-level pure maths skill.

在这个例题中，我们将使用中点、斜率以及负倒数的规则来求一条垂直平分线的方程。你还将看到如何检验一个点是否在给定直线上，这是 AQA AS 与 A-level 纯数学中非常常见的技能。

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。

2. Key Concepts Needed | 所需核心概念

Before calculating, recall the four results that are used in this Example 5.2.1. First, the midpoint of a segment with endpoints (x₁, y₁) and (x₂, y₂) is given by M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2). Second, the gradient of the segment is m = (y₂ − y₁) ÷ (x₂ − x₁). Third, if two non-vertical lines are perpendicular, their gradients multiply to give −1. Fourth, the equation of a line with gradient m passing through (x₁, y₁) can be written as y − y₁ = m(x − x₁).

在计算之前，先回顾本题所用到的四个结论。第一，以 (x₁, y₁) 和 (x₂, y₂) 为端点的线段中点为 M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)。第二，线段的斜率为 m = (y₂ − y₁) ÷ (x₂ − x₁)。第三，如果两条非竖直直线互相垂直，那么它们的斜率乘积等于 −1。第四，斜率为 m 且过点 (x₁, y₁) 的直线方程可以写成 y − y₁ = m(x − x₁)。

• 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 — 斜率公式：m = Δy ÷ Δx
• Perpendicular condition: m₁ × m₂ = −1 — 垂直条件：m₁ × m₂ = −1
• Point-gradient form: y − y₁ = m(x − x₁) — 点斜式：y − y₁ = m(x − x₁)

3. Step 1: Find the Midpoint of AB | 第 1 步：求 AB 的中点

Let A = (2, 5) and B = (−4, −1). Substitute into the midpoint formula. The x-coordinate is (2 + (−4)) ÷ 2 = (−2) ÷ 2 = −1. The y-coordinate is (5 + (−1)) ÷ 2 = 4 ÷ 2 = 2. Therefore the midpoint M is (−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)

4. Step 2: Find the Slope of AB | 第 2 步：求 AB 的斜率

Use the gradient formula with A(2, 5) and B(−4, −1). The change in y is

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