Common Question Types & Solutions in Coordinate Geometry | 坐标几何常见题型与解法

📚 Common Question Types & Solutions in Coordinate Geometry | 坐标几何常见题型与解法

Coordinate geometry, also known as analytic geometry, is a cornerstone of A-Level mathematics. It allows geometric problems to be solved through algebraic methods, and it appears consistently across all major examination boards including Edexcel, CIE, AQA and OCR.

坐标几何又称解析几何，是A-Level数学的基石。它使我们能够通过代数方法解决几何问题，并且它在所有主要考试局（Edexcel、CIE、AQA和OCR）的试卷中持续出现。

This article breaks down the most common coordinate geometry question types, the techniques needed to solve them, and the pitfalls that students frequently encounter in exams.

本文将分解坐标几何最常见的题型、解题所需的方法，以及学生在考试中经常遇到的陷阱。

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1. Distance and Midpoint | 距离与中点

The distance between two points A(x₁, y₁) and B(x₂, y₂) is derived directly from Pythagoras’ theorem:

两点A(x₁, y₁)与B(x₂, y₂)之间的距离直接由勾股定理推导得出：

d = √[(x₂ – x₁)² + (y₂ – y₁)²]

The midpoint M of segment AB is found by averaging the x-coordinates and the y-coordinates separately:

线段AB的中点M通过分别求横坐标与纵坐标的平均值得到：

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Worked example: Given A(3, -2) and B(-1, 4), find the distance AB and the midpoint M.

例题：已知A(3, -2)和B(-1, 4)，求距离AB和中点M。

Distance: d = √[(-1 – 3)² + (4 – (-2))²] = √(16 + 36) = √52 = 2√13.

距离：d = √[(-1 – 3)² + (4 – (-2))²] = √(16 + 36) = √52 = 2√13。

Midpoint: M = ((3 + (-1))/2, (-2 + 4)/2) = (1, 1).

中点：M = ((3 + (-1))/2, (-2 + 4)/2) = (1, 1)。

A common examiner trick is to give one endpoint and the midpoint, then ask for the unknown endpoint. Simply reverse the midpoint formula: if M(2, 3) is the midpoint of A(x, y) and B(4, -1), then 2 = (x + 4)/2 and 3 = (y + (-1))/2, giving x = 0 and y = 7.

一种常见的出题技巧是给出一个端点与中点，然后求另一个端点。只需反向使用中点公式：若M(2, 3)是A(x, y)与B(4, -1)的中点，则2 = (x + 4)/2，3 = (y + (-1))/2，解得x = 0，y = 7。

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2. Gradient of a Line | 直线斜率

The gradient (slope) m of a line passing through A(x₁, y₁) and B(x₂, y₂) measures how steeply the line rises or falls:

经过A(x₁, y₁)与B(x₂, y₂)的直线的斜率m度量其倾斜程度：

m = (y₂ – y₁)/(x₂ – x₁)

A horizontal line has gradient 0, while a vertical line has an undefined gradient. A line rising from left to right has a positive gradient; a line falling from left to right has a negative gradient. The gradient also equals tan θ, where θ is the angle the line makes with the positive x-axis.

水平线的斜率为0，而垂直线的斜率不存在。从左到右上升的直线斜率为正；从左到右下降的直线斜率为负。斜率也等于tan θ，其中θ是直线与x轴正方向所成的角。

Worked example: Find the gradient of the line joining P(2, 5) and Q(6, -3).

例题：求连接P(2, 5)与Q(6, -3)的直线的斜率。

m = (-3 – 5)/(6 – 2) = -8/4 = -2. The negative sign confirms the line slopes downward.

m = (-3 – 5)/(6 – 2) = -8/4 = -2。负号表明该直线向下倾斜。

Be careful when substituting negative coordinates: the most common error in gradient questions is a sign mistake. Write the subtraction step out fully before simplifying.

代入负坐标时务必小心：斜率题最常见的错误就是符号错误。在化简之前，请先完整写出减法步骤。

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3. Equation of a Straight Line | 直线方程

There are several standard forms for the equation of a straight line. Choosing the right form depends on the information given in the question.

直线方程有几种标准形式。根据题目给出的条件选择适当的形式。

Form	Equation	Use when
Point-slope	y – y₁ = m(x – x₁)	Gradient and one point known
Slope-intercept	y = mx + c	Gradient and y-intercept known
Two-point	(y – y₁)/(x – x₁) = (y₂ – y₁)/(x₂ – x₁)	Two points known
General	ax + by + c = 0	Final answer expected in this form

Worked example: Find the equation of the line through (3, -1) with gradient 2.

例题：求经过点(3, -1)且斜率为2的直线方程。

Using the point-slope form: y – (-1) = 2(x – 3), so y + 1 = 2x – 6, hence y = 2x – 7. In general form: 2x – y – 7 = 0.

使用点斜式：y – (-1) = 2(x – 3)，即y + 1 = 2x – 6，因此y = 2x – 7。化为一般式：2x – y – 7 = 0。

Always read carefully whether the question asks for the answer in a specific form. If no form is specified, the slope-intercept or general form is usually acceptable, but check the marks scheme guidance for your board.

始终仔细审题，看题目是否要求特定形式的答案。若未指定形式，斜截式或一般式通常都可以接受，但请查阅你所在考试局的评分方案指引。

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4. Parallel and Perpendicular Lines | 平行与垂直直线

Two lines with gradients m₁ and m₂ are parallel if their gradients are equal.

两条斜率分别为m₁和m₂的直线，若其斜率相等，则两条直线平行。

Parallel: m₁ = m₂

Two lines with gradients m₁ and m₂ are perpendicular if the product of their gradients equals -1.

两条斜率分别为m₁和m₂的直线，若其斜率之积等于-1，则两条直线互相垂直。

Perpendicular: m₁ × m₂ = -1, i.e. m₂ = -1/m₁

Worked example: A line L passes through (2, 3) and is perpendicular to the line y = 2x + 1. Find the equation of L.

例题：直线L经过点(2, 3)且垂直于直线y = 2x + 1。求L的方程。

The given line has gradient 2, so the perpendicular gradient is m = -1/2. Using y – 3 = (-1/2)(x – 2), we obtain 2y – 6 = -x + 2, hence x + 2y – 8 = 0.

已知直线的斜率为2，

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