Coordinate Geometry Key Points for CIE A-Level Mathematics | A-Level CIE 数学：坐标几何考点精讲

📚 Coordinate Geometry Key Points for CIE A-Level Mathematics | A-Level CIE 数学：坐标几何考点精讲

Coordinate geometry, also known as analytic geometry, is a crucial topic in CIE A-Level Mathematics (Pure Mathematics 1). It combines algebra and geometry to solve problems involving points, lines, and circles on a Cartesian plane. Mastering this topic requires a solid understanding of formulas, their derivations, and application techniques. This revision guide covers all essential concepts and exam-style strategies.

坐标几何，也称解析几何，是 CIE A-Level 数学（纯数学 1）中的重要主题。它将代数与几何相结合，解决笛卡尔平面上点、直线和圆的相关问题。要掌握这一主题，需要扎实理解公式、推导及应用技巧。本复习指南涵盖所有核心概念和应试策略。

1. Distance Formula | 距离公式

The distance between two points A(x₁, y₁) and B(x₂, y₂) is derived directly from Pythagoras’ theorem. If you form a right triangle with the line segment AB as the hypotenuse, the legs have lengths |x₂ − x₁| and |y₂ − y₁|. Squaring and adding gives the formula below. It is symmetric and always yields a non-negative result.

两点 A(x₁, y₁) 和 B(x₂, y₂) 之间的距离直接由勾股定理导出。若以线段 AB 为斜边构造直角三角形，两直角边长度分别为 |x₂ − x₁| 和 |y₂ − y₁|。平方相加即得如下公式。该公式具有对称性且结果恒为非负数。

AB = √((x₂ − x₁)² + (y₂ − y₁)²)

Worked example: Find the distance between P(3, −1) and Q(7, 4). Substituting gives √((7−3)² + (4−(−1))²) = √(4² + 5²) = √(16+25) = √41.

示例：求 P(3, −1) 与 Q(7, 4) 的距离。代入得 √((7−3)² + (4−(−1))²) = √(4²+5²) = √(16+25) = √41。

In coordinate geometry, you often need to set up an equation involving an unknown coordinate and a given distance. Square both sides to remove the square root before solving.

在坐标几何中，经常会建立含未知坐标与已知距离的方程。求解时应先将两边平方以消去根号。

2. Midpoint Formula | 中点公式

The midpoint M of the segment joining A(x₁, y₁) and B(x₂, y₂) is simply the average of the x-coordinates and the average of the y-coordinates. This is because the midpoint lies exactly halfway along both axes.

线段 AB 的中点 M 是 x 坐标与 y 坐标各自平均值的点。这是因为中点恰好位于两轴向距离的中点。

M = ((x₁ + x₂)/2 , (y₁ + y₂)/2)

Example: The midpoint of C(5, 2) and D(1, 8) is ((5+1)/2 , (2+8)/2) = (3, 5). This same formula is used to find a missing endpoint when the other endpoint and the midpoint are known.

示例：点 C(5, 2) 与 D(1, 8) 的中点为 ((5+1)/2 , (2+8)/2) = (3, 5)。当已知一个端点和中点时，可用同一公式反求另一个端点。

The midpoint concept also extends to dividing a segment in a given ratio. For A-Level, the internal division formulas are occasionally tested, but the simple midpoint is the most frequent.

中点概念还可推广到按给定比例分割线段。A-Level 阶段偶尔考查内分点公式，但最简单的中间点仍是最常见的考点。

3. Gradient and Angle of Inclination | 斜率与倾斜角

The gradient (or slope) of a line measures its steepness. For two distinct points (x₁, y₁) and (x₂, y₂), the gradient m is the change in y divided by the change in x. If Δx = 0, the line is vertical and its gradient is undefined.

斜率用于衡量直线的倾斜程度。对于两个不同点 (x₁, y₁) 和 (x₂, y₂)，斜率 m 定义为 y 的改变量除以 x 的改变量。若 Δx = 0，直线为竖直线，其斜率无定义。

m = (y₂ − y₁) / (x₂ − x₁)

The gradient also equals the tangent of the angle θ that the line makes with the positive x-axis: m = tan θ. This relationship is useful for finding the angle of inclination or for determining the direction vector.

斜率也等于直线与 x 轴正方向夹角 θ 的正切：m = tan θ。这一关系常用于求倾斜角或确定方向向量。

Example: A line through (2, 0) and (6, 3) has m = (3−0)/(6−2) = 3/4. The angle of inclination is θ = tan⁻¹(3/4) ≈ 36.9°.

示例：过 (2,0) 和 (6,3) 的直线斜率为 m = (3−0)/(6−2) = 3/4，其倾斜角为 θ = tan⁻¹(3/4) ≈ 36.9°。

4. Equations of a Straight Line | 直线方程

There are several equivalent forms for the equation of a straight line. The choice depends on the given information. The most versatile form for exam problems is the point-gradient form, which can be converted to other forms as needed.

直线方程有几种等价形式，选用哪种取决于已知条件。考试中最灵活多用的是点斜式，需要时可转换为其他形式。

Point-gradient form: y − y₁ = m(x − x₁). This is used when you know a point (x₁, y₁) and the gradient m.

点斜式： y − y₁ = m(x − x₁)。当已知一点 (x₁, y₁) 和斜率 m 时使用。

Slope-intercept form: y = mx + c, where c is the y-intercept. This is helpful for quickly identifying gradient and intercept.

斜截式： y = mx + c，其中 c 为 y 截距。该式便于快速识别斜率和截距。

Two-point form: (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁). It is essentially the gradient formula set equal to the gradient of the variable point.

两点式： (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁)。本质上就是让动点与已知点连线的斜率等于已知两点间的斜率。

General form: ax + by + c = 0, where a, b and c are integers. This form is often preferred for writing final answers in exams.

一般式： ax + by + c = 0，其中 a、b、c 为整数。在考试中书写最终答案时通常偏爱这一形式。

Form	Key information needed	Example
Point-gradient	Point and gradient	y − 3 = 2(x + 1)
Slope-intercept	Gradient and y-intercept	y = 2x + 5
General	Any three coefficients	3x + 4y − 12 = 0

You should be able to switch between these forms confidently, especially when solving intersection or distance problems.

你应当能够熟练在不同形式间转换，尤其是在求解交点或距离问题时。

5. Parallel and Perpendicular Lines | 平行与垂直条件

For two distinct lines with gradients m₁ and m₂, parallelism means equal gradients (provided both are defined). For perpendicular lines, the product of their gradients is −1, again assuming neither line is vertical. Vertical lines are parallel to each other and perpendicular to horizontal lines.

对于斜率分别为 m₁ 和 m₂ 的两条不重合直线，平行意味着斜率相等（当两者均有定义时）。对于垂直线，它们的斜率之积为 −1，同样假设无竖直线。竖直线互相平行，并与水平线垂直。

Parallel: m₁ = m₂

Perpendicular: m₁ × m₂ = −1

Example: Line L₁ has gradient 3. A line parallel to L₁ also has gradient 3. A line perpendicular to L₁ has gradient −1/3.

示例：直线 L₁ 斜率为 3。平行于 L₁ 的直线斜率也为 3。垂直于 L₁ 的直线斜率为 −1/3。

These conditions are essential for constructing perpendicular bisectors and tangent lines to circles. Many problems ask you to find the equation of a line passing through a given point that is parallel or perpendicular to another line.

这些条件是构造垂直平分线和圆的切线的基础。许多题目要求找出过已知点且平行或垂直于另一直线的直线方程。

6. Perpendicular Bisector | 垂直平分线

The perpendicular bisector of a segment AB is the line that passes through the midpoint of AB and is perpendicular to AB. Its steps combine the midpoint formula, gradient calculation, and point-gradient form. It is the locus of points equidistant from A and B.

线段 AB 的垂直平分线是过 AB 中点且垂直于 AB 的直线。求作步骤综合了中点公式、斜率计算和点斜式。它是到 A 与 B 距离相等的点的轨迹。

Step 1: Find the midpoint M of AB.

步骤 1： 求 AB 的中点 M。

Step 2: Calculate the gradient of AB, m_AB, then the gradient of the perpendicular bisector is m_perp = −1/m_AB (if m_AB ≠ 0).

步骤 2： 计算 AB 的斜率 m_AB，则垂直平分线的斜率为 m_perp = −1/m_AB（若 m_AB ≠ 0）。

Step 3: Use y − y_M = m_perp(x − x_M) to write the equation.

步骤 3： 运用点斜式 y − y_M = m_perp(x − x_M) 写出方程。

Worked example: Given A(2, −3) and B(4, 5), the midpoint is (3, 1). Gradient of AB = (5−(−3))/(4−2) = 8/2 = 4. Perpendicular gradient = −1/4. Equation: y − 1 = −1/4 (x − 3), i.e. x + 4y = 7 in general form.

示例：已知 A(2, −3) 和 B(4, 5)，中点为 (3, 1)。AB 斜率 = (5−(−3))/(4−2) = 4。垂直斜率为 −1/4。方程：y − 1 = −1/4(x − 3)，化为一般式为 x + 4y = 7。

7. Equation of a Circle (Standard Form) | 圆的标准方程

A circle is the set of all points at a fixed distance (radius r) from a fixed point (centre (a, b)). The standard equation is derived directly from the distance formula. It is important to recognise the centre and radius from this form instantly.

圆是所有到定点（圆心 (a, b)）距离为定长（半径 r）的点的集合。标准方程直接由距离公式导出。能够立即从该形式认出圆心和半径至关重要。

(x − a)² + (y − b)² = r²

For example, (x + 3)² + (y − 2)² = 25 represents a circle with centre (−3, 2) and radius 5. Note that the sign of the centre coordinates is opposite to that in the brackets.

例如，(x + 3)² + (y − 2)² = 25 表示圆心为 (−3, 2)、半径为 5 的圆。注意圆心坐标的符号与括号内相反。

When the centre is at the origin, the equation simplifies to x² + y² = r². Many problems involve finding the radius given the centre and a point on the circumference: simply substitute the point into the standard form to solve for r².

当圆心在原点时，方程简化为 x² + y² = r²。许多题目涉及已知圆心和圆周上一点求半径：只需将点代入标准形式解出 r² 即可。

8. General Form of a Circle | 圆的一般方程

Expanding the standard form leads to the general equation x² + y² + 2gx + 2fy + c = 0. It represents a circle only if g² + f² − c > 0. Completing the square is the key technique to convert general form back to standard form and reveal the centre and radius.

将标准形式展开可得到一般方程 x² + y² + 2gx + 2fy + c = 0。只有当 g² + f² − c > 0 时才表示圆。配方法是从一般式转回标准式并显露出圆心和半径的关键技巧。

Centre: (−g, −f)

圆心： (−g, −f)

Radius: √(g² + f² − c)

半径： √(g² + f² − c)

Example: Convert x² + y² − 6x + 4y − 3 = 0 to standard form. Group x and y terms: (x² − 6x) + (y² + 4y) = 3. Complete squares: (x − 3)² − 9 + (y + 2)² − 4 = 3 → (x − 3)² + (y + 2)² = 16. Centre (3, −2), radius 4.

示例：将 x² + y² − 6x + 4y − 3 = 0 化为标准形。并项：(x² − 6x) + (y² + 4y) = 3。配方：(x − 3)² − 9 + (y + 2)² − 4 = 3 → (x − 3)² + (y + 2)² = 16。圆心 (3, −2)，半径 4。

In exams, you may be asked to determine whether a given quadratic equation represents a circle, a point, or an empty set by checking the sign of g² + f² − c.

在考试中，可能会要求判断给定的二次方程表示圆、点还是空集，方法就是检查 g² + f² − c 的符号。

9. Intersection of a Line and a Circle | 直线与圆的交点

To find the intersection points, substitute the line equation (preferably in the form y = mx + c or x = k) into the circle equation. This yields a quadratic equation in one variable. The discriminant (Δ = b² − 4ac) determines the nature of intersection.

求交点时，将直线方程（最好用 y = mx + c 或 x = k 的形式）代入圆的方程，得到关于一个变量的二次方程。判别式 (Δ = b² − 4ac) 决定相交的情况。

Δ > 0: two distinct intersection points (line cuts the circle).

Δ > 0： 有两个不同交点（直线与圆相交）。

Δ = 0: exactly one intersection point (line is tangent to the circle).

Δ = 0： 有一个交点（直线与圆相切）。

Δ < 0: no real intersection points (line misses the circle).

Δ < 0： 无实数交点（直线与圆不相交）。

Example: Find the intersection of y = x + 2 and x² + y² = 4. Substituting: x² + (x+2)² = 4 → 2x² + 4x + 4 = 4 → 2x² + 4x = 0 → 2x(x+2)=0 → x=0 or x=−2. Points: (0,2) and (−2,0).

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Coordinate Geometry Key Points for CIE A-Level Mathematics | A-Level CIE 数学：坐标几何 考点精讲