IGCSE CIE Maths: Coordinate Geometry | IGCSE CIE 数学：坐标几何考点精讲

📚 IGCSE CIE Maths: Coordinate Geometry | IGCSE CIE 数学：坐标几何考点精讲

Coordinate geometry, also known as analytic geometry, bridges algebra and geometry through the Cartesian coordinate system. In IGCSE CIE Mathematics, mastering coordinates, distances, midpoints, gradients, and equations of straight lines is essential for solving geometric problems algebraically. This article provides a comprehensive revision of all key points, formulas, and exam techniques.

坐标几何，也称解析几何，通过笛卡尔坐标系统将代数与几何联系起来。在IGCSE CIE数学中，掌握坐标、距离、中点、梯度和直线方程对于用代数方法解决几何问题至关重要。本文全面梳理了所有考点、公式和应试技巧。

1. Understanding Coordinates | 理解坐标

The Cartesian plane consists of a horizontal x‑axis and a vertical y‑axis intersecting at the origin (0,0). A point is written as (x, y), where x is the horizontal displacement and y is the vertical displacement.

笛卡尔平面由水平x轴和竖直y轴组成，相交于原点(0,0)。一个点表示为(x,y)，其中x为水平位移，y为竖直位移。

Positive x values are to the right, negative x to the left; positive y upwards, negative y downwards. Understanding the coordinate system is the foundation for all subsequent formulas.

x正值向右，负值向左；y正值向上，负值向下。理解坐标系是所有后续公式的基础。

2. Distance Formula | 距离公式

For two points A(x₁, y₁) and B(x₂, y₂), the distance AB is given by the formula derived from Pythagoras’ theorem:

对于两点A(x₁, y₁)和B(x₂, y₂)，距离AB由毕达哥拉斯定理推导出的公式给出：

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

This formula calculates the length of the straight line segment joining the two points. Always subtract coordinates in the same order and square the differences.

该公式计算连接两点的直线段的长度。始终用相同顺序相减，并平方差值。

Example: Distance between (3,4) and (7,1) is √[(7−3)² + (1−4)²] = √[16+9] = √25 = 5. When the points lie on a horizontal line (same y), the distance simplifies to |x₂ − x₁|; on a vertical line it is |y₂ − y₁|.

例如，点(3,4)和(7,1)的距离是√[(7-3)²+(1-4)²]=√(16+9)=5。当两点在同一水平线上（y相等），距离简化为|x₂−x₁|；在同一竖直线上则为|y₂−y₁|。

3. Midpoint Formula | 中点公式

The midpoint M of segment AB with endpoints A(x₁,y₁) and B(x₂,y₂) is the average of the coordinates:

线段AB（端点A(x₁,y₁)和B(x₂,y₂)）的中点M是坐标的平均值：

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

The midpoint splits the segment into two equal lengths. This is useful for finding the centre of a line segment, or in symmetry and geometry problems where a point is the mirror image of another.

中点将线段分成两条长度相等的线段。它用于求线段中心，或对称与几何问题中一点为另一点的镜像。

4. Gradient (Slope) of a Line | 直线的梯度（斜率）

Gradient m measures the steepness and direction of a line:

梯度m衡量直线的倾斜度和方向：

m = (y₂ − y₁) / (x₂ − x₁)

If m > 0, the line rises from left to right; if m < 0, it falls. A horizontal line has m = 0; a vertical line has undefined gradient (division by zero).

若m>0，直线从左到右上升；若m<0，则下降。水平线的梯度为0；竖直线的梯度未定义（除以零）。

Always keep the same order for y and x differences. For example, between (1,2) and (4,8): m = (8−2)/(4−1) = 6/3 = 2.

务必保持y差和x差相同的相减顺序。例如，(1,2)与(4,8)之间：m=(8−2)/(4−1)=6/3=2。

5. Equation of a Line: y = mx + c | 直线方程：y = mx + c

The slope‑intercept form is y = mx + c, where m is the gradient and c is the y‑intercept (the value where the line crosses the y‑axis).

斜截式为y = mx + c，其中m是梯度，c是y轴截距（直线与y轴交点的值）。

If you know the gradient and the y‑intercept, you can write the equation immediately. For instance, a line with gradient 2 and passing through (0,3) has equation y = 2x + 3.

如果知道梯度和y轴截距，可直接写出方程。例如，梯度为2且经过(0,3)的直线方程为y=2x+3。

To find c given a point (x,y) and gradient m, substitute the coordinates into y = mx + c and solve for c.

已知一点(x,y)和梯度m求c，将坐标代入y=mx+c，解出c。

6. Point‑Slope Form and General Form | 点斜式与一般式

Point‑slope form: y − y₁ = m(x − x₁). This is very useful when you know a point on the line and its gradient.

点斜式：y − y₁ = m(x − x₁)。当已知直线上一点和梯度时非常有用。

General form: ax + by + c = 0 or ax + by = d. It is often needed for further algebraic manipulation, such as finding intersections or rearranging for y.

一般式：ax+by+c=0或ax+by=d。常用于进一步的代数运算，如求交点或变形为y=。

You can convert between forms: starting from y − 1 = 3(x − 2) → y − 1 = 3x − 6 → y = 3x − 5, or 3x − y − 5 = 0 in general form.

可在形式间转换：由y-1=3(x-2)→y-1=3x-6→y=3x-5，或一般式3x-y-5=0。

7. Parallel and Perpendicular Lines | 平行线与垂直线

Parallel lines have equal gradients: m₁ = m₂.

平行线的梯度相等：m₁ = m₂。

Perpendicular lines have gradients that multiply to −1: m₁ × m₂ = −1, provided neither line is vertical. The gradient of a perpendicular line is the negative reciprocal: m₂ = −1/m₁.

垂直线的梯度乘积为−1：m₁×m₂=−1，前提是两条线都不竖直。垂直线的梯度是负倒数：m₂=−1/m₁。

To find the equation of a line parallel to y = 4x + 1 passing through (2,3), keep the gradient 4: y − 3 = 4(x − 2). For a perpendicular line, use gradient −¼: y − 3 = −¼(x − 2).

求过点(2,3)且平行于y=4x+1的直线，保持梯度4：y-3=4(x-2)。求垂直线，用梯度−¼：y-3=−¼(x-2)。

8. Finding the Intersection of Two Lines | 求两直线交点

The intersection point of two lines is the simultaneous solution of their equations. Solve the two linear equations using substitution or elimination.

两直线的交点是它们方程的公共解。用代入法或消元法解两个线性方程。

Example: y = 2x + 1 and y = −x + 4. Set equal: 2x + 1 = −x + 4 → 3x = 3 → x = 1, then y = 2(1)+1 = 3. Intersection (1,3).

例如，y=2x+1和y=−x+4。令相等：2x+1=−x+4→3x=3→x=1，然后y=2(1)+1=3。交点(1,3)。

If the equations are in general form, e.g. 2x + y = 7 and x − y = 2, add them to eliminate y: 3x = 9 → x = 3, back‑substitute to get y = 1.

若方程为一般式，如2x+y=7和x−y=2，相加消去y：3x=9→x=3，回代得y=1。

9. Perpendicular from a Point to a Line | 点到直线的垂线

To find the foot of the perpendicular from a point P(x₀,y₀) to a line L with gradient m, first note that the perpendicular

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IGCSE CIE Maths: Coordinate Geometry | IGCSE CIE 数学：坐标几何 考点精讲