Coordinate Geometry Key Points Revision | 坐标几何考点精讲

📚 Coordinate Geometry Key Points Revision | 坐标几何考点精讲

Coordinate geometry, also known as analytic geometry, unites algebra and geometry by using a coordinate system to describe points, lines, and curves. For IB and CCEA Mathematics, mastering coordinate geometry is essential for solving problems related to straight lines, circles, and their intersections. This revision guide highlights the core concepts, formulas, and techniques, with clear explanations and examples to boost your confidence.

坐标几何，又称解析几何，通过坐标系将代数与几何结合起来，用以描述点、线和曲线。对于 IB 和 CCEA 数学来说，掌握坐标几何是解决直线、圆及其交点问题的关键。本考点精讲突出核心概念、公式和技巧，配合清晰的解释与例题，帮助你增强信心。

1. Distance Between Two Points | 两点间的距离

The distance d between two points (x₁, y₁) and (x₂, y₂) is found using Pythagoras’ theorem. The formula is derived from the right-angled triangle formed by the horizontal and vertical differences.

两点 (x₁, y₁) 和 (x₂, y₂) 之间的距离 d 可通过勾股定理求得。该公式源自水平差和垂直差所形成的直角三角形。

d = √[(x₂ – x₁)² + (y₂ – y₁)²]

For example, the distance between (3, 4) and (7, 1) is √[(7-3)² + (1-4)²] = √[16 + 9] = √25 = 5 units. Always keep the order of points consistent – it doesn’t matter which is first, since squaring eliminates negatives.

例如，(3, 4) 与 (7, 1) 之间的距离为 √[(7-3)² + (1-4)²] = √[16 + 9] = √25 = 5 个单位。保持点的顺序一致即可，哪个点在前无关紧要，因为平方会消去负号。

2. Midpoint of a Line Segment | 线段的中点

The midpoint M of a segment joining (x₁, y₁) and (x₂, y₂) is simply the average of the x-coordinates and the y-coordinates. This gives the point exactly halfway between the two endpoints.

连接 (x₁, y₁) 与 (x₂, y₂) 的线段中点 M 就是 x 坐标和 y 坐标分别取平均值。由此得到恰好位于两端点正中间的点。

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

If the endpoints are (-2, 5) and (6, -3), the midpoint is ( (-2+6)/2 , (5-3)/2 ) = (2, 1). Midpoints are often used to find the centre of a circle when given the endpoints of a diameter.

若端点为 (-2, 5) 和 (6, -3)，中点为 ( (-2+6)/2 , (5-3)/2 ) = (2, 1)。当已知直径两端点时，常用中点公式确定圆心。

3. Gradient of a Straight Line | 直线的斜率

The gradient m measures the steepness and direction of a line. It is defined as the change in y divided by the change in x between any two distinct points on the line.

斜率 m 衡量直线的倾斜程度与方向。其定义为直线上任意两个不同点之间 y 的变化量除以 x 的变化量。

m = (y₂ – y₁) / (x₂ – x₁)

A positive gradient slopes upward from left to right; a negative gradient slopes downward. A horizontal line has gradient 0, and a vertical line has an undefined gradient. The gradient is constant along a straight line, which is the foundation for the equation of a line.

正斜率从左到右向上倾斜；负斜率向下倾斜。水平线的斜率为 0，垂直线的斜率无定义。直线上各点的斜率恒定，这是直线方程的基础。

4. Equation of a Straight Line | 直线方程

There are several forms for the equation of a line. The most versatile is the point-gradient form: y – y₁ = m(x – x₁), where (x₁, y₁) is a known point on the line and m is the gradient.

直线方程有多种形式。最通用的是点斜式：y – y₁ = m(x – x₁)，其中 (x₁, y₁) 为直线上已知点，m 为斜率。

The gradient-intercept form y = mx + c is popular because m is the gradient and c is the y-intercept (where the line crosses the y-axis). The general form ax + by + c = 0 is useful for solving systems and for distances from a point to a line.

斜截式 y = mx + c 因 m 为斜率、c 为 y 轴截距而常用。一般式 ax + by + c = 0 在解方程组和求点到直线距离时非常方便。

To find the equation of the line through (2, 3) with gradient 4, use y – 3 = 4(x – 2), which simplifies to y = 4x – 5. If two points are given, first calculate the gradient, then apply the point-gradient form.

求过点 (2, 3) 且斜率为 4 的直线方程，用 y – 3 = 4(x – 2)，化简得 y = 4x – 5。若已知两点，先求斜率，再代入点斜式。

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

Two lines are parallel if and only if their gradients are equal: m₁ = m₂. Parallel lines never intersect. For perpendicular lines, the product of their gradients is -1, or equivalently, one gradient is the negative reciprocal of the other: m₁ × m₂ = -1.

两条直线平行当且仅当它们的斜率相等：m₁ = m₂。平行线永不相交。对于垂直线，斜率之积为 -1，即一条线的斜率是另一条的负倒数：m₁ × m₂ = -1。

If a line has gradient 3, any line parallel to it also has gradient 3. A line perpendicular to it would have gradient -1/3. Be careful: vertical lines (undefined gradient) are perpendicular to horizontal lines (gradient 0).

若直线斜率为 3，与之平行的直线斜率也为 3。与之垂直的直线斜率则为 -1/3。注意：垂直直线（斜率无定义）与水平直线（斜率为 0）相互垂直。

To find the equation of a line through (5, -2) perpendicular to y = (1/2)x + 4, identify the gradient of the given line as 1/2. The perpendicular gradient is -2. Then use y – (-2) = -2(x – 5) to get y = -2x + 8.

求过点 (5, -2) 且与 y = (1/2)x + 4 垂直的直线方程，先得已知斜率为 1/2，则垂直线斜率为 -2。然后代入 y – (-2) = -2(x – 5)，得到 y = -2x + 8。

6. Perpendicular Bisector | 垂直平分线

The perpendicular bisector of a line segment is the set of points equidistant from the segment’s endpoints. It passes through the midpoint of the segment and is perpendicular to it. Finding its equation combines midpoint and perpendicular gradient calculations.

线段的垂直平分线是到线段两端点距离相等的点的集合。它经过线段的中点并垂直于该线段。求其方程需要结合中点计算和垂直斜率。

For points A(1, 2) and B(5, 4), the midpoint is (3, 3). The gradient of AB is (4-2)/(5-1) = 1/2. The perpendicular gradient is -2. Thus the perpendicular bisector equation is y – 3 = -2(x – 3), or y = -2x + 9.

对于点 A(1, 2) 和 B(5, 4)，中点为 (3, 3)。AB 的斜率为 (4-2)/(5-1) = 1/2。垂直斜率为 -2。因此垂直平分线方程为 y – 3 = -2(x – 3)，即 y = -2x + 9。

This line is often used in circle geometry, as the perpendicular bisector of any chord passes through the centre of the circle.

该线常用于圆的几何问题中，任何弦的垂直平分线都通过圆心。

7. Equation of a Circle | 圆的方程

A circle with centre (a, b) and radius r has the standard equation (x – a)² + (y – b)² = r². This comes directly from the distance formula, expressing that any point on the circle is exactly r units from the centre.

以 (a, b) 为圆心、r 为半径的圆的标准方程为 (x – a)² + (y – b)² = r²。这直接来自距离公式，表明圆上任意点到圆心的距离恰好为 r。

If the centre is at the origin (0, 0), the equation simplifies to x² + y² = r². The general form x² + y² + 2gx + 2fy + c = 0 can be converted to standard form by completing the square for x and y. The centre is (-g, -f) and radius is √(g² + f² – c), provided g² + f² – c > 0.

若圆心在原点 (0, 0)，方程简化为 x² + y² = r²。一般式 x² + y² + 2gx + 2fy + c = 0 可通过配方转化为标准式。圆心为 (-g, -f)，半径为 √(g² + f² – c)，前提是 g² + f² – c > 0。

Example: Convert x² + y² – 6x + 4y – 3 = 0 to standard form. Completing the square: (x – 3)² – 9 + (y + 2)² – 4 – 3 = 0 → (x – 3)² + (y + 2)² = 16. Centre (3, -2), radius 4.

例题：将 x² + y² – 6x + 4y – 3 = 0 化为标准式。配方：(x – 3)² – 9 + (y + 2)² – 4 – 3 = 0 → (x – 3)² + (y + 2)² = 16。圆心 (3, -2)，半径 4。

8. Tangents to a Circle | 圆的切线

A tangent to a circle is a straight line that touches the circle at exactly one point. The radius to the point of tangency is perpendicular to the tangent. This geometric property is the key to finding tangent equations.

圆的切线是一条与圆恰好只有一个交点的直线。过切点的半径垂直于切线。这一几何性质是求切线方程的关键。

To find the tangent of a circle at a given point P on the circle, first find the gradient of the radius OP (where O is the centre). The gradient of the tangent is the negative reciprocal. Then use the point-gradient form.

求圆上给定点 P 处的切线方程，首先求半径 OP 的斜率（O 为圆心）。切线的斜率是其负倒数。然后使用点斜式。

Example: For the circle x² + y² = 25, find the tangent at (3, 4). Centre is (0, 0). Gradient of radius = 4/3. Tangent gradient = -3/4. Tangent equation: y – 4 = -3/4(x – 3) → 3x + 4y = 25.

例题：对于圆 x² + y² = 25，求点 (3, 4) 处的切线。圆心 (0, 0)，半径斜率为 4/3。切线斜率为 -3/4。切线方程：y – 4 = -3/4(x – 3) → 3x + 4y = 25。

If the point is external, you may need to use the condition that the perpendicular distance from the centre to the line equals the radius. Let the line be y = mx + c and solve for c using distance formula.

若点位于圆外，可能需要利用圆心到直线的垂直距离等于半径这一条件。设直线为 y = mx + c，通过距离公式求解 c。

9. Intersection of Lines and Circles | 直线与圆的交点

To find the intersection points between a line and a circle, solve their equations simultaneously. Substitute the linear equation into the circle equation, resulting in a quadratic in x (or y). The discriminant determines the number of intersection points:

求直线与圆的交点，需联立方程求解。将直线方程代入圆的方程，得到关于 x（或 y）的二次方程。其判别式确定交点个数：

If discriminant > 0: two distinct points of intersection (secant). | 若判别式大于 0：两个不同交点（割线）。
If discriminant = 0: one point of intersection (tangent). | 若判别式等于 0：一个交点（切线）。
If discriminant < 0: no intersection (line misses the circle). | 若判别式小于 0：无交点（直线与圆相离）。

For example, find the intersection of the line y = 2x + 1 and the circle x² + y² = 10. Substitute: x² + (2x+1)² = 10 → x² + 4x² + 4x + 1 = 10 → 5x² + 4x – 9 = 0. The discriminant is 4² – 4(5)(-9) = 16 + 180 = 196 > 0, so there are two points. Solve the quadratic to find them.

例如，求直线 y = 2x + 1 与圆 x² + y² = 10 的交点。代入得：x² + (2x+1)² = 10 → 5x² + 4x – 9 = 0。判别式 196 > 0，有两个交点。解二次方程即可求出。

10. Distance from a Point to a Line | 点到直线的距离

The perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is given by a formula derived from geometric considerations. It is particularly useful in coordinate geometry problems involving circles and tangents.

点 (x₁, y₁) 到直线 ax + by + c = 0 的垂直距离公式由几何推导而来，在涉及圆和切线的坐标几何问题中尤其有用。

d = |ax₁ + by₁ + c| / √(a² + b²)

For example, the distance from (2, -3) to the line 3x – 4y + 5 = 0 is |3(2) -4(-3) + 5| / √(3²+(-4)²) = |6+12+5|/5 = 23/5 = 4.6 units. This formula is essential when determining if a line is tangent to a circle; set this distance equal to the radius.

例如，点 (2, -3) 到直线 3x – 4y + 5 = 0 的距离为 |3×2 -4×(-3) +5| / √(3²+(-4)²) = 23/5 = 4.6。在判断直线是否与圆相切时，让该距离等于半径即可。

11. Applications in Coordinate Geometry | 坐标几何的应用

Coordinate geometry frequently appears in exam questions involving triangles, quadrilaterals, and real-world contexts. For instance, you might need to prove that a triangle is right-angled by showing that the product of the gradients of two sides is -1, or that a quadrilateral is a parallelogram by demonstrating that both pairs of opposite sides are parallel.

坐标几何常出现在涉及三角形、四边形及实际背景的考题中。例如，你可能需要通过证明两条边的斜率之积为 -1 来证明三角形为直角三角形，或通过证明两组对边平行来证明四边形为平行四边形。

Problems on loci are common: e.g., find the path of a point that remains a fixed distance from a given line (two parallel lines) or from a point (a circle). The intersection of loci often combines line and circle equations.

关于轨迹的问题也很常见：例如，求到给定直线保持固定距离的点的轨迹（两条平行线），或到定点的轨迹（圆）。轨迹的交点常涉及直线与圆的联立方程。

Another key application is determining the equation of a circle given three points on its circumference. Use the perpendicular bisector method: find the intersection of the perpendicular bisectors of two chords to locate the centre, then calculate the radius.

另一重要应用是已知圆周上三点求圆的方程。使用垂直平分线法：求两条弦的垂直平分线交点确定圆心，再计算半径。

12. Exam Tips and Common Mistakes | 应试技巧与常见错误

When using the gradient formula, ensure you subtract coordinates in the same order for both numerator and denominator. A common slip is writing (y₂ – y₁)/(x₁ – x₂). Always double-check the sign of the gradient when lines are given in a diagram.

使用斜率公式时，确保分子和分母中坐标的相减顺序一致。常见错误是写成 (y₂ – y₁)/(x₁ – x₂)。当题目通过图形给出直线时，务必再次检查斜率的正负。

For circle equations, remember that the radius must be positive. If you obtain the radius squared as a negative number after completing the square, recheck your signs. Also, the point (a, b) in (x – a)² + (y – b)² = r² is the centre; don’t confuse the signs.

对于圆的方程，切记半径必须为正。若配方后半径平方为负数，重新检查符号。此外，(x – a)² + (y – b)² = r² 中的 (a, b) 是圆心，不要搞错符号。

When solving intersection problems, always find the discriminant first to confirm the nature of the intersection. Also, present your final answer in the required form, whether simplified exact values or to a given decimal place. Units of distance are often omitted, but always include them if the question states units.

在求交点问题时，先求判别式以确认交点性质。最终答案按题目要求的形式给出，无论是简化精确值还是指定小数位数。距离单位常被省略，但若题目注明单位则务必写出。

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Coordinate Geometry Key Points Revision | 坐标几何 考点精讲