📚 IGCSE Maths Unit G6 Coordinate Geometry: Teacher's Guide 550 | IGCSE 数学 G6 单元坐标几何：教师用书 550
In IGCSE Mathematics, coordinate geometry connects algebra with geometry by using coordinates to describe points, lines and their properties on the Cartesian plane. This unit focuses on the midpoint, distance, gradient and equation of a straight line, which are core skills for both the Core and Extended papers. For teachers, it is a rare topic where every formula can be proved from a diagram, so it rewards careful build-up rather than memorisation.
在 IGCSE 数学中，坐标几何通过坐标来描述笛卡尔平面上的点、直线及其性质，将代数与几何联系起来。本单元重点学习中点、两点间距离、直线斜率以及直线方程，这些都是 Core 与 Extended 试卷的核心技能。对教师而言，这是一个难得可以"从图形推出所有公式"的单元，因此教学时应循序渐进地推导，而不是让学生死记硬背。
1. The Cartesian Plane and Ordered Pairs | 笛卡尔平面与有序数对
The Cartesian plane is built from a horizontal x-axis and a vertical y-axis that meet at the origin (0, 0). Any point P is written as an ordered pair (x, y), where x is the horizontal position and y is the vertical position. The word "ordered" matters: (3, 5) and (5, 3) are two different points.
笛卡尔平面由水平的 x 轴和竖直的 y 轴组成，两轴相交于原点 (0, 0)。任一点 P 用有序数对 (x, y) 表示，其中 x 是水平位置，y 是竖直位置。"有序"二字非常关键：(3, 5) 与 (5, 3) 是两个不同的点。
The four quadrants are numbered anticlockwise: quadrant I has positive x and positive y, quadrant II has negative x and positive y, quadrant III has negative x and negative y, and quadrant IV has positive x and negative y. Points on the axes belong to no quadrant.
四个象限按逆时针编号：第一象限 x 和 y 均为正，第二象限 x 为负、y 为正，第三象限 x 和 y 均为负，第四象限 x 为正、y 为负。位于坐标轴上的点不属于任何象限。
A useful teaching routine is to ask students to plot four points, name the quadrant of each, and then move one point across an axis so they see the sign change. Scale choice is also examinable: a grid of 1 cm per unit and a grid of 2 units per centimetre give very different-looking lines, so always state the scale.
一个有效的教学流程是：让学生先描出四个点并说出各自所在象限，再把其中一个点沿坐标轴平移，观察符号的变化。坐标轴的刻度也是考点：每格 1 个单位与每格 2 个单位画出的直线形状差别很大，因此作图时务必标出刻度。
2. Midpoint of a Line Segment | 线段的中点
The midpoint M of a line segment joining A(x₁, y₁) and B(x₂, y₂) is found by averaging the x-coordinates and the y-coordinates. Geometrically, M is the point that divides AB into two equal halves.
连接 A(x₁, y₁) 和 B(x₂, y₂) 的线段中点 M，可以通过对 x 坐标和 y 坐标分别求平均得到。从几何上看，M 就是把 AB 分成相等两段的那个点。
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
4, 7). Notice that the midpoint is exactly halfway in both directions, which is why the same averaging step appears twice.
例如，A(2, 3) 与 B(6, 11) 的中点为 ((2 + 6)/2, (3 + 11)/2) = (4, 7)。可以看到，中点在两个方向上分别都处于正中间，这正是公式中要"平均两次"的原因。
Two extensions are worth flagging. First, if you are given one endpoint and the midpoint, you must work backwards: 2M − A gives the missing endpoint. Second, the same averaging idea generalises to a point dividing a segment in a given ratio, and to finding the centre of a circle from the ends of a diameter.
有两个延伸点值得强调。第一，若已知一个端点和中点，需要反向求解：用 2M − A 即可求出另一个端点。第二，同样的"平均"思想可以推广到按给定比例分割线段的点，也用于由直径两端求圆心坐标。
3. Distance Between Two Points | 两点间的距离
The distance d between two points A(x₁, y₁) and B(x₂, y₂) comes from Pythagoras' theorem. You form a right-angled triangle whose horizontal leg is the difference in x and whose vertical leg is the difference in y, then find the length of the hypotenuse.
两点 A(x₁, y₁) 与 B(x₂, y₂) 之间的距离 d 来自勾股定理。你可以构造一个直角三角形，其水平直角边是 x 坐标之差，竖直直角边是 y 坐标之差，然后求出斜边的长度。
d = √[ (x₂ − x₁)² + (y₂ − y₁)² ]
For A(1, 2) and B(4, 6), d = √[(4 − 1)² + (6 − 2)²] = √9 + 16 = √25 about the order of subtraction, so students should not lose marks by checking signs twice.
对于 A(1, 2) 和 B(4, 6)，d = √[(4 − 1)² + (6 − 2)²] = √9 + 16 = √25 = 5。由于平方会自动消除相减顺序的影响，学生不必反复纠结符号，节省答题时间。
In exact-value questions the answer is often a surd, for example d = √50 = 5√2. Give the surd form unless the question says "give your answer correct to 3 significant figures". Interior diagonal checks, such as showing that a rectangle's diagonals are equal, rely on this formula twice.
在要求精确值的题目中，答案常常是根式，例如 d = √50 = 5√2。除非题目要求"保留三位有效数字"，否则应给出根式形式。检验矩形对角线相等这一类问题，需要把这个公式用两次。
4. Gradient of a Straight Line | 直线的斜率
The gradient (slope) of a straight line measures how steeply it rises or falls. It is the change in y divided by the change in x between any two points on the line.
直线的斜率（梯度）用来衡量它上升或下降的陡峭程度，等于直线上任意两点之间 y 的变化量除以 x 的变化量。
m = (y₂ − y₁) / (x₂ − x₁) = Δy / Δx = rise / run
0, and a vertical line has an undefined gradient because Δx = 0 would force division by zero.
对于 A(1, 2) 和 B(4, 6)，m = (6 − 2)/(4 − 1) = 4/3。正斜率从左向右上升，负斜率从左向右下降，水平线斜率为 0，而竖直线的斜率不存在，因为此时 Δx = 0 会导致除以零。
tudents should be able to move between the fraction, the decimal and the angle, and to explain why a line with m = 1 makes 45° with the x-axis.
斜率还等于直线与 x 轴正方向所成角 θ 的正切值，即 m = tan θ，且满足 tan θ = 升高量 / 水平量。学生应能在分数、小数和角度之间自由转换，并能解释为什么 m = 1 的直线与 x 轴成 45°。
5. Equation of a Straight Line: y = mx + c | 直线方程 y = mx + c
m is the gradient and c is the y-intercept, that is, the y-value where the line crosses the y-axis. This is called the gradient-intercept form.
每一条非竖直直线都可以写成 y = mx + c 的形式，其中 m 是斜率，c 是 y 轴截距，也就是直线与 y 轴交点的纵坐标。这个形式称为斜截式。
rcept, set y = 0 and solve: 3x − 4 = 0, so x = 4/3. The x-intercept is therefore (4/3, 0).
例如，直线 y = 3x − 4 的斜率为 3，且经过点 (0, −4)。求 x 轴截距时，令 y = 0 并求解：3x − 4 = 0，得 x = 4/3，因此 x 轴截距为 (4/3, 0)。
2/3)x + 4, so the gradient is −2/3 and the y-intercept is 4. Rearranging without error is a skill in its own right.
直线有时以一般式 ax + by + c = 0 给出。要读出斜率，需要把 y 表示为自变量。例如 2x + 3y = 12 化为 y = −(2/3)x + 4，斜率为 −2/3，y 轴截距为 4。准确变形本身就是一项需要训练的技能。
6. Finding the Equation Through Two Points | 求过两点的直线方程
ne point and the gradient into y = mx + c and solve for c; finally write the equation with the values of m and c.
求过两个已知点的直线方程，分三步：先用 m = (y₂ − y₁)/(x₂ − x₁) 求出斜率；再把其中一个点和斜率代入 y = mx + c，解出 c；最后把 m 与 c 代回写出方程。
A faster route uses the point-gradient form, which is simply the gradient definition rearranged:
更快捷的方法是用点斜式，它本质上就是斜率公式的变形：
y − y₁ = m(x − x₁)
n general form. Both the fraction and the integer form are acceptable unless the question specifies otherwise.
例题：求经过 A(1, 2) 和 B(4, 6) 的直线。斜率为 4/3，于是 y − 2 = (4/3)(x − 1)，整理得 y = (4/3)x + 2/3，一般式为 4x − 3y + 2 = 0。除题目特别说明外，分数形式与整式形式都算正确。
Teachers should insist on a final check: substitute both original points back into the equation. If either fails, the error is almost always in the sign of the gradient or in the arithmetic for c.
教师应要求学生养成最后验算的习惯：把两个原始点分别代回方程。只要有一个不成立，错误几乎总是出在斜率的符号或求 c 时的运算上。
7. Parallel and Perpendicular Lines | 平行线与垂直直线
o their equations differ only in the constant term, for example y = 2x + 1 and y = 2x − 7.
两条直线平行的充要条件是斜率相同：m₁ = m₂。平行线永不相交，因此它们的方程只在常数项上不同，例如 y = 2x + 1 与 y = 2x − 7。
m₁. The gradient of a perpendicular line is the negative reciprocal of the original gradient.
两条直线垂直的充要条件是斜率之积为 −1：m₁ × m₂ = −1，等价地 m₂ = −1/m₁。垂直直线的斜率是原直线斜率的负倒数。
m₁ = m₂ (parallel) · m₁ × m₂ = −1 (perpendicular)
dient 0; they are perpendicular to each other even though the product rule cannot be applied. Examiners like this case because it tests understanding rather than formula recall.
一个常见陷阱涉及竖直线与水平线。像 x = 5 这样的竖直线斜率不存在，而像 y = 3 这样的水平线斜率为 0；虽然无法使用乘积法则，它们仍然互相垂直。考官偏爱这一情形，因为它考查的是理解而非公式记忆。
8. The 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. It has two defining properties: it cuts AB into two equal parts, and it meets AB at a right angle.
线段 AB 的垂直平分线，是经过 AB 中点且与 AB 垂直的直线。它有两个定义性质：把 AB 分成相等的两部分，并与 AB 成直角相交。
endicular gradient, then substitute into y − y₁ = m(x − x₁) with the midpoint coordinates.
求它的方程：先用中点公式求中点，再求 AB 的斜率，取其负倒数作为垂直方向的新斜率，最后把中点坐标代入 y − y₁ = m(x − x₁)。
is −1/2, so the perpendicular bisector is y − 7 = −(1/2)(x − 4), that is y = −(1/2)x + 9.
例题：对于 A(2, 3) 与 B(6, 11)，中点为 (4, 7)，AB 的斜率为 8/4 = 2。垂直方向的斜率为 −1/2，因此垂直平分线为 y − 7 = −(1/2)(x − 4)，即 y = −(1/2)x + 9。
This construction is the geometric heart of the unit. Any point on the perpendicular bisector is equidistant from A and B, which is exactly why it is used to find the centre of a circle through three points: intersect two perpendicular bisectors and you have the circumcentre.
这一作图是本单元的几何核心。垂直平分线上的任意一点到 A 与 B 的距离相等，这正是用它求"过三点的圆的圆心"的原因：两条垂直平分线的交点即为外心。
9. Sketching and Interpreting Straight Lines | 直线的作图与解读
A sketch needs three things: the correct gradient sign and steepness, the y-intercept marked, and at least one more point to fix the line. A reliable method is to plot the two intercepts, since the x-intercept and y-intercept are quick to compute.
画草图需要三点：斜率符号与陡峭程度正确、标出 y 轴截距、再找至少一个点来确定直线。一个可靠的方法是描出两个截距，因为 x 轴截距与 y 轴截距都很容易算出。
) and (4/3, 0). Drawing a straight edge through them gives an accurate sketch. Always label the line with its equation so the marker can see which line is which when several appear on one grid.
对于 y = 3x − 4，两个截距分别是 (0, −4) 和 (4/3, 0)。用直尺连接这两点即可得到准确的草图。务必在直线旁标注其方程，这样当同一网格上出现多条直线时，阅卷人能清楚分辨。
Interpretation questions ask what the gradient and intercept mean in context. If y is the total cost in dollars and x is the number of hours, then the gradient is the hourly rate and the intercept is the fixed call-out charge. Reading a real graph in this way appears regularly in Extended problem-solving papers.
解读类问题会问斜率和截距在情境中的含义。若 y 表示总费用（元），x 表示小时数，则斜率是每小时费率，截距是固定上门费。这种把图象"翻译"回现实情境的能力，在 Extended 应用题中经常出现。
10. Coordinate Geometry in Context | 坐标几何的实际应用
Coordinate geometry is the standard tool for proving properties of shapes placed on a grid. To show a quadrilateral is a parallelogram, verify that both pairs of opposite sides have equal gradients; to show it is a rectangle, add the perpendicular check on adjacent sides.
坐标几何是证明网格上图形性质的标准工具。要证明一个四边形是平行四边形，可验证两组对边的斜率分别相等；要证明它是矩形，还需再验证相邻两边互相垂直。
To show a triangle is isosceles, use the distance formula on all three sides and look for two equal lengths. To show it is right-angled, either show two gradients multiply to −1, or show that Pythagoras' theorem holds for the three side lengths squared.
要证明一个三角形是等腰三角形，可用距离公式计算三边长度，找出两条相等的边。要证明它是直角三角形，可以证明两条边的斜率乘积为 −1，或验证三边长度的平方满足勾股定理。
ide is axis-parallel, the shoelace method or a surrounding rectangle gives the answer. Circles are also common: the equation (x − a)² + (y − b)² = r² has centre (a, b) and radius r, and the distance formula is exactly what proves it.
若三角形有一条水平或竖直的底边，面积很好求：面积 = ½ × 底 × 对应高，可直接从坐标读出。若没有边与坐标轴平行，可用"鞋带公式"或用外接矩形相减求面积。圆也很常见：方程 (x − a)² + (y − b)² = r² 的圆心为 (a, b)，半径为 r，而距离公式正是它的证明依据。
11. Common Errors and Exam Technique | 常见错误与应试技巧
The most frequent loss of marks comes from mixing up the order of operations in the gradient formula. Remember that both differences must follow the same order: if you write (y₂ − y₁) on top, you must write (x₂ − x₁) underneath, never (x₁ − x₂).
最常见的失分点是斜率公式中的相减顺序混乱。记住：两个差必须保持同样的顺序——分子写 (y₂ − y₁)，分母就必须写 (x₂ − x₁)，绝不能写成 (x₁ − x₂)。
A second frequent error is forgetting that the perpendicular gradient is the negative reciprocal, so the sign must flip as well as the fraction. A third is answering the distance question with the squared distance, forgetting the square root at the end.
第二个常见错误是忘记垂直方向的斜率要取负倒数，因此不仅分数要倒过来，符号也必须改变。第三个错误是把"距离的平方"当作距离作答，忘了最后开平方。
- Show your working. Method marks are awarded even when the arithmetic slips, so write the formula, the substitution and the answer on separate lines. / 写出过程。即使算错，方法分仍然可得，因此把公式、代入、答案分行书写。
- Check the form requested. If the question says "give your answer in the form y = mx + c", an answer as 4x − 3y + 2 = 0 may lose the final mark. / 注意题目要求的形式。若要求写成 y = mx + c，答成 4x − 3y + 2 = 0 可能丢掉最后一分。
- Watch the rounding. Never round the gradient before using it; keep the exact fraction until the final line. / 注意取整。切勿在使用前先把斜率四舍五入，应保留精确分数直到最后一步。
- Label coordinates. Writing A(1, 2) rather than A(1, 2, 3) or missing brackets costs accuracy marks. / 规范书写坐标。漏写括号或多写数字都会造成无谓失分。
- Do not forget units. In context questions, the gradient may carry units such as dollars per hour. / 不要漏单位。在情境题中，斜率可能带有单位，例如"元/小时"。
A quick way to see how the two tiers differ in demand is set out below.
下面这张表可以快速看出两个层级在要求上的差异。
| Skill / 技能 | Core / 核心卷 | Extended / 拓展卷 |
|---|---|---|
| Midpoint / 中点 | Direct substitution / 直接代入 | Reverse problems, ratio division / 反向求解、按比例分割 |
| Distance / 距离 | Integer answers usually / 通常为整数答案 | Surds, proofs of shape type / 根式、图形性质证明 |
| Gradient / 斜率 | Read from a grid / 从网格读取 | Algebraic, with unknowns / 含未知数的代数运算 |
| Line equation / 直线方程 | y = mx + c given m or c / 已知 m 或 c | Through two points, perpendiculars / 过两点、垂直平分线 |
12. Practice Questions with Worked Answers | 练习题与详解
−3). Midpoint = ((−2 + 4)/2, (5 + (−3))/2) = (1, 1). Length = √[(4 − (−2))² + (−3 − 5)²] = √36 + 64 = √100 = 10.
第 1 题：求连接 P(−2, 5) 与 Q(4, −3) 的线段中点与长度。中点 = ((−2 + 4)/2, (5 + (−3))/2) = (1, 1)。长度 = √[(4 − (−2))² + (−3 − 5)²] = √36 + 64 = √100 = 10。
1) gives y = (4/3)x + 2/3. Setting y = 0 gives x = −1/2, so the x-intercept is (−1/2, 0).
第 2 题：一条直线经过 (1, 2)，斜率为 4/3。写出它的方程并求 x 轴截距。由 y − 2 = (4/3)(x − 1) 得 y = (4/3)x + 2/3。令 y = 0 得 x = −1/2，故 x 轴截距为 (−1/2, 0)。
4), giving y = −(1/2)x + 9. This is the same perpendicular bisector computed in Section 8.
第 3 题：求与 y = 2x + 1 垂直且经过 (4, 7) 的直线方程。垂直斜率为 −1/2，故 y − 7 = −(1/2)(x − 4)，得 y = −(1/2)x + 9。这与第 8 节算出的垂直平分线完全一致。
stion 4: Show that A(0, 0), B(3, 4), C(7, 1) form a right isosceles triangle. AB = 5, BC = √16 + 9 = 5, AC = √49 + 1 = √50 = BC and AB² + BC² = 25 + 25 = 50 = AC², the triangle is right-angled at B and isosceles.
第 4 题：证明 A(0, 0)、B(3, 4)、C(7, 1) 构成等腰直角三角形。AB = 5，BC = √16 + 9 = 5，AC = √49 + 1 = √50。由于 AB = BC 且 AB² + BC² = 25 + 25 = 50 = AC²，该三角形在 B 处为直角，且为等腰三角形。
Question 5: The line y = kx + 3 passes through (2,
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