📚 GCSE AQA Maths: Coordinate Geometry Key Points Revision | GCSE AQA 数学：坐标几何 考点精讲
Coordinate geometry, also known as straight-line graphs, is a key topic in the AQA GCSE Mathematics syllabus. It combines algebra with geometry and appears on both Foundation and Higher tier papers. Mastering the concepts of distance, midpoint, gradient, and equations of lines will help you solve many exam questions confidently.
坐标几何，也称为直线图，是 AQA GCSE 数学大纲中的重要主题。它结合了代数与几何，并出现在基础与高等两个层次的试卷中。掌握距离、中点、斜率和直线方程的概念，将帮助你自信地解答相关考题。
1. The Coordinate Plane | 坐标系
The coordinate plane is formed by the x-axis (horizontal) and y-axis (vertical), intersecting at the origin (0,0). Points are written as (x, y), where x gives the horizontal position and y the vertical position. The plane is divided into four quadrants: Quadrant I (+,+), Quadrant II (−,+), Quadrant III (−,−), and Quadrant IV (+,−).
坐标平面由 x 轴（水平）和 y 轴（垂直）组成，它们相交于原点 (0,0)。点写作 (x, y)，其中 x 表示水平位置，y 表示垂直位置。平面分为四个象限：第一象限 (+ , +)，第二象限 (− , +)，第三象限 (− , −) 和第四象限 (+ , −)。
2. Distance Between Two Points | 两点间距离
The distance d between two points (x₁, y₁) and (x₂, y₂) is found using Pythagoras' theorem:
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
两点 (x₁, y₁) 和 (x₂, y₂) 之间的距离 d 可用勾股定理求得：
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
For example, the distance between (1, 2) and (4, 6) is √[(4−1)² + (6−2)²] = √9 + 16 = √25 = 5. This formula is essential for finding side lengths in triangles and quadrilaterals.
例如，点 (1, 2) 与 (4, 6) 之间的距离为 √[(4−1)² + (6−2)²] = √9 + 16 = √25 = 5。在求三角形和四边形的边长时，这个公式必不可少。
3. Midpoint of a Line Segment | 线段的中点
The midpoint M of the segment joining (x₁, y₁) and (x₂, y₂) is found by averaging the x‑coordinates and y‑coordinates:
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
dpoints. For instance, the midpoint of (2, 5) and (8, 11) is ((2+8)/2, (5+11)/2) = (5, 8).
连接 (x₁, y₁) 与 (x₂, y₂) 的线段中点 M 可通过平均 x 坐标和 y 坐标求得：
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
它给出恰好位于两端点正中间的点。例如，(2, 5) 与 (8, 11) 的中点为 ((2+8)/2, (5+11)/2) = (5, 8)。
4. Gradient of a Straight Line | 直线的斜率
The gradient m measures the steepness of a line. It is calculated as the change in y divided by the change in x:
m = (y₂ − y₁) / (x₂ − x₁)
vertical line has an undefined gradient. For example, the gradient of the line through (1, 2) and (3, 8) is (8−2)/(3−1) = 6/2 = 3.
斜率 m 衡量直线的倾斜程度，计算方式为 y 的变化量除以 x 的变化量：
m = (y₂ − y₁) / (x₂ − x₁)
正斜率表示直线从左到右向上倾斜；负斜率表示向下倾斜。水平线的斜率为 0，垂直线的斜率未定义。例如，过 (1, 2) 和 (3, 8) 两点直线的斜率为 (8−2)/(3−1) = 6/2 = 3。
5. Equation of a Straight Line: y = mx + c | 直线方程：y = mx + c
f a straight line is usually written as y = mx + c, where m is the gradient and c is the y‑intercept (the point where the line crosses the y‑axis). To write the equation from a graph, identify the y‑intercept and calculate the gradient between two clear points.
直线方程通常写作 y = mx + c，其中 m 为斜率，c 为 y 截距（直线与 y 轴的交点）。要从图形写出方程，先确定 y 截距，再选取两个明显的点计算斜率。
If a line cuts the y‑ 1 unit to the right, then m = 2 and c = 4, giving y = 2x + 4. Always check that c is the y‑coordinate where the line meets the y‑axis.
若一条直线与 y 轴交于 (0, 4)，且每向右移动 1 个单位上升 2 个单位，则 m = 2, c = 4，方程为 y = 2x + 4。务必检查 c 是直线与 y 轴交点的 y 坐标。
6. Finding the Equation from Two Points | 由两点求直线方程
m. Then substitute one point into y = mx + c to solve for c. Alternatively, use the point‑gradient form: y − y₁ = m(x − x₁).
若已知直线上两点，先求斜率 m，再将其中一点代入 y = mx + c 求 c。也可使用点斜式：y − y₁ = m(x − x₁)。
7−3)/(4−2) = 2. Using (2,3): 3 = 2(2) + c → c = −1. So equation is y = 2x − 1. Using point‑gradient: y − 3 = 2(x − 2) → y = 2x − 1. Both methods give the same result.
例子：过 (2, 3) 和 (4, 7) 的直线。m = (7−3)/(4−2) = 2。代入 (2,3)：3 = 2(2) + c → c = −1，故方程为 y = 2x − 1。使用点斜式：y − 3 = 2(x − 2) → y = 2x − 1。两种方法结果一致。
7. Parallel and Perpendicular Lines | 平行线与垂直线
product is −1: m₁ × m₂ = −1, or m₂ = −1/m₁. This is a classic AQA Higher-tier requirement.
平行线具有相同的斜率：m₁ = m₂。垂直线的斜率乘积为 −1：m₁ × m₂ = −1，即 m₂ = −1/m₁。这是 AQA 高等层次考试的经典考点。
| Line 1 gradient (m₁) | Line 2 gradient for perpendicular (m₂) |
| 2 | −½ |
| −3 | ⅓ |
| ½ | −2 |
cular to y = 2x + 5 must have gradient −½. A line parallel to y = −¾x + 1 has gradient −¾.
例如，垂直于 y = 2x + 5 的直线斜率必为 −½。平行于 y = −¾x + 1 的直线斜率为 −¾。
8. Intersection of Two Lines | 两条直线的交点
+ 1 = −x + 4 → 3x = 3 → x = 1, substitute back gives y = 3. The lines intersect at (1, 3).
求两条直线的交点需同时解两个方程。例如，y = 2x + 1 与 y = −x + 4。令两式相等：2x + 1 = −x + 4 → 3x = 3 → x = 1，代回得 y = 3。交点坐标为 (1, 3)。
If the lines are parallel, there will be no solution
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