📚 IGCSE Maths: Straight Line Graphs and Coordinate Geometry | IGCSE 数学：直线图像与坐标几何
s the gradient and c is the y-intercept. You need to be confident with plotting points, finding gradients, interpreting equations, using parallel and perpendicular gradients, and applying midpoint and distance formulas.
在 IGCSE 数学中，直线图像与坐标几何是一个高频考点，它将代数、几何和实际图像联系起来。核心形式是 y = mx + c，其中 m 表示斜率，c 表示 y 轴截距。你需要熟练掌握描点、求斜率、解读方程、使用平行与垂直斜率，以及应用中点与距离公式。
1. Coordinates and the Cartesian plane | 坐标与笛卡尔平面
Every point on a straight line graph can be described by an ordered pair (x, y). The horizontal axis is the x-axis and the vertical axis is the y-axis. They cross at the origin (0, 0). A point such as (3, −2) means move 3 units to the right and 2 units down.
直线图像上的每个点都可以用有序数对 (x, y) 表示。横轴是 x 轴，纵轴是 y 轴。它们相交于原点 (0, 0)。例如点 (3, −2) 表示向右移动 3 个单位，再向下移动 2 个单位。
The coordinate plane is divided into four quadrants. In the first quadrant both coordinates are positive. In the second quadrant x is negative and y is positive; in the third quadrant both are negative; in the fourth quadrant x is positive and y is negative. Being confident with negative coordinates is essential when plotting straight lines.
坐标平面分为四个象限。第一象限中两个坐标均为正；第二象限中 x 为负、y 为正；第三象限中两者均为负；第四象限中 x 为正、y 为负。在绘制直线时，能够熟练处理负坐标非常重要。
2. Gradient as a measure of steepness | 斜率表示直线的倾斜程度
e ÷ run. If a line rises from left to right, m is positive. If it falls from left to right, m is negative.
斜率 m 表示直线的倾斜程度。它是竖直变化量与水平变化量之比：m = 上升高度 ÷ 水平距离。如果直线从左到右向上倾斜，则 m 为正；如果直线从左到右向下倾斜，则 m 为负。
For two points (x₁, y₁) and (x₂, y₂), the gradient is calculated using the change in y divided by the change in x. Always subtract the coordinates in the same order. A horizontal line has gradient 0, while a vertical line has an undefined gradient.
对于两点 (x₁, y₁) 和 (x₂, y₂)，斜率用 y 的变化量除以 x 的变化量来计算。必须始终按照相同的顺序相减。水平线的斜率为 0，竖直线的斜率未定义。
m = (y₂ − y₁) ÷ (x₂ − x₁)
3. The equation y = mx + c | 直线方程 y = mx + c
ent and c is the y-intercept. The y-intercept is the point where the line crosses the y-axis, so its coordinates are (0, c). You can read c directly from the graph when the line is already drawn.
在方程 y = mx + c 中，m 是斜率，c 是 y 轴截距。y 轴截距是直线与 y 轴相交的点，因此其坐标为 (0, c)。当图像已经画出时，你可以直接从图上读出 c。
t (0, 5). If an equation is not in this form, rearrange it first so that y is the subject.
例如，y = 2x + 3 的斜率为 2，并且与 y 轴相交于 (0, 3)。直线 y = −x + 5 的斜率为 −1，并且与 y 轴相交于 (0, 5)。如果方程不是这种形式，应先将其整理成 y 为主语的形式。
4. Finding gradient from two points | 通过两点求斜率
= (1, 2) and (x₂, y₂) = (4, 8). Then m = (8 − 2) ÷ (4 − 1) = 6 ÷ 3 = 2. The gradient is 2.
例题：一条直线经过 A(1, 2) 和 B(4, 8)。设 (x₁, y₁) = (1, 2)，(x₂, y₂) = (4, 8)。则 m = (8 − 2) ÷ (4 − 1) = 6 ÷ 3 = 2。斜率为 2。
−6 ÷ −3 = 2. The sign rules matter: dividing two negative numbers gives a positive result.
另一个例子：经过 (5, 7) 和 (2, 1)，m = (1 − 7) ÷ (2 − 5) = −6 ÷ −3 = 2。符号规则很重要：两个负数相除得到正数。
5. Finding the y-intercept and the full equation | 求 y 轴截距与完整方程
with m = 2, we have 2 = 2(1) + c, so c = 0. The equation of the line is therefore y = 2x.
一旦知道 m，就可以将一个已知点代入 y = mx + c 来求 c。对于点 A(1, 2) 且 m = 2，我们有 2 = 2(1) + c，所以 c = 0。因此这条直线的方程为 y = 2x。
ur final equation works for both given points. This substitution method is reliable and is often the fastest way in exam questions.
如果图像清楚地与 y 轴相交于 3，那么 c = 3。始终检查最终方程是否适用于两个已知点。这种代入方法非常可靠，在考试题目中通常也是最快的方法。
6. Rearranging into y = mx + c | 整理成 y = mx + c
= 6. Subtract 3x: 2y = −3x + 6. Divide by 2: y = −3x/2 + 3. Therefore m = −3/2 and c = 3.
有些考试题目会给出 ax + by = c 或 ax + by + c = 0 的形式。你必须将它整理成 y = mx + c，才能确定斜率和截距。例如：3x + 2y = 6。先减去 3x：2y = −3x + 6。两边除以 2：y = −3x/2 + 3。因此 m = −3/2，c = 3。
ividing by a negative coefficient, because the signs of every term must be reversed correctly.
另一个例子：4y − 8x = 12 可化为 4y = 8x + 12，再得到 y = 2x + 3。当除以负数系数时要特别小心，因为每一项的符号都必须正确变号。
7. Parallel and perpendicular lines | 平行直线与垂直直线
re parallel because both have m = 3. Their y-intercepts are different, so they never meet.
平行直线具有相同的斜率。直线 y = 3x + 1 和 y = 3x − 5 是平行的，因为两者的 m 都等于 3。它们的 y 轴截距不同，因此永不相交。
If a line has gradient 2, a perpendicular line has gradient −1/2 because 2 × (−1/2) = −1.
垂直直线的斜率相乘等于 −1，也就是说一个斜率是另一个的负倒数。如果一条直线的斜率为 2，那么与它垂直的直线斜率为 −1/2，因为 2 × (−1/2) = −1。
Parallel: m₁ = m₂ | Perpendicular: m₁ × m₂ = −1
| Original gradient m | Perpendicular gradient |
| 2 | −1/2 |
| −3 | 1/3 |
| 1/4 | −4 |
8. Midpoint and length of a line segment | 中点与线段长度
The midpoint of two points A(x₁, y₁) and B(x₂, y₂) is found by averaging the x-coordinates and averaging the y-coordinates. This gives the point exactly halfway between A and B.
两点 A(x₁, y₁) 和 B(x₂, y₂) 的中点可以通过分别求 x 坐标的平均值和 y 坐标的平均值得到。这个点正好位于 A 和 B 的中间。
M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
The length of the segment AB is found by Pythagoras' theorem. For A(1, 2) and B(4, 6), AB = √[(4 − 1)² + (6 − 2)²] = √9 + 16 = √25 = 5.
线段 AB 的长度可以用勾股定理求得。对于 A(1, 2) 和 B(4, 6)，AB = √[(4 − 1)² + (6 − 2)²] = √9 + 16 = √25 = 5。
AB = √[(x₂ − x₁)² + (y₂ − y₁)
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