📚 Equations of Lines | 直线方程
In A-Level AQA Mathematics, the equation of a straight line is one of the most fundamental tools in coordinate geometry. It underpins graph sketching, solving simultaneous equations, differentiation (finding tangents and normals), and many examination problems. This article covers the core methods for writing and using line equations with precision and efficiency.
在 A-Level AQA 数学中，直线方程是坐标几何中最基础的工具之一。它是绘制图像、解联立方程、微分（求切线与法线）以及许多考试题目的根基。本文将系统讲解直线方程的书写与运用的核心方法，帮助你在考试中做到精准而高效。
1. The Gradient of a Line | 直线的斜率
The gradient measures how steep a line is. For two points (x₁, y₁) and (x₂, y₂) on a line, the gradient m is defined as the change in y divided by the change in x:
斜率衡量一条直线的倾斜程度。对于直线上的两点 (x₁, y₁) 和 (x₂, y₂)，斜率 m 定义为 y 的变化量除以 x 的变化量：
m = (y₂ − y₁) / (x₂ − x₁)
It is often convenient to write this as m = Δy / Δx, where Δ (delta) means 'change in' downward. A horizontal line has gradient 0, and a vertical line has an undefined gradient.
通常也写成 m = Δy / Δx，其中 Δ（delta）表示“变化量”。斜率为正表示直线从左向右上升；斜率为负表示直线从左向右下降。水平线的斜率为 0，垂直线的斜率不存在（无定义）。
- Positive gradient (m > 0): line increases as x increases. | 正斜率（m > 0）：直线随 x 增大而上升。
- Negative gradient (m < 0): line decreases as x increases. | 负斜率（m < 0）：直线随 x 增大而下降。
- Zero gradient (m = 0): horizontal line. | 零斜率（m = 0）：水平直线。
- Undefined gradient: vertical line. | 斜率无定义：垂直直线。
In AQA examinations, you must take care to substitute coordinates in the correct order and simplify fractions fully.
在 AQA 考试中，务必按正确顺序代入坐标，并将分数化到最简。
2. The Equation y = mx + c | 直线方程 y = mx + c
f a straight-line equation is y = mx + c, where m is the gradient and c is the y-intercept — es the y-axis. This form is ideal when you know the gradient and the y-intercept directly.
最常用的直线方程形式是 y = mx + c，其中 m 为斜率，c 为 y 轴截距，即直线与 y 轴的交点。当已知斜率和 y 轴截距时，这一形式最为直接。
− 2. To check whether a point lies on a line, substitute its coordinates into the equation: if both sides balance, the point is on the line.
例如，斜率为 3、y 轴截距为 −2 的直线方程为 y = 3x − 2。要判断某点是否在直线上，只需将其坐标代入方程：若两边相等，则该点在直线上。
When sketching a line from its equation, find the y-intercept (0, c) and use the gradient to locate a second point; for example, a gradient of 3 means moving 1 unit right and 3 units up.
由方程作图时，先找到 y 轴截距 (0, c)，再利用斜率确定第二个点；例如斜率为 3 表示向右移 1 个单位、再向上移 3 个单位。
3. Finding the Equation from a Point and Gradient | 由一点和斜率求直线方程
If you know the gradient m and a single point (x₁, y₁) on the line, use the point-gradient formula:
若已知斜率 m 和直线上的一点 (x₁, y₁)，可使用点斜式公式：
y − y₁ = m(x − x₁)
gradient. After substituting the known values, expand and rearrange into the required form — usually y = mx + c or ax + by + c = 0.
该公式直接由斜率的定义导出。代入已知值后，展开并整理为题目要求的形式——通常是 y = mx + c 或 ax + by + c = 0。
nt 2 that passes through (3, 5). Solution: y − 5 = 2(x − 3) → y − 5 = 2x − 6 → y = 2x − 1.
例题：求斜率为 2 且经过点 (3, 5) 的直线方程。解：y − 5 = 2(x − 3) → y − 5 = 2x − 6 → y = 2x − 1。
Always test your final equation by substituting the original point to ensure it satisfies the equation.
写出最终方程后，务必代回原定点检验，确保满足方程。
4. Finding the Equation from Two Points | 由两点求直线方程
When two points are given, the strategy is a two-step process: first find the gradient, then use either point in the point-gradient formula.
当给出两个点时，采用两步策略：先求斜率，再用其中任意一点代入点斜式公式。
Given points A(x₁, y₁) and B(x₂, y₂):
已知两点 A(x₁, y₁) 和 B(x₂, y₂)：
m = (y₂ − y₁) / (x₂ − x₁), then y − y₁ = m(x − x₁)
nd (4, 11). First, m = (11 − 2) / (4 − 1) = 9 / 3 = 3. Then y − 2 = 3(x − 1) → y = 3x − 1.
例题：求经过 (1, 2) 和 (4, 11) 两点的直线方程。先求斜率 m = (11 − 2) / (4 − 1) = 9 / 3 = 3。再由 y − 2 = 3(x − 1) → y = 3x − 1。
Be careful with negative coordinates; write out every step to avoid sign errors. AQA examiners frequently award marks for method even when arithmetic slips occur.
注意负坐标的处理；写出完整步骤，避免符号错误。AQA 考官常按方法给分，即使计算有误也能获得过程分。
5. The General Form ax + by + c = 0 | 一般形式 ax + by + c = 0
here a, b and c are integers. This form has the advantage of avoiding fractions and applying uniformly to vertical lines.
AQA 考题有时要求将答案写成一般形式 ax + by + c = 0，其中 a、b、c 为整数。该形式的好处是避免分数，且对垂直直线同样适用。
= 0 or −2x + 3y + 3 = 0. Both are acceptable, but conventionally you may keep a positive.
将 y = mx + c 转化为一般形式时，如有分母先乘以公分母，再将所有项移到同一边。例如 y = (2/3)x − 1 化为 3y = 2x − 3，即 2x − 3y − 3 = 0 或 −2x + 3y + 3 = 0。两种写法均可，但习惯上可保持 a 为正。
is m = −a/b and the y-intercept is −c/b, obtained by rearranging to y = (−a/b)x + (−c/b).
由一般形式整理为 y = (−a/b)x + (−c/b)，可得斜率 m = −a/b，y 轴截距为 −c/b。
6. Parallel and Perpendicular Lines | 平行线与垂直线
Two lines are parallel if they have the same gradient. Two lines are perpendicular if the product of their gradients is −1, provided both gradients are defined. In symbols:
两条直线平行当且仅当它们的斜率相等。两条直线垂直当且仅当它们的斜率之积为 −1（前提是两条直线的斜率均存在）。用符号表示：
m₁ = m₂ (parallel) | m₁ × m₂ = −1 (perpendicular)
he negative reciprocal. For example, a line perpendicular to y = 4x − 7 has gradient −1/4.
若一条直线的斜率为 m，则其垂直线的斜率为 −1/m，通常称为负倒数。例如，与 y = 4x − 7 垂直的直线斜率为 −1/4。
rule m₁ × m₂ = −1 does not apply directly in this case, so remember the geometrical fact.
特殊情况：水平线（m = 0）与垂直线（斜率无定义）相互垂直。此时 m₁ × m₂ = −1 的规则不直接适用，需记住几何事实。
In AQA problems, you will often be asked to find the equation of a line parallel or perpendicular to a given line through a given point. First state the required gradient, then use the point-gradient formula.
在 AQA 题目中，经常要求你求过某点且与已知直线平行或垂直的直线方程。先判断所需斜率，再使用点斜式公式。
7. Intersection of Two Lines | 两直线的交点
The intersection point of two lines is found by solving their equations simultaneously. Since both equations are linear, the solution is unique unless the lines are parallel.
两条直线的交点通过联立方程组求解。由于两个方程均为线性，除非两直线平行，否则解是唯一的。
equation into the other, or use elimination. For example, solve y = 2x + 1 and y = −x + 4
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