📚 4 Equations of Lines | 直线方程
In AQA A-Level Mathematics, the equation of a straight line is a central coordinate geometry topic. You need to move confidently between gradient, intercepts, points, and general form, and apply these ideas to parallel and perpendicular lines, intersections, perpendicular bisectors, and linear modelling. This revision guide explains every key form and method, with worked-style notes and exam tips.
在 AQA A-Level 数学中，直线方程是坐标几何的核心主题。你需要在梯度、截距、点和一般式之间灵活转换，并把这些思想应用到平行线、垂直线、交点、垂直平分线以及线性建模中。本复习指南逐一解释每种重要形式和方法，并配有解题式注释与考试技巧。
1. Gradient of a Straight Line | 直线的梯度
The gradient measures the steepness and direction of a line. For two points (x₁, y₁) and (x₂, y₂) on a non-vertical line, the gradient m is found by dividing the change in y by the change in x. A positive gradient rises from left to right; a negative gradient falls from left to right. A horizontal line has gradient 0, and a vertical line has undefined gradient.
梯度衡量直线的陡峭程度和方向。对于非竖直直线上的两点 (x₁, y₁) 和 (x₂, y₂)，梯度 m 等于 y 的变化量除以 x 的变化量。正梯度从左到右上升，负梯度从左到右下降。水平线的梯度为 0，竖直线的梯度无定义。
m = (y₂ − y₁)/(x₂ − x₁)
. This means that for every 1 unit moved to the right, the line moves up 2 units. The order of subtraction must be consistent: use y₂ − y₁ over x₂ − x₁, not a mixture.
如果给定两点 A(2, 3) 和 B(6, 11)，那么 m = (11 − 3)/(6 − 2) = 8/4 = 2。这意味着每向右移动 1 个单位，直线向上移动 2 个单位。减法顺序必须保持一致：分子用 y₂ − y₁，分母用 x₂ − x₁，不要混用。
2. The Form y = mx + c | 斜截式 y = mx + c
s the y-axis at (0, c). This form is especially useful for sketching, because c gives the starting point and m gives the direction.
斜截式 y = mx + c 通常是已知梯度 m 和 y 轴截距 c 时写出直线的最快方式。y 轴截距是 x = 0 时的 y 值，因此直线与 y 轴交于 (0, c)。这种形式对画图尤其有用，因为 c 给出了起点，m 给出了方向。
y = mx + c
as 2y = 4x + 6, divide through by 2 to get y = 2x + 3 before identifying m = 2 and c = 3.
例如，y = 3x − 2 的梯度为 3，y 轴截距为 −2。该直线经过 (0, −2) 并且上升较陡。如果给出的方程是 2y = 4x + 6，先两边除以 2 得到 y = 2x + 3，再确定 m = 2 和 c = 3。
3. Point-Gradient Form y − y₁ = m(x − x₁) | 点斜式 y − y₁ = m(x − x₁)
iven directly. You can expand and rearrange the result into y = mx + c or ax + by + c = 0.
当你知道直线上的一点 (x₁, y₁) 和梯度 m 时，可以使用点斜式 y − y₁ = m(x − x₁)。当没有直接给出 y 轴截距时，这种形式特别有用。你可以将其展开并整理成 y = mx + c 或 ax + by + c = 0。
y − y₁ = m(x − x₁)
s: y = 4x − 3. The same line can be written as 4x − y − 3 = 0 in general form if required.
示例：求过点 (2, 5) 且梯度为 4 的直线方程。代入得到 y − 5 = 4(x − 2)。展开：y − 5 = 4x − 8。两边加 5：y = 4x − 3。如果需要，同一直线也可写成一般式 4x − y − 3 = 0。
4. Finding the Equation from Two Points | 由两点求直线方程
x₂ − x₁), then substitute one of the points into the point-gradient form. You may choose either point; the final equation will be the same. Always simplify the gradient fully before substituting.
如果已知两点，先使用 m = (y₂ − y₁)/(x₂ − x₁) 计算梯度，然后将其中一个点代入点斜式。你可以选择任意一个点，最终的方程相同。代入前务必先将梯度完全化简。
ore y = 2x. Using point B(5, 10) gives y − 10 = 2(x − 5), which also simplifies to y = 2x.
对于 A(1, 2) 和 B(5, 10)，梯度为 m = (10 − 2)/(5 − 1) = 8/4 = 2。使用点 A(1, 2) 得到 y − 2 = 2(x − 1)，所以 y − 2 = 2x − 2，因此 y = 2x。使用点 B(5, 10) 得到 y − 10 = 2(x − 5)，它同样化简为 y = 2x。
5. General Form ax + by + c = 0 | 一般式 ax + by + c = 0
ne side and, if necessary, multiply through to remove fractions. This form is useful for comparing equations and for some geometric problems.
直线也可以写成一般式 ax + by + c = 0，在大多数 AQA 答案中 a、b 和 c 为整数。要从 y = mx + c 转换，把所有项移到一边，如有必要再乘以某个数消去分数。这种形式便于比较方程以及处理某些几何问题。
ax + by + c = 0
at the coefficient of x is positive in final answers unless the question states otherwise.
例如，y = (2/3)x − 4 可以两边乘以 3 得到 3y = 2x − 12。移项得到 2x − 3y − 12 = 0。这里 a = 2，b = −3，c = −12。除非题目另有说明，最终答案中通常保持 x 的系数为正。
6. Parallel and Perpendicular Lines | 平行线与垂直线
m₂ = −1/m₁. This reciprocal and sign change is a very common source of marks in AQA exams.
两条不同的直线如果梯度相同，则平行：m₁ = m₂。如果两条直线的梯度乘积为 −1，则它们互相垂直，这意味着 m₂ = −1/m₁。这种倒数并变号的操作是 AQA 考试中非常常见的得分点。
Parallel: m₁ = m₂
Perpendicular: m₁ × m₂ = −1 or m₂ = −1/m₁
cept. Before comparing gradients, always rearrange each equation into the form y = mx + c.
示例：与 y = 3x + 1 垂直的直线梯度为 −1/3。与 y = 3x + 1 平行的直线梯度为 3，但 y 轴截距不同。在比较梯度之前，务必先把每个方程化为 y = mx + c 的形式。
7. Midpoint and Distance Between Two Points | 中点与两点间距离
Midpoint and distance are not equations of a line themselves, but they are frequently needed in line-equation problems, especially perpendicular bisector questions. For points A(x₁, y₁) and B(x₂, y₂), the midpoint M is the average of the x-coordinates and the average of the y-coordinates.
中点和距离本身不是直线方程，但在直线方程问题中经常用到，尤其是在垂直平分线问题中。对于点 A(x₁, y₁) 和 B(x₂, y₂)，中点 M 是 x 坐标的平均值和 y 坐标的平均值。
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
The distance AB is found using Pythagoras' theorem. In coordinate form, it is the square root of the sum of the squared differences in x and y. This gives the length of the line segment joining the two points.
距离 AB 通过勾股定理求得。在坐标形式中，它是 x 差值的平方与 y 差值的平方之和的平方根。这给出了连接两点的线段长度。
AB = √[(x₂ − x₁)² + (y₂ − y₁)²]
For a perpendicular bisector of AB, first find the midpoint M and the gradient of AB. The perpendicular bisector passes through M and has gradient −1/mAB. Then use y − y₁ = m(x − x₁) with M as the known point.
对于 AB 的垂直平分线，首先求出中点 M 和 AB 的梯度。垂直平分线经过 M，并且梯度为 −1/mAB。然后将 M 作为已知点，使用 y − y₁ = m(x − x₁) 求方程。
8. Intersection of Lines by Solving Simultaneous Equations | 直线的交点与联立方程
set the right-hand sides equal to each other. This gives a single equation in x, which you can solve, then substitute back to find y.
要求两条直线的交点，需要联立它们的方程。如果两个方程都是 y = mx + c 的形式，则令右边的表达式相等。这样得到只含 x 的方程，求出 x 后再代回求 y。
6 and x = 2. Substitute x = 2 into y = 2x + 1 to get y = 5. The lines intersect at (2, 5).
对于 y = 2x + 1 和 y = −x + 7，令 2x + 1 = −x + 7。两边加 x 得到 3x + 1 = 7，所以 3x = 6，x = 2。将 x = 2 代入 y = 2x + 1 得到 y = 5。两直线交于 (2, 5)。
+ 1 into the second equation to obtain 3x + 2x + 1 = 11, so 5x = 10 and x = 2, then y = 5.
如果其中一个方程是一般式，可以先化为 y = mx + c 或使用代入法。例如，对于 y = 2x + 1 和 3x + y = 11，将 y = 2x + 1 代入第二个方程得到 3x + 2x + 1 = 11，所以 5x = 10，x = 2，然后 y = 5。
9. Modelling with Linear Equations | 线性方程建模
Linear models appear in AQA questions as relationships such as cost, temperature, or simple motion. You may be given a rate of change, which is the gradient, and a fixed starting value, which is the y-intercept. Alternatively, you may be given two data points and must construct the equation from them.
线性模型在 AQA 题目中常以成本、温度或简单运动等关系出现。题目可能给出变化率，即梯度，以及固定的初始值，即 y 轴截距。也可能给出两个数据点，需要由它们建立方程。
For instance, a taxi fare has a fixed charge of £2 and charges £ is the cost per mile, and the intercept 2 is the fixed charge. This linear model can predict the cost for different distances.
例如，某出租车收费有 £2 的固定费用，每英里收费 £3。行驶 d 英里的总费用 C 为 C = 3d + 2。梯度 3 是每英里的费用，截距 2 是固定费用。这个线性模型可以预测不同距离的费用。
In modelling questions, interpret the gradient as the rate of change of the dependent variable with respect to the independent variable. Be careful when extrapolating far outside the given data range, because real-world behaviour may not remain linear.
在建模题中，应将梯度解释为因变量相对于自变量的变化率。在给定数据范围之外进行过度外推时要谨慎，因为现实世界中的行为可能不再保持线性。
10. Exam Tips and Common Mistakes | 考试技巧与常见错误
x + c; confusing the y-intercept with the x-intercept; and making sign errors when finding a perpendicular gradient. These errors can be avoided by checking your rearranged equation and substituting the given point.
直线方程题中的常见错误包括：忘记竖直线方程是 x = k，而不是 y = mx + c；混淆 y 轴截距与 x 轴截距；在求垂直梯度时出现符号错误。通过检查整理后的方程并代入给定点，可以避免这些错误。
- Vertical line: x = k; horizontal line: y = c.
- Perpendicular gradient: m₂ = −1/m₁.
- If the question requests the form ax + by + c = 0, rearrange fully and remove fractions.
- Always test a known point in your final equation to confirm it is correct.
In the exam, show each step clearly: state the gradient formula, substitute values, simplify, and write the final equation in the requested form. This earns method marks even if a small arithmetic error occurs later.
考试中要清晰展示每一步：写出梯度公式、代入数值、化简，并按题目要求的形式写出最终方程。即使后面出现小的算术错误，这样做也能获得方法分。
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