📚 Equations of Lines | 直线方程
In AQA A-Level Mathematics, the equations of straight lines are foundational. You will use them in coordinate geometry, calculus, modelling and many mixed problems. This revision guide covers gradient, standard forms, parallel and perpendicular lines, midpoints, intersections, distances, sketching and common exam pitfalls.
在 AQA A-Level 数学中，直线方程是基础内容。你在坐标几何、微积分、建模以及许多综合题中都会用到它。本复习指南涵盖斜率、标准形式、平行线与垂直线、中点、交点、距离、绘图以及常见考试失分点。
1. Gradient of a straight line | 直线斜率
The gradient, often denoted by m, measures how steep a line is. For two points (x₁, y₁) and (x₂, y₂), the gradient is the change in y divided by the change in x:
斜率通常记作 m，用来衡量直线的倾斜程度。对于两点 (x₁, y₁) 和 (x₂, y₂)，斜率等于 y 的变化量除以 x 的变化量：
m = (y₂ − y₁) / (x₂ − x₁)
A positive gradient means the line rises from left to right. A negative gradient means it falls. A horizontal line has gradient 0, and a vertical line has an undefined gradient because the denominator would be zero.
斜率为正表示直线从左到右上升；斜率为负表示直线下降。水平线的斜率为 0；竖直线的斜率不存在，因为分母为零。
2. Forms of the equation of a line | 直线方程的形式
e gradient-intercept form is y = mx + c, where m is the gradient and c is the y-intercept.
你需要熟练掌握三种主要形式。斜截式为 y = mx + c，其中 m 是斜率，c 是直线在 y 轴上的截距。
y = mx + c
= m(x − x₁). It is especially useful when you know one point on the line and the gradient.
点斜式为 y − y₁ = m(x − x₁)。当你知道直线上一点和斜率时，这种形式特别有用。
y − y₁ = m(x − x₁)
a, b and c are integers in many exam answers. You may be asked to rearrange into this form.
一般式为 ax + by + c = 0，考试答案中 a、b、c 通常要求为整数。你可能需要把方程整理成这种形式。
ax + by + c = 0
3. Finding the equation from two points | 已知两点求方程
using m = (y₂ − y₁)/(x₂ − x₁). Then substitute one of the points into y − y₁ = m(x − x₁).
要求经过两点的直线方程，先利用 m = (y₂ − y₁)/(x₂ − x₁) 计算斜率，然后把其中一个点代入 y − y₁ = m(x − x₁)。
the gradient is (8 − 2)/(3 − 1) = 3. Using point A gives y − 2 = 3(x − 1), so y = 3x − 1.
例如，已知 A(1, 2) 和 B(3, 8)，斜率为 (8 − 2)/(3 − 1) = 3。代入点 A 得 y − 2 = 3(x − 1)，整理后为 y = 3x − 1。
m = (8 − 2) / (3 − 1) = 3
n by substituting the second point. For B(3, 8), 3(3) − 1 = 8, so the equation is correct.
最后务必把第二个点代入检验。对于 B(3, 8)，3(3) − 1 = 8，所以方程正确。
4. Parallel and perpendicular lines | 平行线与垂直线
es have the same gradient. If line L₁ has gradient m₁ and line L₂ has gradient m₂, then L₁ ∥ L₂ means m₁ = m₂.
平行直线的斜率相等。如果直线 L₁ 的斜率为 m₁，直线 L₂ 的斜率为 m₂，那么 L₁ ∥ L₂ 意味着 m₁ = m₂。
m₁ = m₂
−1. This is the negative reciprocal relationship: if one gradient is m, the perpendicular gradient is −1/m.
垂直直线满足 m₁ × m₂ = −1。这就是负倒数关系：如果一条直线斜率为 m，那么与它垂直的直线斜率为 −1/m。
m₁ × m₂ = −1
Example: A line with gradient 2 is perpendicular to a line with gradient −1/2. Be careful with signs when you form the negative reciprocal.
例如，斜率为 2 的直线与斜率为 −1/2 的直线垂直。在求负倒数时要注意符号。
5. Midpoints and perpendicular bisectors | 中点与垂直平分线
The midpoint of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). This gives the average of the x-coordinates and the average of the y-coordinates.
两点 (x₁, y₁) 和 (x₂, y₂) 的中点坐标是 ((x₁ + x₂)/2, (y₁ + y₂)/2)，即两个 x 坐标的平均值和两个 y 坐标的平均值。
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
A perpendicular bisector passes through the midpoint of a segment and is perpendicular to that segment. First find the midpoint, then the gradient of the original line, and finally use the negative reciprocal gradient with the point-gradient form.
垂直平分线经过线段的中点，并且与线段垂直。解题时先求中点，再求原线段所在直线的斜率，最后用负倒数作斜率代入点斜式。
is 1/2, so the perpendicular bisector has gradient −2. Its equation is y − 3 = −2(x − 3).
例如，已知 A(1, 2) 和 B(5, 4)，中点为 (3, 3)。AB 的斜率为 1/2，所以垂直平分线的斜率为 −2。其方程为 y − 3 = −2(x − 3)。
6. Intersection of two lines | 两直线交点
The point where two lines intersect satisfies both equations. You can solve the simultaneous equations by substitution or elimination.
两条直线的交点同时满足两个方程。你可以通过代入法或消元法解联立方程组。
1 = −x + 7, so 3x = 6 and x = 2. Substitute back to get y = 5. The intersection is (2, 5).
例如，解 y = 2x + 1 和 y = −x + 7。令两者相等：2x + 1 = −x + 7，所以 3x = 6，x = 2。代回得 y = 5。交点为 (2, 5)。
2x + 1 = −x + 7 ⇒ x = 2
Always state the intersection as a coordinate pair. Writing only "x = 2" does not fully answer the question.
交点务必用坐标表示。只写 "x = 2" 不能完整回答问题。
7. Distance between two points | 两点间的距离
The distance between two points (x₁, y₁) and (x₂, y₂) is given by Pythagoras' theorem: d = √[(x₂ − x₁)² + (y₂ − y₁)²].
两点 (x₁, y₁) 和 (x₂, y₂) 之间的距离由勾股定理给出：d = √[(x₂ − x₁)² + (y₂ − y₁)²]。
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
Example: For A(1, 2) and B(4, 6), the distance is √[(4 − 1)² + (6 − 2)²] = √9 + 16 = √25 = 5.
例如，已知 A(1, 2) 和 B(4, 6)，距离为 √[(4 − 1)² + (6 − 2)²] = √9 + 16 = √25 = 5。
This formula is frequently combined with gradient or circle questions, so keep it at the front of your working.
该公式经常与斜率或圆的题目结合，因此要熟练掌握。
8. Distance from a point to a line | 点到直线的距离
The perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is d = |a x₁ + b y₁ + c| / √a² + b². The numerator uses the absolute value because distance cannot be negative.
点 (x₁, y₁) 到直线 ax + by + c = 0 的垂直距离为 d = |a x₁ + b y₁ + c| / √a² + b²。分子使用绝对值，因为
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