Straight Line Graphs | 直线图

📚 Straight Line Graphs | 直线图

Straight line graphs are a core part of the Edexcel A-Level Mathematics specification. They bring together algebra, coordinate geometry and problem solving, and they appear in both pure mathematics and applied contexts such as kinematics. This revision guide covers gradient, different forms of the equation of a line, parallel and perpendicular conditions, intersections, distance and midpoint, inequalities, and common exam techniques.

直线图是 Edexcel A-Level 数学考纲中的核心内容。它们将代数、坐标几何和问题求解结合在一起，在纯数学以及运动学等应用背景中都会出现。本复习指南涵盖斜率、直线方程的不同形式、平行与垂直条件、交点、距离与中点、不等式以及常见考试技巧。

1. Gradient of a Straight Line | 直线斜率

The gradient m of a straight line joining two points (x₁, y₁) and (x₂, y₂) is given by the change in y divided by the change in x: m = (y₂ − y₁) ÷ (x₂ − x₁). Gradient measures steepness and direction; a positive gradient slopes upwards from left to right, while a negative gradient slopes downwards.

连接两点 (x₁, y₁) 和 (x₂, y₂) 的直线斜率 m 等于 y 的变化量除以 x 的变化量：m = (y₂ − y₁) ÷ (x₂ − x₁)。斜率衡量倾斜程度和方向；正斜率从左到右向上倾斜，负斜率从左到右向下倾斜。

If the denominator x₂ − x₁ is zero, the line is vertical and its gradient is undefined. A horizontal line has gradient 0. For Edexcel questions, you may be asked to find the gradient from a diagram, from two coordinates, or from an equation by rearranging into the form y = mx + c.

如果分母 x₂ − x₁ 为零，则直线是竖直的，其斜率无定义。水平线的斜率为 0。在 Edexcel 考题中，你可能需要从图像、两个坐标或通过将方程改写为 y = mx + c 的形式来求斜率。

Example: For A(2, 3) and B(5, 11), m = (11 − 3) ÷ (5 − 2) = 8 ÷ 3 = 8/3. The line rises 8 units for every 3 units horizontally.

例如：对于 A(2, 3) 和 B(5, 11)，m = (11 − 3) ÷ (5 − 2) = 8 ÷ 3 = 8/3。该直线每水平移动 3 个单位就上升 8 个单位。

2. Forms of the Equation of a Straight Line | 直线方程的形式

The most common form is y = mx + c, where m is the gradient and c is the y-intercept, the value of y when x = 0. This is often called the slope-intercept form. It is useful for sketching graphs quickly and for identifying parallel lines.

最常见的形式是 y = mx + c，其中 m 是斜率，c 是 y 轴截距，即 x = 0 时的 y 值。这通常称为斜截式。它有助于快速画出图像并识别平行直线。

Another important form is the point-slope form: y − y₁ = m(x − x₁). This is used when you know the gradient and one point on the line. A third form is the general form ax + by + c = 0, where a, b and c are integers. Edexcel often expects answers in the form ax + by + c = 0 with integer coefficients.

另一种重要形式是点斜式：y − y₁ = m(x − x₁)。当你知道斜率和直线上的一点时使用这种形式。第三种形式是一般式 ax + by + c = 0，其中 a、b 和 c 是整数。Edexcel 通常要求答案写成 ax + by + c = 0 的形式，且系数为整数。

You should be able to convert between these forms. For example, y = 2x − 3 can be written as 2x − y − 3 = 0. Similarly, 3x + 4y − 12 = 0 can be rearranged to y = −3/4 x + 3, showing gradient −3/4 and y-intercept 3.

你应该能够在这些形式之间转换。例如，y = 2x − 3 可以写成 2x − y − 3 = 0。同样，3x + 4y − 12 = 0 可以改写为 y = −3/4 x + 3，显示出斜率为 −3/4，y 轴截距为 3。

3. Finding the Equation from Two Points | 由两点求直线方程

To find the equation of a line through two given points, first calculate the gradient using m = (y₂ − y₁) ÷ (x₂ − x₁). Then substitute one of the points into y − y₁ = m(x − x₁). Finally rearrange into the required form, such as ax + by + c = 0.

要求经过两个已知点的直线方程，首先用 m = (y₂ − y₁) ÷ (x₂ − x₁) 计算斜率。然后将其中一个点代入 y − y₁ = m(x − x₁)。最后整理为所需形式，如 ax + by + c = 0。

Worked example: Find the line through P(1, 4) and Q(3, 10). The gradient is (10 − 4) ÷ (3 − 1) = 6 ÷ 2 = 3. Using P(1, 4): y − 4 = 3(x − 1). Expanding gives y − 4 = 3x − 3, so y = 3x + 1. In general form this is 3x − y + 1 = 0.

例题：求经过 P(1, 4) 和 Q(3, 10) 的直线。斜率为 (10 − 4) ÷ (3 − 1) = 6 ÷ 2 = 3。使用 P(1, 4)：y − 4 = 3(x − 1)。展开得 y − 4 = 3x − 3，所以 y = 3x + 1。写成一般式为 3x − y + 1 = 0。

Be careful when the gradient is a fraction. For example, through (2, 5) and (6, 7), m = 2 ÷ 4 = 1/2. Using (2, 5): y − 5 = 1/2 (x − 2). Multiplying both sides by 2 gives 2y − 10 = x − 2, so x − 2y + 8 = 0.

当斜率是分数时要小心。例如，经过 (2, 5) 和 (6, 7)，m = 2 ÷ 4 = 1/2。使用 (2, 5)：y − 5 = 1/2 (x − 2)。两边乘以 2 得 2y − 10 = x − 2，所以 x − 2y + 8 = 0。

4. Parallel and Perpendicular Lines | 平行与垂直直线

Two distinct lines are parallel if and only if they have the same gradient. For example, y = 2x + 1 and y = 2x − 5 are parallel because both have m = 2. In general form, the lines ax + by + c = 0 and ax + by + d = 0 are parallel if c ≠ d.

两条不同的直线平行当且仅当它们的斜率相同。例如，y = 2x + 1 和 y = 2x − 5 平行，因为两者的斜率都为 m = 2。在一般形式中，直线 ax + by + c = 0 和 ax + by + d = 0 在 c ≠ d 时平行。

Two lines are perpendicular if the product of their gradients is −1. If one line has gradient m, a perpendicular line has gradient −1/m. For example, if a line has gradient 3, any perpendicular line has gradient −1/3. Vertical and horizontal lines are also perpendicular: a vertical line x = k is perpendicular to a horizontal line y = c.

两条直线垂直，如果它们的斜率乘积为 −1。如果一条直线的斜率是 m，那么与之垂直的直线斜率为 −1/m。例如，如果一条直线的斜率是 3，任何与之垂直的直线斜率都是 −1/3。竖直线和水平线也互相垂直：竖直线 x = k 垂直于水平线 y = c。

Typical Edexcel question: The line L has equation 4x − 2y + 7 = 0. Find the gradient of a line perpendicular to L. First rewrite as y = 2x + 7/2, so m = 2. Perpendicular gradient is −1/2.

典型的 Edexcel 题目：直线 L 的方程为 4x − 2y + 7 = 0。求与 L 垂直的直线的斜率。首先改写为 y = 2x + 7/2，所以 m = 2。垂直直线的斜率为 −1/2。

5. Midpoint and Distance Between Two Points | 中点与两点间距离

The midpoint M of two points (x₁, y₁) and (x₂, y₂) has coordinates ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2). This is the average of the x-coordinates and the average of the y-coordinates. Midpoints are often used in coordinate geometry proofs and when finding the equation of a perpendicular bisector.

两点 (x₁, y₁) 和 (x₂, y₂) 的中点 M 的坐标为 ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)。这是 x 坐标的平均值和 y 坐标的平均值。中点在坐标几何证明和求垂直平分线方程时经常使用。

The distance d between two points is found using Pythagoras’ theorem: d = √[(x₂ − x₁)² + (y₂ − y₁)²]. This formula gives the length of the straight line segment joining the points. It is essential when working with circles or calculating lengths in geometric problems.

两点之间的距离 d 用勾股定理求得：d = √[(x₂ − x₁)² + (y₂ − y₁)²]。该公式给出连接两点的直线段的长度。在处理圆或计算几何问题中的长度时，这是必不可少的。

Example: For A(1, 2) and B(5, 10), midpoint is ((1 + 5) ÷ 2, (2 + 10) ÷ 2) = (3, 6). Distance AB = √[(5 − 1)² + (10 − 2)²] = √(16 + 64) = √80 = 4√5.

例如：对于 A(1, 2) 和 B(5, 10)，中点为 ((1 + 5) ÷ 2, (2 + 10) ÷ 2) = (3, 6)。距离 AB = √[(5 − 1)² + (10 − 2)²] = √(16 + 64) = √80 = 4√5。

A perpendicular bisector of a line segment passes through the midpoint and has gradient equal to the negative reciprocal of the segment’s gradient. This combines midpoint and perpendicular gradient skills frequently tested in Edexcel papers.

线段的垂直平分线经过中点，且其斜率等于该线段斜率的负倒数。这结合了中点和垂直斜率技能，在 Edexcel 试卷中经常考查。

6. Intersection of Two Lines | 两直线交点

To find where two straight lines intersect, solve their equations simultaneously. If the lines are given in the form y = m₁x + c₁ and y = m₂x + c₂, set them equal: m₁x + c₁ = m₂x + c₂. Solve for x, then substitute back to find y.

要求两条直线的交点，需要联立求解它们的方程。如果直线以 y = m₁x + c₁ 和 y = m₂x + c₂ 的形式给出，令它们相等：m₁x + c₁ = m₂x + c₂。解出 x，然后代回求出 y。

If the lines are in general form ax + by = c, you may use substitution or elimination. The intersection point (x, y) must satisfy both equations. If the two lines are parallel, there is no intersection; if they are the same line, there are infinitely many intersection points.

如果直线是一般式 ax + by = c，可以使用代入法或消元法。交点 (x, y) 必须同时满足两个方程。如果两条直线平行，则没有交点；如果它们是同一条直线，则有无穷多个交点。

Example: Find the intersection of y = 2x + 1 and y = −x + 7. Set 2x + 1 = −x + 7, so 3x = 6, x = 2. Substitute into either equation: y = 2(2) + 1 = 5. The intersection is (2, 5).

例如：求 y = 2x + 1 和 y = −x + 7 的交点。令 2x + 1 = −x + 7，得 3x = 6，x = 2。代入任一方程：y = 2(2) + 1 = 5。交点为 (2, 5)。

7. Inequalities and Regions | 不等式与区域

A linear inequality such as y > 2x + 1 describes a region on one side of the boundary line y = 2x + 1. The boundary is dashed if the inequality is strict (> or <) and solid if it includes equality (≥ or ≤). To identify the correct side, test a point that is not on the line, usually (0, 0) if possible.

线性不等式如 y > 2x + 1 描述的是边界线 y = 2x + 1 一侧的区域。如果不等式是严格的（> 或 <），边界线为虚线；如果包含等号（≥ 或 ≤），则为实线。要确定正确的一侧，可以测试一个不在直线上的点，通常如果可能，选择 (0, 0)。

Systems of linear inequalities represent the intersection of several regions. In Edexcel questions, you may be asked to shade the region satisfying multiple inequalities, or to write down inequalities from a shaded graph. Always label the boundary lines and test points carefully.

线性不等式组表示多个区域的交集。在 Edexcel 题目中，你可能需要给满足多个不等式的区域涂上阴影，或者根据阴影图像写出不等式。务必仔细标记边界线并测试点。

Example: For y ≥ x − 2 and y < −x + 4, the boundary y = x − 2 is solid, and y = −x + 4 is dashed. The region satisfying both lies between the two lines, including the first boundary but not the second.

例如：对于 y ≥ x − 2 和 y < −x + 4，边界 y = x − 2 是实线，y = −x + 4 是虚线。满足两者的区域位于两条直线之间，包括第一条边界但不包括第二条边界。

8. Modelling with Straight Lines | 直线模型

Straight line graphs are used to model real-world situations where one quantity changes at a constant rate with respect to another. Examples include temperature conversion, simple interest, distance-time graphs for constant speed, and linear cost functions. The gradient represents the rate of change, and the intercept represents the initial value.

直线图用于对一个量相对于另一个量以恒定速率变化的现实情境进行建模。例子包括温度换算、单利、匀速运动的距离-时间图以及线性成本函数。斜率表示变化率，截距表示初始值。

In a distance-time graph, the gradient gives speed. A steeper line means a higher speed. In a cost model C = 50 + 20x, the fixed cost is 50 and each unit costs 20. Understanding the meaning of gradient and intercept in context is a key Edexcel skill.

在距离-时间图中，斜率表示速度。直线越陡，速度越快。在成本模型 C = 50 + 20x 中，固定成本为 50，每单位成本为 20。在具体情境中理解斜率和截距的含义是 Edexcel 的关键技能。

You may be given two data points and asked to build a linear model. Find the gradient from the two points, then use point-slope form to write the equation. Always interpret the gradient and intercept using the units given in the question.

你可能会得到两个数据点，并被要求建立线性模型。根据两点求斜率，然后用点斜式写出方程。务必使用题目中给出的单位来解释斜率和截距。

9. Common Mistakes and Exam Tips | 常见错误与考试技巧

Confusing the sign of gradient when rearranging equations. Always rewrite in the form y = mx + c to identify m correctly.

在整理方程时混淆斜率符号。务必改写成 y = mx + c 的形式以正确识别 m。
Forgetting that perpendicular gradient is the negative reciprocal, not simply the negative. For example, perpendicular to 2 is −1/2, not −2.

忘记垂直斜率是负倒数，而不仅仅是负数。例如，与 2 垂直的是 −1/2，而不是 −2。
Using the wrong point in y − y₁ = m(x − x₁). Either point works, but all coordinates must belong to the same point.

在 y − y₁ = m(x − x₁) 中使用错误的点。任一点都可以，但所有坐标必须属于同一个点。
Leaving answers as fractions when the question asks for integer coefficients in ax + by + c = 0. Multiply through by the common denominator.

当题目要求 ax + by + c = 0 中系数为整数时，却将答案留作分数。应该乘以公分母。
Drawing solid lines for strict inequalities. Use dashed lines when the boundary is not included.

对于严格不等式画实线。当边界不包括在内时，应使用虚线。

10. Worked Example | 综合例题

Question: The points A(−2, 1) and B(4, 9) lie on a straight line L. Find the equation of L in the form ax + by + c = 0, where a, b and c are integers. Then find the equation of the line perpendicular to L passing through the midpoint of AB.

题目：点 A(−2, 1) 和 B(4, 9) 在直线 L 上。求 L 的方程，写成 ax + by + c = 0 的形式，其中 a、b 和 c 为整数。然后求经过 AB 中点且垂直于 L 的直线方程。

Solution: First find the gradient of L: m = (9 − 1) ÷ (4 − (−2)) = 8 ÷ 6 = 4/3. Using point A(−2, 1): y − 1 = 4/3 (x + 2). Multiply by 3: 3y − 3 = 4x + 8. Rearrange: 4x − 3y + 11 = 0. So a = 4, b = −3, c = 11.

解答：首先求 L 的斜率：m = (9 − 1) ÷ (4 − (−2)) = 8 ÷ 6 = 4/3。使用点 A(−2, 1)：y − 1 = 4/3 (x + 2)。乘以 3：3y − 3 = 4x + 8。整理得：4x − 3y + 11 = 0。所以 a = 4，b = −3，c = 11。

The midpoint of AB is ((−2 + 4) ÷ 2, (1 + 9) ÷ 2) = (1, 5). The gradient perpendicular to L is −3/4. Using the midpoint: y − 5 = −3/4 (x − 1). Multiply by 4: 4y − 20 = −3x + 3. Rearrange: 3x + 4y − 23 = 0.

AB 的中点是 ((−2 + 4) ÷ 2, (1 + 9) ÷ 2) = (1, 5)。垂直于 L 的斜率为 −3/4。使用中点：y − 5 = −3/4 (x − 1)。乘以 4：4y − 20 = −3x + 3。整理得：3x + 4y − 23 = 0。

The final answers are L: 4x − 3y + 11 = 0 and the perpendicular line through the midpoint: 3x + 4y − 23 = 0. Always verify by substituting the original points back into the equation.

最终答案是 L：4x − 3y + 11 = 0，经过中点的垂线：3x + 4y − 23 = 0。务必通过将原始点代回方程来验证。

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Straight Line Graphs | 直线图