Straight Line Graphs: Parallel and Perpendicular Lines, Length and Area | 直线图：平行线与垂直线、长度与面积

📚 Straight Line Graphs: Parallel and Perpendicular Lines, Length and Area | 直线图：平行线与垂直线、长度与面积

Straight line graphs are one of the most important foundation topics in Edexcel A-Level Mathematics. They bring together coordinates, gradients, equations, distances, midpoints and geometric conditions such as parallel and perpendicular lines. A clear understanding of this topic is essential before moving on to circles, differentiation and integration.

直线图是 Edexcel A-Level 数学中最重要的基础主题之一。它将坐标、斜率、方程、距离、中点以及平行线和垂直线等几何条件整合在一起。在进入圆、微分和积分之前，透彻理解这一主题至关重要。

1. The Gradient of a Straight Line | 直线的斜率

The gradient measures how steep a line is. For two points A(x₁, y₁) and B(x₂, y₂), the gradient m is calculated by dividing the change in y by the change in x.

斜率用于衡量一条直线的倾斜程度。对于两点 A(x₁, y₁) 和 B(x₂, y₂)，斜率 m 的计算方法是用 y 的变化量除以 x 的变化量。

m = (y₂ – y₁) / (x₂ – x₁)

If the y-value increases as the x-value increases, the gradient is positive. If the y-value decreases as the x-value increases, the gradient is negative. A horizontal line has gradient 0, and a vertical line has an undefined gradient.

如果 y 值随 x 值增大而增大，则斜率为正；如果 y 值随 x 值增大而减小，则斜率为负。水平线的斜率为 0，竖直线的斜率未定义。

Positive gradient: m > 0 | 正斜率：m > 0
Negative gradient: m < 0 | 负斜率：m < 0
Horizontal line: m = 0 | 水平线：m = 0
Vertical line: undefined gradient | 竖直线：斜率未定义

2. Equations of a Straight Line | 直线方程

The most common form of a straight line equation is y = mx + c, where m is the gradient and c is the y-intercept. This form is useful when the gradient and the point where the line crosses the y-axis are known.

直线方程最常见的形式是 y = mx + c，其中 m 是斜率，c 是 y 轴截距。当已知斜率和直线与 y 轴的交点时，这种形式非常有用。

y = mx + c

Another important form is the point-gradient form. If a line has gradient m and passes through the point (x₁, y₁), its equation can be written as y – y₁ = m(x – x₁). This is especially useful when the intercept is not given directly.

另一种重要形式是点斜式。如果一条直线的斜率为 m 且经过点 (x₁, y₁)，则其方程可以写成 y – y₁ = m(x – x₁)。当截距没有直接给出时，这种形式特别有用。

y – y₁ = m(x – x₁)

You should also be able to rearrange between different forms, including the general form ax + by + c = 0. In the Edexcel exam, answers are sometimes requested in a specific form, so always read the question carefully.

你还应该能够在不同形式之间进行转换，包括一般式 ax + by + c = 0。在 Edexcel 考试中，有时会要求以特定形式作答，因此一定要仔细审题。

3. Parallel Lines | 平行直线

Two straight lines are parallel if and only if they have the same gradient. This is one of the most frequently tested facts in coordinate geometry.

两条直线平行当且仅当它们具有相同的斜率。这是坐标几何中最常考查的结论之一。

m₁ = m₂

For example, the lines y = 3x + 2 and y = 3x – 5 are parallel because both have gradient 3. Parallel lines never meet, so a system formed by two parallel lines with different intercepts has no solution.

例如，直线 y = 3x + 2 和 y = 3x – 5 是平行的，因为两者的斜率都为 3。平行线永不相交，因此由两条截距不同的平行线组成的方程组没有解。

When finding the equation of a line parallel to a given line, simply keep the gradient the same and substitute the new point to find the intercept.

当求一条与已知直线平行的直线方程时，只需保持斜率不变，再代入新点求出截距即可。

4. Perpendicular Lines | 垂直直线

Two straight lines are perpendicular if the product of their gradients is -1. This means that the gradient of one line is the negative reciprocal of the gradient of the other line.

两条直线垂直的条件是它们的斜率乘积为 -1。这意味着一条直线的斜率是另一条直线斜率的负倒数。

m₁ × m₂ = -1

Equivalently, if one line has gradient m, then a line perpendicular to it has gradient -1 / m. For example, a line with gradient 2 is perpendicular to a line with gradient -1/2.

等价地，如果一条直线的斜率为 m，那么与它垂直的直线斜率为 -1 / m。例如，斜率为 2 的直线与斜率为 -1/2 的直线垂直。

A special case must be remembered: a horizontal line, which has gradient 0, is perpendicular to a vertical line, which has an undefined gradient. In this case the product rule cannot be applied because the gradient of the vertical line is undefined.

必须记住一个特殊情况：斜率为 0 的水平线与斜率未定义的竖直线垂直。在这种情况下，由于竖直线的斜率未定义，乘积法则无法使用。

5. Length Between Two Points | 两点间距离

The distance between two points A(x₁, y₁) and B(x₂, y₂) is found using Pythagoras’ theorem. The horizontal difference and vertical difference form the two shorter sides of a right-angled triangle.

两点 A(x₁, y₁) 和 B(x₂, y₂) 之间的距离可以用勾股定理求得。水平差和竖直差构成直角三角形的两条直角边。

d = √[(x₂ – x₁)² + (y₂ – y₁)²]

For example, the distance between (1, 2) and (4, 6) is √[(4 – 1)² + (6 – 2)²] = √(9 + 16) = √25 = 5. This distance formula is widely used in circle geometry and optimisation problems.

例如，点 (1, 2) 与点 (4, 6) 之间的距离为 √[(4 – 1)² + (6 – 2)²] = √(9 + 16) = √25 = 5。该距离公式广泛用于圆的几何和优化问题中。

Always write the square root in simplified form where possible. In exact value questions, leave your answer in surd form unless the question asks for a decimal approximation.

在可能的情况下，尽量将平方根化为最简形式。在精确值题目中，除非题目要求保留小数近似值，否则答案应保留根号形式。

6. Midpoint of a Line Segment | 线段中点

The midpoint of a line segment joining A(x₁, y₁) and B(x₂, y₂) is found by averaging the x-coordinates and averaging the y-coordinates.

连接点 A(x₁, y₁) 与点 B(x₂, y₂) 的线段中点，可以通过分别对 x 坐标和 y 坐标求平均得到。

M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

For example, the midpoint of (3, 5) and (7, -1) is ((3 + 7) / 2, (5 + (-1)) / 2) = (5, 2). This formula is especially useful when solving problems involving symmetry or when constructing the perpendicular bisector of a line segment.

例如，点 (3, 5) 与点 (7, -1) 的中点是 ((3 + 7) / 2, (5 + (-1)) / 2) = (5, 2)。该公式在涉及对称性或作线段垂直平分线的问题中尤其有用。

7. Area of Triangles and Polygons | 三角形与多边形面积

In coordinate geometry, the area of a triangle can be found by treating one side as the base and then calculating the perpendicular height from the third vertex to that base. This often involves using the length formula and the perpendicular distance from a point to a line.

在坐标几何中，三角形的面积可以将一条边作为底，然后计算第三个顶点到该底的垂直高度。这通常需要用到距离公式以及点到直线的垂直距离。

Area = ½ × base × height

For a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃), a more direct coordinate formula can also be used. The shoelace formula gives the area as half the absolute value of the sum of diagonal products.

对于顶点为 (x₁, y₁)、(x₂, y₂) 和 (x₃, y₃) 的三角形，也可以使用更直接的坐标公式。鞋带公式给出的面积为对角线乘积之和的绝对值的一半。

Area = ½ | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |

This method is efficient when all three vertices are known and no obvious base-height pair is given. It can also be extended to polygons by listing the vertices in order.

当三个顶点均已知且没有明显易求的底和高时，这种方法非常高效。通过按顺序列出顶点，它还可以推广到多边形面积的计算。

8. Modelling with Straight Lines | 直线建模

Straight line models are often used in real-world contexts such as cost analysis, temperature conversion, speed-time relationships and simple supply-demand problems. In these questions, two variables are connected by a linear equation of the form y = mx + c.

直线模型常用于现实情境中，例如成本分析、温度转换、速度-时间关系以及简单的供求问题。在这些题目中，两个变量通过形如 y = mx + c 的线性方程联系在一起。

For example, a taxi company may charge a fixed call-out fee of £2 plus £1.50 per mile. This can be modelled by the equation C = 1.5d + 2, where C is the total cost and d is the distance travelled in miles.

例如，一家出租车公司可能收取 2 英镑的固定出车费，再加上每英里 1.50 英镑。这可以用方程 C = 1.5d + 2 来建模，其中 C 是总费用，d 是行驶距离（英里）。

In modelling questions, interpret the gradient as the rate of change and the y-intercept as the initial fixed value. You may also need to evaluate the model for given inputs or solve for an unknown input using the linear equation.

在建模题中，将斜率解释为变化率，将 y 轴截距解释为初始固定值。你可能还需要根据给定的输入计算模型结果，或利用线性方程求解未知输入。

9. Common Exam Pitfalls | 常见考试失分点

Many candidates lose marks by mixing up the gradient formula or by substituting coordinates in the wrong order. Always label your points clearly and write the formula first before substituting values.

许多考生由于混淆斜率公式或将坐标代入顺序弄错而失分。一定要清楚标出点的坐标，并在代入数值之前先写出公式。

Another common error is forgetting that perpendicular gradient is the negative reciprocal, not simply the reciprocal. For example, if m = 3, the perpendicular gradient is -1/3, not 1/3 or -3.

另一个常见错误是忘记垂直斜率是负倒数，而不仅仅是倒数。例如，若 m = 3，则垂直斜率为 -1/3，而不是 1/3 或 -3。

When calculating distances, make sure you subtract coordinates in the same order for both x and y, but note that squaring removes sign differences. When finding a midpoint, do not forget to divide the sum by 2 for both coordinates.

计算距离时，请确保 x 坐标和 y 坐标的相减顺序一致，但注意平方会消除符号差异。求中点时，不要忘记两个坐标的和都要除以 2。

10. Worked Example | 例题精讲

Example: Points A(2, 3) and B(6, 11) are given. Find the gradient of AB, the equation of the line through A and B, and the gradient of a line perpendicular to AB.

例题：已知点 A(2, 3) 和 B(6, 11)。求 AB 的斜率、经过 A 和 B 的直线方程，以及与 AB 垂直的直线斜率。

First, find the gradient using m = (y₂ – y₁) / (x₂ – x₁). Substituting (2, 3) and (6, 11) gives m = (11 – 3) / (6 – 2) = 8 / 4 = 2.

首先，利用公式 m = (y₂ – y₁) / (x₂ – x₁) 求斜率。代入 (2, 3) 和 (6, 11) 得到 m = (11 – 3) / (6 – 2) = 8 / 4 = 2。

Using the point-gradient form with A(2, 3), the equation is y – 3 = 2(x – 2). Expanding gives y – 3 = 2x – 4, so the final equation is y = 2x – 1.

使用点斜式，以 A(2, 3) 为已知点，可得方程 y – 3 = 2(x – 2)。展开得 y – 3 = 2x – 4，因此最终方程为 y = 2x – 1。

The gradient of a line perpendicular to AB is the negative reciprocal of 2. Therefore the perpendicular gradient is mₚ = -1 / 2.

与 AB 垂直的直线斜率是 2 的负倒数。因此垂直斜率 mₚ = -1 / 2。

11. Practice Style Questions | 练习型题目

Try these exam-style questions to strengthen your understanding. Work through each step without looking at the solution too quickly.

尝试以下考试型题目，以巩固理解。逐步完成每一问，不要过早查看答案。

Find the gradient of the line joining (1, 4) and (5, 12). | 求连接 (1, 4) 和 (5, 12) 的直线斜率。
Write the equation of the line with gradient -3 passing through (2, 7). | 写出斜率为 -3 且经过 (2, 7) 的直线方程。
Determine whether the lines y = 2x + 5 and 2y + x = 6 are perpendicular. | 判断直线 y = 2x + 5 与 2y + x = 6 是否垂直。
Find the distance and midpoint between (-2, 1) and (4, 9). | 求 (-2, 1) 与 (4, 9) 之间的距离和中点。
A line passes through (3, 2) and is parallel to y = 5x – 1. Find its equation. | 一条直线经过 (3, 2) 且平行于 y = 5x – 1。求其方程。

12. Summary Checklist | 总结清单

Make sure you can confidently recall and use the following key facts before your Edexcel A-Level Mathematics exam.

在参加 Edexcel A-Level 数学考试之前，请确保你能够熟练回忆并运用以下关键知识点。

Gradient formula: m = (y₂ – y₁) / (x₂ – x₁) | 斜率公式：m = (y₂ – y₁) / (x₂ – x₁)
Equation forms: y = mx + c and y – y₁ = m(x – x₁) | 方程形式：y = mx + c 和 y – y₁ = m(x – x₁)
Parallel lines: m₁ = m₂ | 平行线：m₁ = m₂
Perpendicular lines: m₁ × m₂ = -1 | 垂直线：m₁ × m₂ = -1
Distance formula: d = √[(x₂ – x₁)² + (y₂ – y₁)²] | 距离公式：d = √[(x₂ – x₁)² + (y₂ – y₁)²]
Midpoint formula: M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2) | 中点公式：M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)
Triangle area using base and height or shoelace formula | 三角形面积使用底与高或鞋带公式

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Straight Line Graphs: Parallel and Perpendicular Lines, Length and Area | 直线图：平行线与垂直线、长度与面积