📚 Worked Example 5.4.1: Equation of a Parallel Line through a Point | 例题 5.4.1：求过一点且平行于已知直线的方程
and A-Level papers, often worth 3 to 4 marks. The example uses the point A(2, 5) and the line y = 3x - 4, but the method generalises to any linear equation.
在这个 AQA A-Level 数学例题中，我们讲解一个标准的坐标几何问题：求一条过已知点、且平行于指定直线的直线方程。这类题型在 AQA 纯数学 AS 与 A-Level 试卷中频繁出现，通常占 3 到 4 分。例题使用点 A(2, 5) 与直线 y = 3x - 4，但该方法可推广到任何线性方程。
1. The Problem Statement | 题目陈述
through the point A(2, 5) and is parallel to l₁. Give your answer in the form y = mx + c.
已知直线 l₁ 的方程为 y = 3x - 4，求过点 A(2, 5) 且平行于 l₁ 的直线 l₂ 的方程，并将答案写成 y = mx + c 的形式。
This example is typical of the questions found in the coordinate geometry section of the AQA specification. It tests your ability to interpret the gradient of a line and to apply the point-gradient formula efficiently.
该例题是 AQA 考试大纲中坐标几何部分的典型问题。它考查你解读直线斜率以及高效应用点斜式公式的能力。
2. Key Concept: Parallel Lines Share the Same Gradient | 关键概念：平行直线斜率相同
so its gradient m₁ is 3. Therefore any line parallel to it must also have gradient m₂ = 3.
在坐标几何中，两条非竖直直线平行，当且仅当它们的斜率相同。直线 y = 3x - 4 已是斜截式，因此其斜率 m₁ 为 3。所以任何与它平行的直线，斜率 m₂ 也必须为 3。
pt form, you should rearrange it into the form y = mx + c before reading off the gradient.
认识到这一事实是第一步，也是最重要的一步。如果给定直线不是斜截式，你应该先将其整理成 y = mx + c 的形式，再读取斜率。
3. Using the Point-Gradient Form | 使用点斜式
ne and m is the gradient. Substituting the point A(2, 5) and m = 3 gives y - 5 = 3(x - 2).
直线的点斜式为 y - y₁ = m(x - x₁)，其中 (x₁, y₁) 是直线上已知的点，m 是斜率。代入点 A(2, 5) 与 m = 3，得到 y - 5 = 3(x - 2)。
This formula is especially useful because it directly links the gradient to a specific point, so there is no need to guess the y-intercept at this stage.
这个公式尤其有用，因为它直接将斜率与一个具体点联系起来，因此在这一阶段无需猜测 y 轴截距。
4. Expanding and Simplifying | 展开与化简
both sides to isolate y: y = 3x - 1. This is the required equation in the form y = mx + c.
展开等号右边：y - 5 = 3x - 6。两边同时加 5，解出 y：y = 3x - 1。这就是所求的直线方程，形式为 y = mx + c。
Always check that the final line has the same gradient as the original line. Here both have gradient 3, and the y-intercept is now -1 rather than -4.
务必检查最终直线是否与原直线具有相同的斜率。这里两者的斜率都为 3，而 y 轴截距现在是 -1 而不是 -4。
5. Alternative Method: Direct Substitution into y = mx + c | 替代方法：直接代入 y = mx + c
= 3(2) + c, so 5 = 6 + c, giving c = -1. Hence y = 3x - 1, confirming the previous result.
由于新直线的斜率必须为 3，我们可以直接写成 y = 3x + c。为求 y 轴截距 c，代入点 A(2, 5) 的坐标：5 = 3(2) + c，即 5 = 6 + c，得到 c = -1。因此 y = 3x - 1，与前面结果一致。
a different form, such as ax + by + c = 0, you can rearrange the final result at the end.
对许多学生来说，这种方法更快，但要求直线写成斜截式。如果题目要求答案写成其他形式，例如 ax + by + c = 0，你可以在最后将结果移项整理。
6. Extending to Perpendicular Lines | 推广到垂直直线
gradient is -1/3. Using y - 5 = -1/3(x - 2), the equation simplifies to y = -1/3 x + 17/3.
常见的延伸问题是求过同一点且垂直于 y = 3x - 4 的直线。垂直直线的斜率乘积为 -1，因此垂直直线斜率为 -1/3。利用 y - 5 = -1/3(x - 2)，化简得到 y = -1/3 x + 17/3。
Remember that parallel lines share the same gradient, while perpendicular lines have gradients that are negative reciprocals of each other. Mixing up these two rules is a common source of error.
记住：平行直线斜率相同，而垂直直线的斜率互为负倒数。混淆这两条规则是常见的错误来源。
7. Checking Your Answer Graphically | 用图形检验答案
uting x = 2 gives y = 3(2) - 1 = 5, which matches. This quick check confirms the solution.
画出原直线 y = 3x - 4 与新直线 y = 3x - 1。它们应当永不相交且斜率相同。点 (2, 5) 应落在新直线上：代入 x = 2 得 y = 3(2) - 1 = 5，与坐标符合。这个快速检验可确认解的正确性。
| Line | Gradient | y-intercept |
| y = 3x - 4 | 3 | -4 |
| y = 3x - 1 | 3 | -1 |
This comparison shows that the two lines differ only in the y-intercept, which is exactly what we expect for parallel lines.
这一比较表明，两条直线仅在 y 轴截距上不同，这正是平行直线应有的特征。
8. Common Mistakes and How to Avoid Them | 常见错误与避免方法
g because it does not pass through (2, 5). Another error is forgetting that parallel lines must have exactly the same gradient, not the reciprocal. Always identify the gradient from the given line first and then apply the point-gradient form.
一个常见错误是直接使用原直线的 y 轴截距，而未代入给定点。例如写成 y = 3x - 4 是错误的，因为该直线不经过 (2, 5)。另一个错误是忘记平行直线必须具有完全相同的斜率，而不是倒数。务必先从给定直线中识别斜率，再应用点斜式。
ng by 2 gives y = 3x - 4, so the gradient is still 3. Take care with every algebraic step.
一些考生在移项时还会混淆符号。如果给定直线是 2y = 6x - 8，两边除以 2 得到 y = 3x - 4，因此斜率仍然是 3。每个代数步骤都要小心。
9. AQA Exam Technique and Mark Allocation | AQA 考试技巧与分值分配
In AQA mark schemes, this type of question typically awards one mark for stating or using the correct parallel gradient, one mark for substituting the point into the line equation, and one final mark for the correct simplified equation. Showing all steps clearly is essential, even when the answer can be found mentally.
在 AQA 的评分标准中，这类题通常会给 1 分用于写出或使用正确的平行斜率，1 分用于将点代入直线方程，最后 1 分用于写出正确的化简方程。清晰展示所有步骤至关重要，即使答案可以心算得出。
s check the question wording for the required form, such as y = mx + c or ax + by + c = 0.
如果最终答案不是题目要求的形式，你可能会失去最后的准确分。务必检查题目要求的形式，例如 y = mx + c 或 ax + by + c = 0。
10. Practice Variation: Changing the Point and Line | 练习变式：改变点与直线
ine as y = -2x + 4, so the gradient is -2. Then use y - 4 = -2(x + 1), giving y = -2x + 2.
为增强信心，可用不同数值重复该方法。例如，求过点 B(-1, 4) 且平行于 2y + 4x = 8 的直线。先将给定直线改写为 y = -2x + 4，因此斜率为 -2。然后使用 y - 4 = -2(x + 1)，得到 y = -2x + 2。
on, such as 3x - 2y = 6, will also prepare you for the range of formats used in AQA exams.
用不同形式的原方程进行练习，例如 3x - 2y = 6，也有助于你应对 AQA 考试中出现的各种格式。
11. Key Takeaways | 核心要点
To solve any parallel-line equation problem: (1) identify the gradient of the given line; (2) use the same gradient for the new line; (3) substitute the known point to find the intercept; (4) write the final equation in the requested form. This structured approach minimises errors and earns full marks in exams.
解决任何平行直线方程问题的步骤为：(1) 识别给定直线的斜率；(2) 新直线使用相同斜率；(3) 代入已知点求截距；(4) 按照题目要求的形式写出最终方程。这种结构化方法能减少错误，并在考试中拿到满分。
These steps also apply when the point is given as an intersection of two lines or when the parallel line is a tangent to a circle, so the skill has wide applications across the A-Level syllabus.
当给定点是两条直线的交点，或者平行线是圆的切线时，这些步骤同样适用。因此这一技能在 A-Level 大纲中应用广泛。
12. Summary | 总结
This worked example demonstrates that finding a parallel line through a point is a straightforward application of gradient facts and the point-gradient formula. Mastery of this skill supports more advanced AQA topics such
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