📚 Example 5.3.2: Coordinate Geometry - Parallel and Perpendicular Lines | 例 5.3.2 坐标几何：平行与垂直直线
In AQA A-Level Mathematics, coordinate geometry brings together algebra and geometry on the Cartesian plane. One of the most frequently examined skills is identifying when two straight lines are parallel or perpendicular, then using that condition to find unknown equations. This article works through a typical Example 5.3.2 style problem and explains the key ideas clearly.
在 AQA A-Level 数学中，坐标几何将代数与平面几何结合在笛卡尔坐标系中。最常考查的技能之一是判断两条直线何时平行或垂直，并利用该条件求未知直线方程。本文讲解一道典型的例 5.3.2 型题目，并清晰解释核心概念。
1. Gradient and Straight Line Equations | 斜率与直线方程
The gradient m of a straight line measures its steepness and direction. For any two points A(x₁, y₁) and B(x₂, y₂) on the line, the gradient is given by the change in y divided by the change in x.
直线的斜率 m 衡量其倾斜程度和方向。对于直线上任意两点 A(x₁, y₁) 和 B(x₂, y₂)，斜率等于 y 的变化量除以 x 的变化量。
m = (y₂ - y₁) ÷ (x₂ - x₁)
+ c = 0, which is often used in exam questions because it avoids fractional coefficients.
直线方程最常见的形式是 y = mx + c，其中 c 是 y 轴截距。另一种通用形式为 ax + by + c = 0，考试题目中常使用这种形式，因为它可以避免分数系数。
2. Parallel Lines Condition | 平行直线的条件
s gradient m₁ and line L₂ has gradient m₂, then L₁ is parallel to L₂ exactly when m₁ = m₂.
两条不同的直线平行当且仅当它们具有相同的斜率。也就是说，如果直线 L₁ 的斜率为 m₁，直线 L₂ 的斜率为 m₂，那么 L₁ 平行于 L₂ 当且仅当 m₁ = m₂。
mx + c. For example, y = 3x + 1 and y = 3x - 4 are parallel because both have gradient 3.
平行直线永不相交，当写成 y = mx + c 形式时，它们可以有不同的 y 轴截距，但 x 的系数相同。例如 y = 3x + 1 与 y = 3x - 4 平行，因为两者的斜率均为 3。
3. Perpendicular Lines Condition | 垂直直线的条件
ndicular to L₂ exactly when m₁ × m₂ = -1, provided neither line is vertical or horizontal.
两条直线垂直当且仅当它们斜率的乘积为 -1。如果直线 L₁ 的斜率为 m₁，直线 L₂ 的斜率为 m₂，那么 L₁ 垂直于 L₂ 当且仅当 m₁ × m₂ = -1，前提是两条直线均非竖直或水平。
m₁ × m₂ = -1
This relationship is often remembered as 'the negative reciprocal': if one gradient is m, then the gradient of a perpendicular line is -1 ÷ m. For example, if a line has gradient 2, a perpendicular line has gradient -1/2.
这一关系常被记为“负倒数”：如果一条直线的斜率为 m，则与其垂直的直线斜率为 -1 ÷ m。例如，若一条直线的斜率为 2，则垂线的斜率为 -1/2。
4. Finding a Line Through a Given Point | 求过给定点的直线
To find the equation of a straight line when you know its gradient m and a point (x₁, y₁) that lies on it, use the point-slope formula. This formula is extremely useful in coordinate geometry problems involving parallel or perpendicular lines.
当已知一条直线的斜率 m 及其上一点 (x₁, y₁) 时，可使用点斜式公式求其方程。该公式在涉及平行或垂直直线的坐标几何问题中极为有用。
y - y₁ = m(x - x₁)
+ by + c = 0 as required. Always check that the given point satisfies the final equation.
然后可以根据题目要求将其整理为 y = mx + c 或 ax + by + c = 0 的形式。务必检验给定点是否满足最终方程。
5. Worked Example 5.3.2 | 示例 5.3.2 详解
4x + 5. Then find the equation of the line through P that is perpendicular to y = 4x + 5.
让我们来看一道典型的例 5.3.2 题目。求经过点 P(3, -2) 且平行于直线 y = 4x + 5 的直线方程。然后再求经过点 P 且垂直于 y = 4x + 5 的直线方程。
ne, the gradient is the same, so m = 4. Using the point-slope formula with P(3, -2) gives:
给定直线 y = 4x + 5 的斜率为 m = 4。对于平行直线，斜率相同，所以 m = 4。使用点斜式公式并代入点 P(3, -2)：
y - (-2) = 4(x - 3)
om both sides to obtain y = 4x - 14. Therefore, the required parallel line is y = 4x - 14.
逐步化简该方程。首先 y + 2 = 4x - 12。然后两边同时减去 2，得到 y = 4x - 14。因此所求的平行直线为 y = 4x - 14。
For the perpendicular line, the gradient must be the negative reciprocal of 4, which is -1/4. Applying the point-slope formula with P(3, -2) gives:
对于垂直直线，其斜率必须是 4 的负倒数，即 -1/4。将点 P(3, -2) 代入点斜式公式：
y - (-2) = -1/4 × (x - 3)
orm yields x + 4y - 5 = 0. The perpendicular line can also be written as y = -1/4 x + 5/4.
两边同时乘以 4 以消去分数：4(y + 2) = -(x - 3)。展开得 4y + 8 = -x + 3。整理为一般式可得 x + 4y - 5 = 0。该垂直直线也可写作 y = -1/4 x + 5/4。
6. Checking Your Answers | 检验你的答案
After finding a line equation, always verify that it satisfies the given conditions. Substitute the coordinates of the given point into the final equation and check that the gradient matches the required value.
求出直线方程后，务必检验其是否满足给定条件。将给定点的坐标代入最终方程，并检查斜率是否符合要求。
- so the point (3, -2) lies on the line. The gradient is 4, so it is parallel to y = 4x + 5.
- 对于 y = 4x - 14，代入 x = 3 得 y = 4(3) - 14 = -2，所以点 (3, -2) 在该直线上。斜率为 4，因此它平行于 y = 4x + 5。
- 5 = 3 - 8 - 5 = -10, which is not zero. Wait - this suggests an error. Let us re-evaluate.
- 对于 x + 4y - 5 = 0，代入 x = 3, y = -2 得 3 + 4(-2) - 5 = 3 - 8 - 5 = -10，并非零。等等——这表明有错误。让我们重新检查。
is still -1/4. Substituting (3, -2) gives 3 + 4(-2) + 5 = 3 - 8 + 5 = 0, which is correct.
错误出在整理过程中。由 4y + 8 = -x + 3，两边同时加上 x 并减去 3：x + 4y + 5 = 0。我们来修正这一点。正确的一般式为 x + 4y + 5 = 0，其斜率仍为 -1/4。代入 (3, -2) 得 3 + 4(-2) + 5 = 3 - 8 + 5 = 0，正确无误。
This checking step is essential because small sign errors in rearranging are very common in exam conditions. Always test your final equation with the given point.
这一检验步骤至关重要，因为在考试环境中整理方程时出现符号错误非常常见。务必用给定点测试最终方程。
7. Special Cases: Horizontal and Vertical Lines | 特殊情况：水平与竖直直线
es have undefined gradient and their equation is of the form x = h, where h is a constant.
水平直线的斜率为 0，其方程形式为 y = k，其中 k 为常数。竖直直线的斜率不存在，其方程形式为 x = h，其中 h 为常数。
d. In such cases, simply state the relationship directly: y = k is perpendicular to x = h.
水平直线垂直于竖直直线，但由于其中一个斜率不存在，无法计算它们斜率的乘积。在这种情况下，直接说明关系即可：y = k 垂直于 x = h。
Also, two horizontal lines are parallel to each other, and two vertical lines are parallel to each other. Make sure you recognise these special cases before applying the gradient product rule.
此外，两条水平直线互相平行，两条竖直直线也互相平行。在应用斜率乘积法则之前，务必识别这些特殊情况。
8. Solving Problems with Simultaneous Equations | 运用联立方程解决问题
Some coordinate geometry problems ask you to find the intersection point of two lines, then use that point to find a third line parallel or perpendicular to one of them. Start by solving the two original line equations simultaneously.
有些坐标几何问题要求先求出两条直线的交点，然后利用该点求第三条与其中一条平行或垂直的直线。首先需要联立求解两条原始直线方程。
s gives 3x = 9, so x = 3. Substituting back gives y = 1. The intersection point is (3, 1).
例如，若 L₁: 2x + y = 7，L₂: x - y = 2，则可用代入法或消元法求解。将两式相加得 3x = 9，所以 x = 3。代回得 y = 1。交点为 (3, 1)。
ular gradient is 1/2. Use the point-slope formula with (3, 1) to get y - 1 = (1/2)(x - 3).
然后，若需要一条经过该交点且垂直于 L₁ 的直线，首先将 L₁ 改写为斜截式：y = -2x + 7，故 m₁ = -2。垂直直线的斜率为 1/2。利用点斜式公式并代入 (3, 1) 得 y - 1 = (1/2)(x - 3)。
this example, multiplying by 2 gives 2y - 2 = x - 3, so x - 2y + 1 = 0 after rearranging.
务必将最终答案化简为题目要求的形式，例如 ax + by + c = 0。对于此例，两边同乘以 2 得 2y - 2 = x - 3，整理后为 x - 2y + 1 = 0。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
l. For example, a line perpendicular to y = (2/3)x + 1 has gradient -3/2, not 3/2 or -2/3.
一个常见错误是忘记垂直直线的斜率是负倒数，而不仅仅是倒数。例如，垂直于 y = (2/3)x + 1 的直线斜率为 -3/2，而不是 3/2 或 -2/3。
tes. Write the formula with brackets first, then substitute carefully: y - y₁ = m(x - x₁).
另一个常见错误是在应用点斜式公式时混淆 x₁ 和 y₁，或错误处理负坐标。先用括号写出公式，然后仔细代入：y - y₁ = m(x - x₁)。
| Given line | Gradient m₁ | Perpendicular gradient m₂ |
| y = 5x - 2 | 5 | -1/5 |
| y = -3x + 4 | -3 | 1/3 |
| y = (1/2)x - 6 | 1/2 | -2 |
| y = -4/5 x + 1 | -4/5 | 5/4 |
This table shows the correct negative reciprocal relationships. Practise converting between parallel and perpendicular gradients until it becomes automatic.
上表展示了正确的负倒数关系。练习在平行与垂直斜率之间进行转换，直到能够自动完成。
10. Exam Tips for AQA Coordinate Geometry | AQA 坐标几何考试技巧
In AQA A-Level Mathematics exams, coordinate geometry questions often appear as part of a larger problem involving circles, differentiation, or integration. Always read the question carefully to identify what form your answer should take.
在 AQA A-Level 数学考试中，坐标几何题目常常作为涉及圆、微分或积分的更大问题的一部分出现。务必仔细阅读题目，明确答案应采用何种形式。
Show your method clearly when finding gradients and using the point-slope formula. Marks are awarded for correct working even if the final answer has a small arithmetic slip. Write down the parallel or perpendicular condition explicitly before calculating.
求斜率和使用点斜式公式时，要清晰地展示解题过程。即使最终答案有小的算术错误，正确的步骤也能得分。在计算之前，明确写出平行或垂直的条件。
Finally, manage your time well. These problems are typically worth 4 to 6 marks, so spend around 5 minutes on each. Check your final equation by substituting the given point.
最后，合理分配时间。这类题目通常分值为 4 到 6 分，因此每题大约用 5 分钟。通过代入给定点来检查最终方程。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图