Example 5.3.2: Perpendicular Line Equations | 示例5.3.2：垂直直线方程

📚 Example 5.3.2: Perpendicular Line Equations | 示例5.3.2：垂直直线方程

In this article, we explore Example 5.3.2, a classic problem from the AQA A-Level Mathematics syllabus covering straight-line graphs. The focus is on finding the equation of a line that is perpendicular to a given line and passes through a specific point.

在本文中，我们解析 AQA A-Level 数学大纲中关于直线图形的经典例 5.3.2。本节重点在于求与已知直线垂直且经过特定点的直线方程。

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1. The Perpendicular Gradient Rule | 垂直梯度法则

For two straight lines to be perpendicular, the product of their gradients must equal -1. If line A has gradient m₁ and line B has gradient m₂, then they are perpendicular only when m₁ × m₂ = -1. In other words, the gradient of one line is the negative reciprocal of the other: m₂ = -1/m₁ (provided m₁ ≠ 0).

对于两条垂直的直线，其梯度的乘积必须等于 -1。若直线 A 的梯度为 m₁，直线 B 的梯度为 m₂，那么它们垂直当且仅当 m₁ × m₂ = -1。换言之，一条直线的梯度是另一条直线梯度的负倒数：m₂ = -1/m₁（前提是 m₁ ≠ 0）。

m₁ × m₂ = -1

This rule is fundamental in coordinate geometry and appears frequently in AQA exams, usually as part of a multi-step problem.

这一法则是坐标几何的基础，在 AQA 考试中频繁出现，通常作为多步骤问题的一部分。

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2. The Example Problem | 示例问题

Consider the following problem from Example 5.3.2:

考虑示例 5.3.2 中的以下问题：

Line L has equation 3x + 2y = 6. Line M is perpendicular to L and passes through the point (1, -2). Find the equation of M.

We will solve this step by step, explaining the reasoning at each stage. This type of problem tests your ability to rearrange equations, apply the perpendicular gradient rule, and substitute coordinates to find the y-intercept.

我们将逐步求解，并说明每一步的推理。此类问题考查你整理方程、应用垂直梯度法则以及代入坐标求 y 截距的能力。

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3. Step 1 – Find the Gradient of L | 第一步：求 L 的梯度

First, rearrange the equation of line L into the slope-intercept form y = mx + c. This allows us to read off the gradient directly.

首先，将直线 L 的方程整理为斜截式 y = mx + c。这样可以直接读出梯度。

Starting from 3x + 2y = 6, subtract 3x from both sides:

从 3x + 2y = 6 出发，两边减去 3x：

2y = -3x + 6

Now divide every term by 2:

接着每一项除以 2：

y = -3/2 x + 3

Therefore, the gradient of L is m_L = -3/2, and the y-intercept is 3. In AQA exams, always check whether the equation is in the form y = mx + c before extracting the gradient.

因此，L 的梯度为 m_L = -3/2，y 截距为 3。在 AQA 考试中，务必先确认方程是否已整理为 y = mx + c 的形式，再提取梯度。

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4. Step 2 – Apply the Perpendicular Rule | 第二步：应用垂直规则

Since M is perpendicular to L, the gradient of M, denoted m_M, must satisfy m_L × m_M = -1. Using the reciprocal rule:

由于 M 垂直于 L，M 的梯度（记为 m_M）必须满足 m_L × m_M = -1。根据负倒数法则：

m_M = -1 ÷ m_L = -1 ÷ (-3/2) = 2/3

Notice that the negative sign is preserved, and the fraction is inverted. A common mistake is to forget the negative sign or to take the reciprocal without inverting. Here, m_M = 2/3, which is indeed the negative reciprocal of -3/2.

注意负号保留，分数要取倒数。常见错误是忘记负号，或只取倒数而不变号。这里 m_M = 2/3，确实是 -3/2 的负倒数。

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5. Step 3 – Use the Point (1, -2) | 第三步：使用点 (1, -2)

Now that we know M has gradient 2/3, we use the given point (1, -2) to determine the y-intercept c. Substitute x = 1 and y = -2 into the general equation y = mx + c.

既然已知 M 的梯度为 2/3，我们利用已知点 (1, -2) 来求 y 截距 c。将 x = 1 和 y = -2 代入一般式 y = mx + c。

-2 = (2/3)(1) + c

Simplify the right-hand side:

化简右边：

-2 = 2/3 + c

Subtract 2/3 from both sides:

两边减去 2/3：

c = -2 – 2/3 = -8/3

Thus the y-intercept is -8/3. Be careful when subtracting fractions; convert -2 to -6/3 if needed.

因此 y 截距为 -8/3。注意分数减法的运算，必要时将 -2 化为 -6/3。

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6. Step 4 – Write the Final Equation | 第四步：写出最终方程

With m_M = 2/3 and c = -8/3, the equation of M is:

由于 m_M = 2/3，c = -8/3，直线 M 的方程为：

y = 2/3 x – 8/3

It is often preferable to express the equation without fractions. Multiply every term by 3:

通常我们更愿意把方程化为无分数的形式。每项乘以 3：

3y = 2x – 8

Rearranging gives the standard form:

移项得到标准形式：

2x – 3y = 8

This is a clean, AQA-friendly final answer. Both the slope-intercept form and the standard form are acceptable, but the standard form is often easier to verify.

这是一个简洁且符合 AQA 要求的最终答案。斜截式与标准形式均可接受，但标准形式更便于验证。

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7. Verification | 验证

It is always wise to check that the point (1, -2) lies on M. Substitute x = 1 and y = -2 into 2x – 3y = 8:

验证点 (1, -2) 是否在 M 上总是明智的。将 x = 1 和 y = -2 代入 2x – 3y = 8：

2(1) – 3(-2) = 2 + 6 = 8 ✓

Now verify perpendicularity. The gradient of L is -3/2 and the gradient of M is 2/3. Their product is:

再验证垂直性。L 的梯度为 -3/2，M 的梯度为 2/3。它们的乘积为：

(-3/2) × (2/3) = -1 ✓

Both checks pass, confirming that our answer is correct.

两项检查均通过，确认我们的答案正确。

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8. Common Pitfalls | 常见错误

Students often make the following mistakes when dealing with perpendicular lines:

学生在处理垂直直线时经常会犯以下错误：

• Forgetting to invert the fraction: Using 3/2 instead of 2/3 as the perpendicular gradient.

  忘记分数取倒数：误用 3/2 而不是 2/3 作为垂直梯度。

• Ignoring the negative sign: Using 3/2 when the original gradient is -3/2, which gives a product of -9/4, not -1.

  忽略负号：当原梯度为 -3/2 时却用 3/2，导致乘积为 -9/4，而不是 -1。

• Substituting the point before rearranging the given line: Always find m first, then c.

  在整理已知直线前就代入点：应当先求 m，再求 c。

• Sign errors when moving terms: For example, writing 2y = 3x + 6 instead of 2y = -3x + 6.

  移项时符号错误：例如误写 2y = 3x + 6 而不是 2y = -3x + 6。

Avoid these by writing out every step and checking your arithmetic.

避免这些错误的方法是写出每一步并检查计算过程。

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9. Alternative Method – Point-Slope Form | 替代方法：点斜式

Instead of using y = mx + c, we can use the point-slope form of a line:

除了使用 y = mx + c，我们还可以使用直线的点斜式：

y – y₁ = m(x – x₁)

For our point (1, -2) and gradient 2/3, we write:

对于点 (1, -2) 和梯度 2/3，我们写：

y – (-2) = 2/3 (x – 1)

Simplify the left side:

化简左边：

y + 2 = 2/3 x – 2/3

Subtract 2 from both sides, remembering that 2 = 6/3:

两边减去 2，注意 2 = 6/3：

y = 2/3 x – 8/3

This gives the same equation as before. The point-slope form is especially useful when the given point is not the y-intercept.

这与之前得到的方程相同。当已知点不是 y 截距时，点斜式尤其有用。

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10. Practice Question | 练习题

Try this AQA-style question on your own:

请自己尝试这道 AQA 风格练习题：

Find the equation of the line perpendicular to 4x – 3y = 10 that passes through (-2, 5). Express your answer in the form ax + by = c, where a, b and c are integers.

Solution outline: First rearrange 4x – 3y = 10 to y = 4/3 x – 10/3, so m₁ = 4/3. The perpendicular gradient is m₂ = -3/4. Then use (-2, 5) to find c: 5 = (-3/4)(-2) + c → c = 5 – 3/2 = 7/2. Thus y = -3/4 x + 7/2. Multiply by 4: 4y = -3x + 14, or 3x + 4y = 14.

解题概述：先将 4x – 3y = 10 整理为 y = 4/3 x – 10/3，所以 m₁ = 4/3。垂直梯度为 m₂ = -3/4。然后代入 (-2, 5) 求 c：5 = (-3/4)(-2) + c → c = 5 – 3

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