📚 Example 5.3.1: Finding the Equation of a Straight Line | 例题 5.3.1：求直线方程
This article explains Example 5.3.1 from the AQA A-Level Mathematics Pure Year 1 specification, which comes from the chapter on straight line graphs. This worked example teaches a core coordinate geometry skill: how to find the equation of a straight line when you know its gradient and one point on the line.
本文讲解 AQA A-Level 数学纯数第一年大纲中直线图一章里的例题 5.3.1。本题旨在训练一项核心坐标几何技能：已知一条直线的斜率及其上一点的坐标，如何求出该直线的方程。
1. The Form y = mx + c | 直线方程的标准形式 y = mx + c
he gradient (slope) of the line, and c represents the y-intercept, which is the y-coordinate of the point where the line crosses the y-axis.
直线方程最常见的形式为 y = mx + c。其中 m 代表直线的斜率（坡度），c 代表 y 轴截距，即直线与 y 轴交点的纵坐标。
The gradient tells us how steep the line is. A positive gradient means the line rises as it moves from left to right; a negative gradient means the line falls as it moves from left to right.
斜率描述直线的倾斜程度。斜率为正时，直线从左向右上升；斜率为负时，直线从左向右下降。
y = mx + c
tion x = k, where k is a constant, and it has no gradient because the change in x is zero.
所有非竖直直线都可以写成这种形式。竖直直线的方程为 x = k（k 为常数），由于 x 的变化量为零，竖直直线没有斜率。
2. The Point-Gradient Form | 点斜式公式
When you are given the gradient m and a single point (x₁, y₁) on the line, the most direct way to find the equation is to use the point-gradient form.
当题目给出斜率 m 和直线上一点 (x₁, y₁) 时，最直接的方法就是使用点斜式公式。
y − y₁ = m(x − x₁)
This formula is extremely useful because it works with any given point, even when that point is not the y-intercept. It also works when the coordinates are negative, fractional, or involve decimals.
这个公式非常实用，因为它适用于任意给定点，即使该点不是 y 轴截距。当坐标为负数、分数或小数时，公式同样适用。
we substitute m = 2, x₁ = 3 and y₁ = 7 into this formula to obtain the required equation.
在例题 5.3.1 中，我们将 m = 2、x₁ = 3、y₁ = 7 代入此公式，即可求出所求方程。
3. Worked Example 5.3.1 | 例题 5.3.1 详解
Problem statement: and passes through the point (3, 7). Find the equation of the line in the form y = mx + c.
题目条件：一条直线的斜率为 2，且经过点 (3, 7)。求该直线的方程，并将答案写成 y = mx + c 的形式。
tep 1: Identify the known quantities. From the question, we have m = 2, x₁ = 3 and y₁ = 7.
步骤一：确定已知量。由题意可知 m = 2，x₁ = 3，y₁ = 7。
Step 2: Substitute the known values into the point-gradient formula.
步骤二：将已知数值代入点斜式公式。
y − 7 = 2(x − 3)
Step 3: Expand the bracket on the right-hand side. Be careful to multiply both terms inside the bracket by 2.
步骤三：展开右侧括号。注意要用 2 乘括号内的每一项。
y − 7 = 2x − 6
Step 4: Add 7 to both sides to make y the subject.
步骤四：方程两边同时加 7，将 y 表示为等式的主项。
y = 2x + 1
Therefore, the equation of the straight line is y = 2x + 1, so m = 2 and c = 1.
因此，该直线的方程为 y = 2x + 1，即 m = 2，c = 1。
4. Why the Point-Gradient Method Works | 点斜式方法的推导原理
The point-gradient formula is not a random rule; it follows directly from the definition of gradient. For a line containing two points (x₁, y₁) and (x, y), the gradient is the change in y divided by the change in x.
点斜式公式并非凭空而来，而是由斜率的定义直接推出。对于直线上两点 (x₁, y₁) 与 (x, y)，斜率等于 y 的变化量除以 x 的变化量。
m = (y − y₁) ÷ (x − x₁)
Multiply both sides of this equation by (x − x₁), and we obtain the point-gradient form.
将等式两边同时乘以 (x − x₁)，便得到点斜式公式。
y − y₁ = m(x − x₁)
This derivation shows why the formula is valid for any point (x, y) on the line, not just for the particular point (x₁, y₁).
这一推导说明了公式为何对直线上任意一点 (x, y) 都成立，而不仅仅对给定点 (x₁, y₁) 成立。
5. Checking Your Answer | 检验答案
A quick check ensures that our answer is correct. Substitute x = 3 into y = 2x + 1.
快速检验可以确认答案是否正确。将 x = 3 代入 y = 2x + 1。
y = 2 × 3 + 1 = 6 + 1 = 7
Since the point (3, 7) satisfies the equation, the line we found does indeed pass through the given point. The gradient also matches, because the coefficient of x is 2.
由于点 (3, 7) 满足方程，说明我们求出的直线确实经过给定点。同时，x 的系数为 2，斜率也与题意相符。
1). From (0, 1) to (3, 7), the rise is 6 and the run is 3, giving a gradient of 6 ÷ 3 = 2.
另一种检验方法是直接读出 y 轴截距。令 x = 0，得 y = 1，故直线与 y 轴交于 (0, 1)。从 (0, 1) 到 (3, 7)，上升量为 6，水平距离为 3，斜率为 6 ÷ 3 = 2。
6. Alternative Method: Two Points on the Line | 另一种方法：已知直线上两点
Sometimes a question does not give the gradient directly. Instead, it gives two points (x₁, y₁) and (x₂, y₂) on the line. In that case, the gradient must be calculated first.
有时题目不直接给出斜率，而是给出直线上两点 (x₁, y₁) 与 (x₂, y₂)。这时需要先计算斜率。
m = (y₂ − y₁) ÷ (x₂ − x₁)
the point-gradient form with (1, 2), we get y − 2 = 2(x − 1), which simplifies to y = 2x.
例如，一条直线经过 (1, 2) 和 (4, 8)。斜率为 m = (8 − 2) ÷ (4 − 1) = 6 ÷ 3 = 2。然后利用点斜式，以 (1, 2) 代入，得 y − 2 = 2(x − 1)，化简得 y = 2x。
arrange, the final answer is again in the form y = mx + c, even if c turns out to be zero.
注意，展开并整理之后，最终答案仍然是 y = mx + c 的形式，即使 c 恰好为零。
7. Writing the Equation in General Form ax + by + c = 0 | 将方程化为一般式 ax + by + c = 0
ed to give your answer in the general form ax + by + c = 0, where a, b and c are integers.
在 AQA 考试中，有时题目会要求将答案写成一般式 ax + by + c = 0，其中 a、b、c 均为整数。
r worked example, we have y = 2x + 1. Subtract y from both sides to obtain 2x − y + 1 = 0.
以本题为例，已知 y = 2x + 1。两边同时减去 y，得 2
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