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Straight Line Graphs | 直纹图

2026-08-30 04:49 · 本文部分内容由AI辅助创作

📚 Straight Line Graphs | 直纹图

Straight line graphs form the backbone of coordinate geometry and are a guaranteed topic in every IGCSE Mathematics paper. Understanding how to find the gradient, intercepts, and equation of a line is essential for solving problems in both Paper 2 and Paper 4.

直纹图(线性函数图像)是坐标几何的基石,也是每次 IGCSE 数学考试中必定出现的考点。掌握如何求梯度、截距和直线方程,对于解答 Paper 2 和 Paper 4 中的题目至关重要。

1. The Basic Form y = mx + c | 基础形式 y = mx + c

The standard equation of a straight line is y = mx + c, where 'm' represents the gradient (slope) of the line, and 'c' represents the y-intercept (where the line crosses the y-axis).

直线的一般方程为 y = mx + c,其中“m”代表直线的梯度(斜率),“c”代表 y 轴截距(直线与 y 轴交点的纵坐标)。

y = m·x + c

For example, in the equation y = 3x - 2, the gradient is 3, and the y-intercept is -2.

例如,在方程 y = 3x - 2 中,梯度为 3,y 轴截距为 -2。


2. Calculating the Gradient | 计算梯度

The gradient measures the steepness of a line. Given two points (x₁, y₁) and (x₂, y₂), the gradient 'm' is calculated using the formula:

梯度用于衡量直线的陡峭程度。已知两点 (x₁, y₁) 和 (x₂, y₂),梯度'm'可通过以下公式计算:

m = (y₂ - y₁) / (x₂ - x₁) = Δy / Δx

A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards. A horizontal line has a gradient of 0, and a vertical line has an undefined gradient.

正梯度从左到右向上倾斜,负梯度从左到右向下倾斜。水平线的梯度为 0,垂直线的梯度不存在(未定义)。

Example: Find the gradient of the line passing through (1, 2) and (4, 8).

示例:求经过点 (1, 2) 和 (4, 8) 的直线的梯度。

m = (8 - 2) / (4 - 1) = 6 / 3 = 2


3. Finding the Intercepts | 求截距

crosses the y-axis (x = 0). The x-intercept is where the line crosses the x-axis (y = 0).

y 轴截距是直线与 y 轴交点的纵坐标(此时 x = 0)。x 轴截距是直线与 x 轴交点的横坐标(此时 y = 0)。

For y = mx + c, the y-intercept is simply the constant term 'c'. To find the x-intercept, substitute y = 0 and solve for x.

对于 y = mx + c,y 轴截距就是常数项“c”。要求 x 轴截距,可令 y = 0,然后解方程求出 x。

4, the y-intercept is -4. Setting y = 0 gives 0 = 2x - 4, so x = 2. The x-intercept is 2.

示例:对于直线 y = 2x - 4,y 轴截距为 -4。令 y = 0,得到 0 = 2x - 4,解得 x = 2。因此 x 轴截距为 2。


4. Plotting the Graph | 绘制直线图

To plot a straight line graph, you can either use a table of values or use the gradient and y-intercept. The most reliable method in an exam is to substitute x-values into the equation to find corresponding y-values.

绘制直线图时,可以使用数值表,也可以利用梯度和 y 轴截距。在考试中,最可靠的方法是选取 x 值代入方程,求出对应的 y 值。

x0123
y = 2x + 11357

Plot these points on a coordinate grid and draw a straight line through them.

在坐标网格上标出这些点,并画一条直线穿过它们。


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5. Rearranging ax + by = c | 方程形式 ax + by = c

= c. To find the gradient and y-intercept, you must rearrange it into the form y = mx + c.

直线方程通常以 ax + by = c 的形式给出。为了求出梯度和 y 轴截距,必须将其改写成 y = mx + c 的形式。

Example: Rearrange 3x + y = 9 to find the gradient.

示例:将 3x + y = 9 改写,并求出梯度。

y = -3x + 9

Thus, the gradient m = -3 and the y-intercept is 9.

因此,梯度 m = -3,y 轴截距为 9。


6. Parallel and Perpendicular Lines | 平行线与垂直线

arallel lines have the same gradient. For example, y = 2x + 1 and y = 2x - 5 are parallel.

平行线的梯度相同。例如,y = 2x + 1 和 y = 2x - 5 是平行的。

Perpendicular lines have gradients that multiply to -1. If one line has gradient m, the perpendicular line has gradient -1/m.

互相垂直的直线,其梯度相乘等于 -1。如果一条直线的梯度为 m,则与之垂直的直线梯度为 -1/m(负倒数)。

m₁ × m₂ = -1

le: A line is perpendicular to y = 3x + 2. The gradient of the perpendicular line is -1/3.

示例:一条直线与 y = 3x + 2 垂直。该垂直线的梯度为 -1/3。


7. Finding the Intersection of Two Lines | 求两条直线的交点

The point of intersection is where two lines cross. At this point, the x and y values satisfy both equations. You can solve the equations simultaneously using substitution or elimination.

交点就是两条直线相交的地方。在该点,x 和 y 的值同时满足两个方程。可以通过代入法或消元法联立求解两个方程。

Example: Find the intersection of y = 2x + 1 and y = x + 3.

示例:求直线 y = 2x + 1 和 y = x + 3 的交点。

Set the equations equal to each other:

令两个方程相等:

2x + 1 = x + 3

x = 2

Substitute x = 2 into y = x + 3, giving y = 5. The intersection point is (2, 5).

将 x = 2 代入 y = x + 3,得到 y = 5。所以交点坐标为 (2, 5)。


8. Finding the Equation of a Line | 求直线方程

Given the gradient 'm' and a point (x₁, y₁) on the line, the equation can be found using the formula:

已知梯度“m”和直线上一点 (x₁, y₁),可以使用以下公式求出直线方程:

y - y₁ = m(x - x₁)

Example: Find the equation of the line with gradient 2 passing through (3, 4

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