本讲义涵盖主题 3,坐标几何(Coordinate geometry)。标记 (Extended) 的部分只在拓展卷上考查;其他一切对两个层次都适用。
坐标几何
IGCSE 数学 · 第 3 主题
3.1
坐标
大纲
| Subject content | Notes and examples |
|---|---|
| Use and interpret Cartesian coordinates in two dimensions. |
来源:剑桥国际大纲
图上的一个点由它的坐标(coordinates)描述,有时称为笛卡尔坐标,写成 $(x, y)$。第一个数是横向的值而第二个是纵向的值。
- 这两条数轴是坐标轴(axes):水平(horizontal)$x$ 轴和竖直(vertical)$y$ 轴。
- 它们在原点(origin)相交,点 $(0, 0)$。
- 坐标轴把网格分成四个象限(quadrants)。
所以点 $(3, -2)$ 通过向右 $3$ 和向下 $2$ 找到。
探索
The coordinate plane
y = mx + c
Every point has an (x, y) coordinate. A straight line is the set of points where y depends on x in a fixed way.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| coordinates | 坐标 | zuò biāo |
| axes | 坐标轴 | zuò biāo zhóu |
| horizontal | 水平 | shuǐ píng |
| vertical | 竖直 | shù zhí |
| origin | 原点 | yuán diǎn |
| quadrants | 象限 | xiàng xiàn |
3.5
一次函数图像的方程
大纲
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph in the form $y = mx + c$. | Questions may: • use and request lines in the forms $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with the equation $y = 6x + 3$. Candidates are expected to give equations of a line in a fully simplified form. |
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph. | Questions may: • use and request lines in different forms, e.g. $ax + by = c$, $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with equation $5x + 4y = 8$. Candidates are expected to give equations of a line in a fully simplified form. |
来源:剑桥国际大纲
大多数直线能被写成
$$y = mx + c,$$
其中 $m$ 是斜率(gradient)(陡度)而 $c$ 是截距(intercept)——线穿过 $y$ 轴处的 $y$ 值。
- 一条像 $x = k$(例如 $x = 3$)的线是竖直的。
- 一条像 $y = k$(例如 $y = 3$)的线是水平的。
一条线也可能作为 $ax + by = c$ 给出。把它重新排列成 $y = mx + c$ 以读出斜率和截距。
Worked example. 求 $5x + 4y = 8$ 的斜率和 $y$ 截距。
$$4y = -5x + 8 \;\Rightarrow\; y = -\tfrac{5}{4}x + 2.$$
所以斜率是 $-\tfrac{5}{4}$ 而 $y$ 截距是 $2$。
探索
y = mx + c
y = ax + b
Drag the gradient and the intercept. a is the gradient (steepness) and b is where the line crosses the y-axis.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| gradient | 斜率 | xié lǜ |
| intercept | 截距 | jié jù |
3.3
一次函数图像的斜率
大纲
| Subject content | Notes and examples |
|---|---|
| Find the gradient of a straight line. | From a grid only. |
| Subject content | Notes and examples |
|---|---|
| 1 Find the gradient of a straight line. | |
| 2 Calculate the gradient of a straight line from the coordinates of two points on it. |
来源:剑桥国际大纲
斜率(gradient)衡量一条线有多陡:
$$m = \frac{\text{change in } y}{\text{change in } x} = \frac{\text{rise}}{\text{run}}.$$
一个正斜率向右上升;一个负斜率向右下降。
Worked example. 求过 $(1, 2)$ 和 $(4, 11)$ 的线的斜率。
$$m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3.$$
探索
Gradient
y = ax + b
The gradient a measures steepness — rise over run.
3.2
绘制一次函数图像
大纲
| Subject content | Notes and examples |
|---|---|
| Draw straight-line graphs for linear equations. | Equations will be given in the form $y = mx + c$ (e.g. $y = -2x + 5$), unless a table of values is given. |
| Subject content | Notes and examples |
|---|---|
| Draw straight-line graphs for linear equations. | Examples include: • $y = -2x + 5$ • $y = 7 - 4x$ • $3x + 2y = 5$. |
来源:剑桥国际大纲
要画 $y = mx + c$,最快的方式是:
- 在 $y$ 轴上标记截距 $c$。
- 用斜率步进到更多的点(对于 $m = 3$,向右 $1$ 和向上 $3$)。
- 用一条直线连接这些点。
你也能做一个小的数值表(table of values):选择两个或三个 $x$ 值、算出 $y$,并描出这些点。
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| table of values | 数值表 | shù zhí biǎo |
3.5
一次函数图像的方程
大纲
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph in the form $y = mx + c$. | Questions may: • use and request lines in the forms $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with the equation $y = 6x + 3$. Candidates are expected to give equations of a line in a fully simplified form. |
| Subject content | Notes and examples |
|---|---|
| Interpret and obtain the equation of a straight-line graph. | Questions may: • use and request lines in different forms, e.g. $ax + by = c$, $y = mx + c$, $x = k$ • involve finding the equation when the graph is given • ask for the gradient or $y$-intercept of a graph from an equation, e.g. find the gradient and $y$-intercept of the graph with equation $5x + 4y = 8$. Candidates are expected to give equations of a line in a fully simplified form. |
来源:剑桥国际大纲
若你知道斜率 $m$ 和线上的一个点,把这个点放进 $y = mx + c$ 以求 $c$。
Worked example. 一条线有斜率 $3$ 并通过 $(1, 2)$。求它的方程。
$$y = 3x + c, \qquad 2 = 3(1) + c, \qquad c = -1.$$
所以方程是 $y = 3x - 1$。(若你被给出两个点,先求斜率,然后做这个。)
3.4
长度与中点
大纲
| Subject content | Notes and examples |
|---|---|
| 1 Calculate the length of a line segment. | |
| 2 Find the coordinates of the midpoint of a line segment. |
来源:剑桥国际大纲
一条线段(line segment)是两个点之间的直的部分。
要求 $(x_1, y_1)$ 和 $(x_2, y_2)$ 之间线段的长度(length),对水平和竖直的间隙用勾股定理(Pythagoras' theorem):
$$\text{length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$
要求中点(midpoint)(恰好在中间的点),对坐标取平均:
$$\text{midpoint} = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right).$$
Worked example. 求从 $(1, 2)$ 到 $(4, 6)$ 的线段的长度和中点。
$$\text{length} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$
$$\text{midpoint} = \left( \frac{1 + 4}{2}, \; \frac{2 + 6}{2} \right) = (2.5, \, 4).$$
探索
Length and midpoint lab
midpoint is halfway between endpoints
Move along a line segment and see midpoint as halfway.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| line segment | 线段 | xiàn duàn |
| length | 长度 | cháng dù |
| Pythagoras' theorem | 勾股定理 | gōu gǔ dìng lǐ |
| midpoint | 中点 | zhōng diǎn |
3.6
平行线
大纲
| Subject content | Notes and examples |
|---|---|
| Find the gradient and equation of a straight line parallel to a given line. | e.g. find the equation of the line parallel to $y = 4x - 1$ that passes through $(1, -3)$. |
来源:剑桥国际大纲
平行(parallel)线从不相遇,所以它们有相同的斜率。
Worked example. 求平行于 $y = 4x - 1$ 并通过 $(1, -3)$ 的线的方程。
斜率也是 $4$。把点放进去:
$$-3 = 4(1) + c \;\Rightarrow\; c = -7,$$
所以线是 $y = 4x - 7$。
探索
Parallel & perpendicular
y = ax + b
Parallel lines share a gradient; perpendicular gradients multiply to −1.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| parallel | 平行 | píng xíng |
3.7
垂直线
大纲
| Subject content | Notes and examples |
|---|---|
| Find the gradient and equation of a straight line perpendicular to a given line. | Examples include: • find the gradient of a line perpendicular to $2y = 3x + 1$ • find the equation of the perpendicular bisector of the line joining the points $(-3, 8)$ and $(9, -2)$. |
来源:剑桥国际大纲
两条线垂直(perpendicular)若它们以一个直角(right angle)相遇。它们的斜率相乘得 $-1$:
$$m_1 \times m_2 = -1, \qquad \text{so} \qquad m_2 = -\frac{1}{m_1}.$$
用文字说:翻转分数并改变符号。
Worked example. 求垂直于 $2y = 3x + 1$ 的一条线的斜率。
重新排列:$y = \tfrac{3}{2}x + \tfrac{1}{2}$,所以斜率是 $\tfrac{3}{2}$。垂直斜率是 $-\tfrac{2}{3}$。
Perpendicular bisector
一条线段的垂直平分线(perpendicular bisector)以一个直角把它切成两半。要求它的方程:得到中点,然后用过那个中点的垂直斜率。
Worked example. 求连接 $(-3, 8)$ 和 $(9, -2)$ 的线段的垂直平分线。
- 中点:$\left( \frac{-3 + 9}{2}, \frac{8 + (-2)}{2} \right) = (3, 3)$。
- 线段的斜率:$\frac{-2 - 8}{9 - (-3)} = \frac{-10}{12} = -\tfrac{5}{6}$。
- 垂直斜率:$\frac{6}{5}$。
过 $(3, 3)$:$\; 3 = \tfrac{6}{5}(3) + c \Rightarrow c = 3 - \tfrac{18}{5} = -\tfrac{3}{5}$。所以平分线是
$$y = \tfrac{6}{5}x - \tfrac{3}{5}.$$
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| perpendicular | 垂直 | chuí zhí |
| right angle | 直角 | zhí jiǎo |
| perpendicular bisector | 垂直平分线 | chuí zhí píng fēn xiàn |
3.7
考试技巧
- 直线是 $y = mx + c$:$m$ 是斜率而 $c$ 是线穿过 $y$ 轴的地方。
- 斜率 = ($y$ 的变化) ÷ ($x$ 的变化)。把这两个坐标在上部和底部保持相同的顺序。
- 平行线有相同的斜率;垂直线有相乘得 $-1$ 的斜率(负倒数)。
- 中点是坐标的平均;两个点之间的距离来自对差的勾股定理。
本主题的互动课程
逐步学习,并即时检测练习。