Lines, curves and coordinates
| English | Chinese | Pinyin |
|---|---|---|
| gradient | 斜率 | xié lǜ |
| y-intercept | y 轴截距 | y zhóu jié jù |
| parallel | 平行 | píng xíng |
| perpendicular | 垂直 | chuí zhí |
| distance | 距离 | jù lí |
| midpoint | 中点 | zhōng diǎn |
| parabola | 抛物线 | pāo wù xiàn |
| vertex | 顶点 | dǐng diǎn |
| roots | 根 | gēn |
| graphically | 用图像求解 | yòng tú xiàng qiú jiě |
An equation you can see
- $y = 2x - 3$ is a rule. Drawn, it is a straight line, and the line answers questions the rule hides.
- Where does it cross the axis? Where does it meet another line? A picture answers both at a glance.
- Coordinate geometry is how you measure that picture instead of guessing from it.
Gradient and intercept
- In $y = mx + c$, $m$ is the gradient 斜率 and $c$ the y-intercept y 轴截距.
- The gradient is rise over run: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.
- Parallel 平行 lines have equal gradients. Perpendicular 垂直 lines have gradients multiplying to $-1$.
Change the gradient and the intercept
$y = mx + c$
See what m and c each control before you calculate them.
Find the gradient of the line through (2, 1) and (6, 9).
Rise over run: (9 − 1) ÷ (6 − 2) = 8 ÷ 4 = 2.
That line is y = 2x + c. Find c.
Substitute (2, 1): 1 = 4 + c, so c = −3. Check with (6, 9): 2(6) − 3 = 9 ✓.
Which gradient is perpendicular to a line of gradient 4?
The gradients must multiply to −1, and 4 × (−1/4) = −1.
Find the line through $(2, 1)$ and $(6, 9)$.
$m = \dfrac{9 - 1}{6 - 2} = 2$.
Put $(2, 1)$ into $y = 2x + c$: $1 = 4 + c$, so $c = -3$.
The line is $y = 2x - 3$. Check the other point: $2(6) - 3 = 9$ ✓ — always check with the point you did not use.
Find the distance between (1, 2) and (4, 6).
Differences 3 and 4, so d = √(9 + 16) = √25 = 5 — the 3-4-5 triangle.
Distance and midpoint
- The midpoint 中点 is the average of the endpoints: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
- The distance 距离 comes from Pythagoras: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
- Both are used in the module's project work, where a "shape" is just a list of coordinates.
What are the roots of a quadratic, geometrically?
The roots solve y = 0, and y = 0 is the x-axis. The turning point is the vertex.
Put the steps of finding a line through two points in order.
Checking with the unused point is the step that catches an arithmetic slip.
Curves and intersections
- A quadratic draws a parabola 抛物线. Its vertex 顶点 is the turning point.
- The roots 根 are where it crosses the $x$-axis — the solutions of $y = 0$.
- To solve two equations graphically 用图像求解, draw both and read the intersection.
A graphical solution is only as accurate as the drawing. Use it to see how many solutions exist and roughly where — then solve algebraically for the value you write down.
A gradient of $-\frac{1}{2}$ and a gradient of $2$ are perpendicular, because $-\frac{1}{2} \times 2 = -1$. Students often test whether the gradients look opposite instead of multiplying them. Multiply; the sign is half the answer.