Coordinate geometry · 坐标几何
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| gradient/ˈɡreɪdɪənt/ | 斜率 | xié lǜ |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | 垂直 | chuí zhí |
| midpoint/ˈmɪdpɔɪnt/ | 中点 | zhōng diǎn |
| perpendicular bisector/ˌpɜːpənˈdɪkjʊlə baɪˈsektə/ | 垂直平分线 | chuí zhí píng fēn xiàn |
| tangent/ˈtændʒənt/ | 切线 | qiè xiàn |
The mathematics of location
- GPS satellites pinpoint your location to within a few metres using coordinate geometry.
- Every map, every blueprint, every computer graphic relies on the same core ideas: lines, circles, and the distances between points.
位置的数学
- GPS 卫星用坐标几何把你的位置精确到几米以内。
- 每张地图、每张蓝图、每个计算机图形都依赖同样的核心想法:线、圆,以及点之间的距离。
Straight lines
- Gradient 斜率 of the line through $(x_1, y_1)$ and $(x_2, y_2)$: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.
- Line equations: $y = mx + c$ (slope-intercept form), or $y - y_1 = m(x - x_1)$ (point-slope form).
- Parallel lines have equal gradients; perpendicular 垂直 lines have gradients that multiply to $-1$.
Worked example. Line through $(2, 3)$ and $(6, 11)$: $m = \dfrac{11-3}{6-2} = \dfrac{8}{4} = 2$. Equation: $y - 3 = 2(x - 2) \Rightarrow y = 2x - 1$.
Perpendicular gradient trap. If a line has gradient $2$, the perpendicular gradient is $-\dfrac{1}{2}$, not $-2$. You must flip the fraction and change the sign.
直线
- 过 $(x_1, y_1)$ 和 $(x_2, y_2)$ 的线的斜率(gradient):$m = \dfrac{y_2 - y_1}{x_2 - x_1}$。
- 直线方程:$y = mx + c$(斜截式),或 $y - y_1 = m(x - x_1)$(点斜式)。
- 平行(parallel)线有相等的斜率;垂直(perpendicular)线的斜率相乘得 $-1$。
算例。 过 $(2, 3)$ 和 $(6, 11)$ 的线:$m = \dfrac{11-3}{6-2} = \dfrac{8}{4} = 2$。方程:$y - 3 = 2(x - 2) \Rightarrow y = 2x - 1$。
垂直斜率陷阱。 如果一条线的斜率是 $2$,垂直斜率是 $-\dfrac{1}{2}$,不是 $-2$。你必须翻转分数并改变符号。
The straight line · 直线
y = ax + b
The gradient a tilts the line; the intercept b slides it up and down. · 斜率 a 倾斜这条线;截距 b 上下滑动它。
What is the gradient of the line through (2, 3) and (6, 11)? · 过 (2, 3) 和 (6, 11) 的线的斜率是多少?
m = (11 − 3) / (6 − 2) = 8/4 = 2. · m = (11 − 3) / (6 − 2) = 8/4 = 2。
A line has gradient 2. What is the gradient of a line perpendicular to it? · 一条线的斜率是 2。垂直于它的一条线的斜率是多少?
Perpendicular gradients multiply to −1, so the gradient is −1/2 = −0.5. · 垂直斜率相乘得 −1,所以斜率是 −1/2 = −0.5。
Midpoints 中点 and parallel lines
- The midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.
- The perpendicular bisector 垂直平分线 passes through the midpoint with the perpendicular gradient.
中点与平行线
- $(x_1, y_1)$ 和 $(x_2, y_2)$ 的中点(midpoint)是 $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$。
- 垂直平分线(perpendicular bisector)过中点,带垂直斜率。
The midpoint of (2, 3) and (8, 7) has x-coordinate: · (2, 3) 和 (8, 7) 的中点的 x 坐标是:
Midpoint x = (2 + 8)/2 = 5. · 中点 x = (2 + 8)/2 = 5。
Circles
- A circle, centre $(a, b)$, radius $r$: $(x - a)^2 + (y - b)^2 = r^2$.
- An expanded form (e.g. $x^2 + y^2 - 6x + 10y - 27 = 0$) → complete the square in $x$ and $y$ to find the centre and radius.
The equation $(x-3)^2 + (y+2)^2 = 25$ describes a circle with centre $(3, -2)$ and radius $5$.
圆
- 一个圆,圆心 $(a, b)$,半径 $r$:$(x - a)^2 + (y - b)^2 = r^2$。
- 一个展开的形式(例如 $x^2 + y^2 - 6x + 10y - 27 = 0$)→ 在 $x$ 和 $y$ 中配方来找圆心和半径。
方程 $(x-3)^2 + (y+2)^2 = 25$ 描述一个圆心 $(3, -2)$、半径 $5$ 的圆。
The circle (x − 3)² + (y + 2)² = 25 has what radius? · 圆 (x − 3)² + (y + 2)² = 25 的半径是多少?
r² = 25, so r = 5; the centre is (3, −2). · r² = 25,所以 r = 5;圆心是 (3, −2)。
The circle x² + y² − 6x + 10y − 27 = 0 has centre: · 圆 x² + y² − 6x + 10y − 27 = 0 的圆心是:
Complete the square: (x−3)² + (y+5)² = 9 + 25 + 27 = 61. Centre is (3, −5). · 配方:(x−3)² + (y+5)² = 9 + 25 + 27 = 61。圆心是 (3, −5)。
Tangents 切线 to circles
- A tangent touches the circle once and is perpendicular to the radius there.
- This right-angle fact solves most circle problems: find the radius gradient, then the tangent gradient is $-\dfrac{1}{m}$.
圆的切线
- 一条切线(tangent)接触圆一次,并在那里垂直于半径。
- 这个直角事实解决大多数圆的问题:找到半径斜率,然后切线斜率是 $-\dfrac{1}{m}$。
A tangent to a circle is perpendicular to the radius at the point where it touches. · 一个圆的切线在它接触的点垂直于半径。
This right-angle fact solves most circle problems. · 这个直角事实解决大多数圆的问题。
You've got it
- gradient $m = \dfrac{y_2 - y_1}{x_2 - x_1}$; perpendicular gradients multiply to $-1$
- circle: $(x-a)^2 + (y-b)^2 = r^2$, centre $(a,b)$, radius $r$
- a tangent is perpendicular to the radius at the point of contact
- midpoint $= \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
你掌握了
- 斜率 $m = \dfrac{y_2 - y_1}{x_2 - x_1}$;垂直斜率相乘得 $-1$
- 圆:$(x-a)^2 + (y-b)^2 = r^2$,圆心 $(a,b)$,半径 $r$
- 一条切线在接触点垂直于半径
- 中点 $= \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$