Length and midpoint · 长度与中点
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| line segment/laɪn ˈseɡmənt/ | 线段 | xiàn duàn |
| hypotenuse/haɪˈpɒtənjuːs/ | 斜边 | xié biān |
| Pythagoras/paɪˈθæɡərəs/ | 勾股定理 | gōu gǔ dìng lǐ |
| midpoint/ˈmɪdpɔɪnt/ | 中点 | zhōng diǎn |
How far, really?
- Your phone says the café is 1.2 km away — but that's the straight-line distance, not the zigzag walk along streets.
- In coordinate geometry we often need the exact straight-line distance between two points.
- The tool? A 2500-year-old theorem.
真的有多远?
- 你的手机说咖啡馆在 1.2 km 外——但那是直线距离,不是沿街道的曲折步行。
- 在坐标几何中我们常常需要两点之间的精确直线距离。
- 工具?一个 2500 年前的定理。
Length and midpoint lab · 长度与中点实验
midpoint is halfway between endpoints · 中点位于两个端点的正中间
Move along a line segment and see midpoint as halfway. · 沿线段移动，理解中点即为中点位置。
Length of a segment (Extended)
- A line segment 线段 is the straight piece between two points $(x_1, y_1)$ and $(x_2, y_2)$.
- Draw a right triangle with the segment as the hypotenuse 斜边:
- The horizontal gap is $x_2 - x_1$ and the vertical gap is $y_2 - y_1$.
- By Pythagoras 勾股定理, the length is:
The gap across (3) and the gap up (4) give the length by Pythagoras; the midpoint 中点 is the average of the coordinates
一条线段的长度(扩展)
- 一条线段(line segment)是两点 $(x_1, y_1)$ 和 $(x_2, y_2)$ 之间的直的那一段。
- 画一个以该线段为斜边的直角三角形:
- 水平间隙是 $x_2 - x_1$,竖直间隙是 $y_2 - y_1$。
- 由勾股定理(Pythagoras),长度是:
横向的间隙(3)和向上的间隙(4)由勾股定理给出长度;中点是坐标的平均
Worked example — length
- Find the length from $(1, 2)$ to $(4, 6)$.
- Horizontal gap $= 4 - 1 = 3$; vertical gap $= 6 - 2 = 4$.
- Length $= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
- The classic 3-4-5 right triangle makes a clean answer.
Pythagoras finds the length (hypotenuse); averaging finds the midpoint (star).
From the origin. Length from $(0, 0)$ to $(3, 4)$: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Same triangle, different starting point.
例题——长度
- 求从 $(1, 2)$ 到 $(4, 6)$ 的长度。
- 水平间隙 $= 4 - 1 = 3$;竖直间隙 $= 6 - 2 = 4$。
- 长度 $= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$。
- 经典的 3-4-5 直角三角形给出一个干净的答案。
勾股定理找到长度(斜边);求平均找到中点(星)。
从原点。 从 $(0, 0)$ 到 $(3, 4)$ 的长度:$\sqrt{3^2 + 4^2} = \sqrt{25} = 5$。相同的三角形,不同的起点。
Find the length of the segment from (1, 2) to (4, 6). · 求从 (1, 2) 到 (4, 6) 的线段长度。
√((4−1)² + (6−2)²) = √(9 + 16) = √25 = 5.
Find the length of the segment from (0, 0) to (3, 4). · 求从 (0, 0) 到 (3, 4) 的线段长度。
√(3² + 4²) = √(9 + 16) = √25 = 5. The classic 3-4-5 triangle. · √(3² + 4²) = √(9 + 16) = √25 = 5。经典的 3-4-5 三角形。
Find the length of the segment from (1, 5) to (7, 5). · 求从 (1, 5) 到 (7, 5) 的线段长度。
Both points have y = 5 (horizontal segment). Length = 7 − 1 = 6. · 两点 y 坐标均为 5（水平线段）。长度 = 7 − 1 = 6。
Midpoint of a segment (Extended)
- The midpoint is the point exactly halfway between the two endpoints.
- Find it by averaging each coordinate:
- It's the "halfway" $x$ paired with the "halfway" $y$.
一条线段的中点(扩展)
- 中点(midpoint)是恰好在两个端点中间的点。
- 通过求平均每个坐标找到它:
- 它是"中间的" $x$ 与"中间的" $y$ 配对。
Worked example — midpoint
- Find the midpoint from $(1, 2)$ to $(4, 6)$.
- Mid-$x = \dfrac{1 + 4}{2} = \dfrac{5}{2} = 2.5$.
- Mid-$y = \dfrac{2 + 6}{2} = \dfrac{8}{2} = 4$.
- Midpoint $= (2.5,\, 4)$ — exactly halfway along.
Average, don't add. The midpoint of $(1, 2)$ and $(4, 6)$ is $(2.5, 4)$, not $(5, 8)$. You must divide by 2 — averaging, not summing.
例题——中点
- 求从 $(1, 2)$ 到 $(4, 6)$ 的中点。
- 中 $x = \dfrac{1 + 4}{2} = \dfrac{5}{2} = 2.5$。
- 中 $y = \dfrac{2 + 6}{2} = \dfrac{8}{2} = 4$。
- 中点 $= (2.5,\, 4)$——恰好在沿途的一半。
求平均,不要相加。 $(1, 2)$ 和 $(4, 6)$ 的中点是 $(2.5, 4)$,不是 $(5, 8)$。你必须除以 2——求平均,不是求和。
The midpoint of (1, 2) and (4, 6) is (a, 4). What is a? · (1, 2) 和 (4, 6) 的中点是 (a, 4)。a 是多少？
Average the x-values: (1 + 4)/2 = 2.5. · 取 x 值的平均数：(1 + 4)/2 = 2.5。
The midpoint of (2, 3) and (8, 9) is (5, b). What is b? · (2, 3) 和 (8, 9) 的中点是 (5, b)。b 是多少？
Average the y-values: (3 + 9)/2 = 6. · 取 y 值的平均数：(3 + 9)/2 = 6。
The x-coordinate of the midpoint of (x₁, y₁) and (x₂, y₂) is: · (x₁, y₁) 和 (x₂, y₂) 的中点的 x 坐标为：
The midpoint is the average of each coordinate: (x₁ + x₂)/2. Choice 2 forgets to divide; choice 3 gives half the gap, not the position. · 中点是每个坐标的平均值：(x₁ + x₂)/2。选项 2 忘记除以；选项 3 给出的是间距的一半，而不是位置。
Complete: the midpoint x-coordinate = (x₁ + ______) / 2. · 填空：中点的 x 坐标 = (x₁ + ______) / 2。
The midpoint x-coordinate is the average of the two x-values: (x₁ + x₂)/2. · 中点的 x 坐标是两个 x 值的平均数：(x₁ + x₂)/2。
The 3-4-5 triangle family
- The triangle with sides $3, 4, 5$ is the smallest right triangle with whole-number sides.
- Multiples work too: $6, 8, 10$ and $9, 12, 15$ are also right triangles.
- You'll see this family a lot in exam questions — recognise it and save time.
3-4-5 三角形家族
- 边为 $3, 4, 5$ 的三角形是有整数边的最小直角三角形。
- 倍数也有效:$6, 8, 10$ 和 $9, 12, 15$ 也是直角三角形。
- 你会在考题中经常看到这个家族——识别它并节省时间。
For a horizontal segment (both y-values the same), the length is simply the difference of the x-values. · 对于水平线段（y 值相同），长度即为 x 值之差。
When y₁ = y₂, the formula simplifies to √((x₂−x₁)²) = |x₂ − x₁|. · 当 y₁ = y₂ 时，公式简化为 √((x₂−x₁)²) = |x₂ − x₁|。
Match each segment to its length. · 将每条线段与其长度匹配。
(0,0)–(5,0) is horizontal, length 5. (1,1)–(1,4) is vertical, length 3. (0,0)–(6,8): √(36+64) = √100 = 10. (2,3)–(5,7): √(9+16) = √25 = 5. · (0,0)–(5,0) 是水平线，长度为 5。(1,1)–(1,4) 是垂直线，长度为 3。(0,0)–(6,8)：√(36+64) = √100 = 10。(2,3)–(5,7)：√(9+16) = √25 = 5。
You've got it
- length $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ (Pythagoras on the horizontal and vertical gaps)
- midpoint $= \left( \dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2} \right)$ (average each coordinate)
- $(1,2)$ to $(4,6)$: length $5$, midpoint $(2.5,\, 4)$
- the 3-4-5 right triangle appears often — spot it early
你掌握了
- 长度 $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$(对水平和竖直间隙用勾股定理)
- 中点 $= \left( \dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2} \right)$(求平均每个坐标)
- $(1,2)$ 到 $(4,6)$:长度 $5$,中点 $(2.5,\, 4)$
- 3-4-5 直角三角形经常出现——尽早发现它