Course: 11.11 远距离公式

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距离公式
The Distance Formula
::距离公式

In the last section, we saw how to use the Pythagorean Theorem to find lengths. In this section, you’ll learn how to use the Pythagorean Theorem to find the distance between two coordinate points.
::在最后一节,我们看到如何使用毕达哥里安神话来寻找长度。在本节中,你将学会如何使用毕达哥里安神话来找到两个坐标点之间的距离。

Finding the Distance Between Two Points
::寻找两个点之间的距离

1. Find the distance between points $\begin{align*}A = (1, 4)\end{align*}$ and $\begin{align*}B = (5, 2)\end{align*}$ .
::1. 找出A=(1,4)和B=(5,2)之间的距离。

Plot the two points on the coordinate plane .
::在坐标飞机上标出两点

In order to get from point $\begin{align*}A = (1, 4)\end{align*}$ to point $\begin{align*}B = (5, 2)\end{align*}$ , we need to move 4 units to the right and 2 units down. These lines make the legs of a right triangle .
::为了从A=(1,4)点到B=(5,5,2)点,我们需要将4个单元移到右边,并将2个单元移到下方。这些线将右三角形的腿划入右三角形。

To find the distance between $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ we find the value of the hypotenuse , $\begin{align*}d\end{align*}$ , using the Pythagorean Theorem.
::为了找到A和B之间的距离我们用毕达哥伦神话来找到下限值 d

$\begin{aligned} d^{2} & = 2^{2} + 4^{2} = 20 \\ d & = \sqrt{20} = 2 \sqrt{5} = 4.47 \end{aligned}$
$\begin{align*}d^2 &= 2^2+4^2=20\\ d &= \sqrt{20}=2 \sqrt{5}=4.47\end{align*}$
::d2=22+42=20d=20=25=4.47

2. Find the distance between points $\begin{align*}C = (2, -1)\end{align*}$ and $\begin{align*}D = (-3, -4)\end{align*}$ .
::2. 找出C=(2,-1)和D=(-3,-4)之间的距离。

We plot the two points on the graph above.
::我们在上图上绘制两点。

In order to get from point $\begin{align*}C\end{align*}$ to point $\begin{align*}D\end{align*}$ , we need to move 3 units down and 5 units to the left.
::为了从C点到D点我们需要把3个单位往下移动 5个单位往左移动

We find the distance from $\begin{align*}C\end{align*}$ to $\begin{align*}D\end{align*}$ by finding the length of $\begin{align*}d\end{align*}$ with the Pythagorean Theorem.
::我们发现从C到D的距离通过找到与毕达哥里安神话的距离

$\begin{aligned} d^{2} & = 3^{2} + 5^{2} = 34 \\ d & = \sqrt{34} = 5.83 \end{aligned}$
$\begin{align*}d^2 &= 3^2+5^2=34\\ d &= \sqrt{34}=5.83\end{align*}$
::d2=32+52=34d=34=5.83

The Distance Formula
::距离公式

The procedure we just used can be generalized by using the Pythagorean Theorem to derive a formula for the distance between any two points on the coordinate plane.
::我们刚才使用的程序可以通过使用毕达哥里安理论来得出坐标平面上任何两点之间距离的公式,从而实现普遍化。

Let’s find the distance between two general points $\begin{align*}A=(x_1,y_1)\end{align*}$ and $\begin{align*}B=(x_2,y_2)\end{align*}$ .
::让我们找到两个一般点A=(x1,y1)和B=(x2,y2)之间的距离。

Start by plotting the points on the coordinate plane:
::首先绘制坐标平面上的点数 :

In order to move from point $\begin{align*}A\end{align*}$ to point $\begin{align*}B\end{align*}$ in the coordinate plane, we move $\begin{align*}x_2 - x_1\end{align*}$ units to the right and $\begin{align*}y_2 - y_1\end{align*}$ units up.
::为了从坐标平面的A点移到B点,我们把X2-x1单位移到右边,然后将 y2-y1 单位向上移动。

We can find the length $\begin{align*}d\end{align*}$ by using the Pythagorean Theorem:
::使用《毕达哥里安理论论》可以找到长度(d) :

$\begin{aligned} d^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} \end{aligned}$
$\begin{align*}d^2=(x_1-x_2)^2+(y_1-y_2)^2\end{align*}$
::d2=(x1-x2)2+(y1-y2)2

Therefore , $\begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\end{align*}$ . This is called the Distance Formula. More formally:
::因此, d= (x1- x2) 2+(y1-y2) 2。这称为距离公式。更正式地说 :

Given any two points $\begin{align*}(x_1, y_1)\end{align*}$ and $\begin{align*}(x_2, y_2)\end{align*}$ , the distance between them is $\begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.\end{align*}$
::根据任何两个点(x1,y1)和(x2,y2),它们之间的距离是 d=(x1-x2)2+(y1-y2)。

We can use this formula to find the distance between any two points on the coordinate plane. Notice that the distance is the same whether you are going from point $\begin{align*}A\end{align*}$ to point $\begin{align*}B\end{align*}$ or from point $\begin{align*}B\end{align*}$ to point $\begin{align*}A\end{align*}$ , so it does not matter which order you plug the points into the distance formula.
::我们可以使用此公式来找到坐标平面上任何两个点之间的距离。请注意, 无论您从 A 点到 B 点, 还是从 B 点到 A 点, 距离是一样的, 因此命令您将点插入距离公式的顺序并不重要。

Let’s now apply the distance formula to the following examples.
::让我们现在对以下例子应用距离公式。

Using the Distance Formula
::使用距离公式

Plug the values of the two points into the distance formula. Be sure to simplify if possible.
::将两个点的值插入距离公式中。请尽可能简化。

a) (-3, 5) and (4, -2)
::a) (3,5)和(4,2)

$\begin{align*}d=\sqrt{(-3-4)^2+(5-(-2))^2}=\sqrt{49+49}=\sqrt{98}=7 \sqrt{2}\end{align*}$
::d=(-3-4)2+(5-2)2=49+49=98=72

b) (12, 16) and (19, 21)
:b) (12、16)和(19、21)

$\begin{align*}d=\sqrt{(12-19)^2+(16-21)^2}=\sqrt{49+25}=\sqrt{74}\end{align*}$
::d=(12-19)2+(16-21)2=49+25=74

c) (11.5, 2.3) and (-4.2, -3.9)
:c) (11.5、2.3)和(-4.2、-3.9)

$\begin{align*}d=\sqrt{(11.5+4.2)^2+(2.3+3.9)^2}=\sqrt{284.93}=16.88\end{align*}$
::d=(11.5+4.2)2+(2.3+3.9.9)2=284.93=16.88

Applications Using Distance and Midpoint Formulas
::使用距离和中点公式的应用

The distance and are useful in geometry situations where we want to find the distance between two points or the point halfway between two points.
::在几何情况中,我们想要找到两点之间的距离或两点之间的中间点,这种距离是有用的。

Proving that a Triangle is Isosceles
::证明三角是伊索塞尔

Plot the points $\begin{align*}A = ( 4, -2), B = (5, 5)\end{align*}$ , and $\begin{align*}C = (-1, 3)\end{align*}$ and connect them to make a triangle. Show that the triangle is isosceles.
::绘制点 A = (4, - 2, B = (5, 5, 5) 和 C = (- 1, 3) , 并将其连接成三角形。显示三角形为等分。

Let’s start by plotting the three points on the coordinate plane and making a triangle:
::让我们首先在坐标平面上绘制三点图,

We use the distance formula three times to find the lengths of the three sides of the triangle.
::我们用距离公式三次来找到三角形三边的长度

$\begin{aligned} A B & = \sqrt{(4 - 5)^{2} + (- 2 - 5)^{2}} = \sqrt{(- 1)^{2} + (- 7)^{2}} = \sqrt{50} = 5 \sqrt{2} \\ B C & = \sqrt{(5 + 1)^{2} + (5 - 3)^{2}} = \sqrt{(6)^{2} + (2)^{2}} = \sqrt{40} = 2 \sqrt{10} \\ A C & = \sqrt{(4 + 1)^{2} + (- 2 - 3)^{2}} = \sqrt{(5)^{2} + (- 5)^{2}} = \sqrt{50} = 5 \sqrt{2} \end{aligned}$
$\begin{align*}AB &= \sqrt{(4-5)^2+(-2-5)^2}=\sqrt{(-1)^2+(-7)^2}=\sqrt{50}=5 \sqrt{2}\\ BC &= \sqrt{(5+1)^2+(5-3)^2}=\sqrt{(6)^2+(2)^2}=\sqrt{40}=2 \sqrt{10}\\ AC &= \sqrt{(4+1)^2+(-2-3)^2}=\sqrt{(5)^2+(-5)^2}=\sqrt{50}=5 \sqrt{2}\end{align*}$
::AB=(4-5-5)2+(-2-2)2+(-2-5)2=(-1)2+(-7)2=50=52BC=(5+1)2+(5-3)2=(6)2+(2)2=(6)2+(2)2=40=210AC=(4+1)2+(-2-3)2+(-2-3)2=(5)2+(-5)2=50=52

Notice that $\begin{align*}AB = AC\end{align*}$ , therefore triangle $\begin{align*}ABC\end{align*}$ is isosceles.
::请注意AB=AC, 因此三角ABC是等分形。

Real-World Application: Walking Speed
::真实世界应用程序:行走速度

At 8 AM one day, Amir decides to walk in a straight line on the beach. After two hours of making no turns and traveling at a steady rate , Amir is two miles east and four miles north of his starting point . How far did Amir walk and what was his walking speed ?
::一天早上8点,阿米尔决定沿着直线走在海滩上。经过两小时的不转弯和稳步旅行后,阿米尔在起点以东两英里、以北四英里处。阿米尔走多远以及他的行走速度如何?

Let’s start by plotting Amir’s route on a coordinate graph. We can place his starting point at the origin: $\begin{align*}A = (0, 0)\end{align*}$ . Then his ending point will be at $\begin{align*}B = (2, 4)\end{align*}$ .
::让我们先在坐标图上绘制埃米尔的路线。我们可以把他的起点放在起点: A=(0), 然后他的终点在B=(2, 4) 。

The distance can be found with the distance formula:
::可用距离公式找到距离 :

$\begin{aligned} d & = \sqrt{(2 - 0)^{2} + (4 - 0)^{2}} = \sqrt{(2)^{2} + (4)^{2}} = \sqrt{4 + 16} = \sqrt{20} \\ d & = \underline{\underline{4.47 m i l e s}} \end{aligned}$
$\begin{align*}d &= \sqrt{(2-0)^2+(4-0)^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\\ d &= \underline{\underline{4.47 \ miles}}\end{align*}$
::d=(2-0)2+(4-0)2=(2)2+(4)2=4+16=20d=4.47英里__

Since Amir walked 4.47 miles in 2 hours, his speed is $\begin{align*}s=\frac{4.47 \ miles}{2 \ hours}=\underline{\underline{2.24 \ mi/h}}\end{align*}$ .
::由于Amir在2小时内行走4.47英里,他的速度是S=4.47英里2小时=2.24米/小时。

Example
::示例示例示例示例

Example 1
::例1

Find all points on the line $\begin{align*}y = 2\end{align*}$ that are exactly 8 units away from the point (-3, 7).
::查找线y=2上的所有点,即离点(-3、7)不远的8个单位。

Let’s make a sketch of the given situation.
::让我们来勾画一下特定情况。

Draw line segments from the point (-3, 7) to the line $\begin{align*}y = 2\end{align*}$ .
::从点( 3, 7) 到 y=2 线条绘制线条段。

Let $\begin{align*}k\end{align*}$ be the missing value of $\begin{align*}x\end{align*}$ we are seeking.
::k 成为我们所寻求的 x 的缺失值。

$\begin{aligned} Let's use the distance formula: & 8 & = \sqrt{(- 3 - k)^{2} + (7 - 2)^{2}} \\ Square both sides of the equation: & 64 & = (- 3 - k)^{2} + 25 \\ Therefore: & 0 & = 9 + 6 k + k^{2} - 39 or 0 = k^{2} + 6 k - 30 \\ Use the quadratic formula: & k & = \frac{- 6 \pm \sqrt{36 + 120}}{2} = \frac{- 6 \pm \sqrt{156}}{2} \\ Therefore: & k & = 3.24 or k = - 9.24 \end{aligned}$
$\begin{align*}\text{Let's use the distance formula:} && 8& =\sqrt{(-3-k)^2+(7-2)^2}\\ \text{Square both sides of the equation:} && 64& =(-3-k)^2+25\\ \text{Therefore:} && 0& =9+6k+k^2-39 \ \text{or} \ 0=k^2+6k-30\\ \text{Use the quadratic formula:} && k& =\frac{-6 \pm \sqrt{36 + 120}}{2}=\frac{-6 \pm \sqrt{156}}{2}\\ \text{Therefore:} && k& =3.24 \ \text{or} \ k=-9.24\end{align*}$
::让我们使用距离公式: 8=(- 3- k) 2+( 7- 2) 2Square 等式的两边 : 64= (- 3- k) 2+ 25 因此: 0= 9+6k+k2- 39 或 0= k2+6k- 30 使用四方形公式: k @ 6+36+1202\\\\ 61562

The points are (-9.24, 2) and (3.24, 2).
::要点是(9.24,2)和(3.24,2)。

Review
::回顾

Find the distance between the two points.
::查找两个点之间的距离。

(3, -4) and (6, 0)
:3,4)和(6,0)

(-1, 0) and (4, 2)
:-1,0)和(4,2)

(-3, 2) and (6, 2)
:-3,2)和(6,2)

(0.5, -2.5) and (4, -4)
:0.5,-2.5)和(4,4,4)

(12, -10) and (0, -6)
:12,-10)和(0,-6)

(-5, -3) and (-2, 11)
:-5、3-3)和(-2、11)

(2.3, 4.5) and (-3.4, -5.2)
:2.3,4.5)和(3.4,-5.2)

Find all points having an $\begin{align*}x-\end{align*}$ coordinate of -4 whose distance from the point (4, 2) is 10.
::查找与点(4,2)的距离为10的 x - 坐标 - 4 的所有点。

Find all points having a $\begin{align*}y-\end{align*}$ coordinate of 3 whose distance from the point (-2, 5) is 8.
::查找与点(-2, 5) 相距为 8 的 3 的 Y - 坐标的所有点。

Find three points that are each 13 units away from the point (3, 2) but do not have an $\begin{align*}x-\end{align*}$ coordinate of 3 or a $\begin{align*}y-\end{align*}$ coordinate of 2.
::查找三个点,即距离点(3,2)的13个单位各有3个点,但没有3个的x-坐标或2个的y-坐标。

Plot the points $\begin{align*}A = (1, 0), B = (6, 4), C = (9, -2)\end{align*}$ and $\begin{align*}D = (-6, -4), E = (-1, 0), F = (2, -6)\end{align*}$ . Prove that triangles $\begin{align*}ABC\end{align*}$ and $\begin{align*}DEF\end{align*}$ are congruent.
::绘制点A=(1,0,B=(6,4,4),C=(9,-2)和D=(-6,4),E=(-1,0),F=(2,6),证明ABC和DEF三角是相同的。

Plot the points $\begin{align*}A = (4, -3), B = (3, 4), C = (-2, -1), D = (-1, -8).\end{align*}$ Show that $\begin{align*}ABCD\end{align*}$ is a rhombus (all sides are equal)
::绘制点 A = (4) 3, B = (3, 4), C = (-2, - 1), D = (-1), D = (-1) 。显示ABCD 是一种暴风( 各方相等) 。

Plot points $\begin{align*}A = (-5, 3), B = (6, 0), C = (5, 5).\end{align*}$ Find the length of each side. Show that $\begin{align*}ABC\end{align*}$ is a right triangle. Find its area.
::绘图点 A = (- 5, 3), B = (6, 0, C = (5, 5, 5) 。查找每一边的长度。显示ABC 是右三角形。查找其区域。

Find the area of the circle with center (-5, 4) and a point on the circle (3, 2).
::在圆上找到圆区域,以中间( 5, 4) 和圆上的一个点( 3, 2) 。

Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles and then she rides in a new direction for a while longer. When she stops Michelle is 2 miles south and 10 miles west from her starting point. Find the total distance that Michelle covered from her starting point.
::米歇尔决定有一天骑自行车。首先,她骑着从南到南12英里的自行车,然后骑着新方向再长一阵子。当她阻止米歇尔时,她从起点往南2英里,向西10英里。找到米歇尔从起点所覆盖的总距离。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

距离公式

The Distance Formula ::距离公式

Finding the Distance Between Two Points ::寻找两个点之间的距离

The Distance Formula ::距离公式

Using the Distance Formula ::使用距离公式

Applications Using Distance and Midpoint Formulas ::使用距离和中点公式的应用

Proving that a Triangle is Isosceles ::证明三角是伊索塞尔

Real-World Application: Walking Speed ::真实世界应用程序:行走速度

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

The Distance Formula
::距离公式

Finding the Distance Between Two Points
::寻找两个点之间的距离

The Distance Formula
::距离公式

Using the Distance Formula
::使用距离公式

Applications Using Distance and Midpoint Formulas
::使用距离和中点公式的应用

Proving that a Triangle is Isosceles
::证明三角是伊索塞尔

Real-World Application: Walking Speed
::真实世界应用程序:行走速度

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)