课程： 3.12 垂直直线 | The Way To Learn

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直直直线
In most cities, it is common for the streets to be laid out in a grid format like the example below. Do any of the streets intersect at right angles? In this section, we consider pairs of lines with that property .

Perpendicular Lines
::直直直线

Two lines are perpendicular if they intersect at a right or 90 degree angle. Notice that the two lines below have slopes that are different signs—the blue line has a positive slope and the red line has a negative slope . This is one aspect of the relationship between slopes of perpendicular lines .
::两条线如果在右角或90度角交叉,是垂直的。请注意,以下两条线的斜坡有不同的标志——蓝线有正斜坡,红线有负斜坡。这是垂直线斜坡之间关系的一个方面。

Now, let's calculate the slopes of these lines. The blue line goes through the origin and (1,2). The red line goes through (0,1) and (1,-1).
::现在,让我们来计算这些线的斜坡。蓝线通过起源(1, 2) 。红线通过( 0, 1) 和(1, 1) 。

$\begin{aligned} Blue Line & Red Line \\ m = \frac{1 - 0}{2 - 0} & m = \frac{1 - (- 1)}{0 - 1} \\ = \frac{1}{2} & = \frac{2}{- 1} = - 2 \end{aligned}$

::蓝线线=1-02-0-0m=1-(-1-1)-0-1=12=2-12

In addition to the fact that the slopes have the opposite signs, notice that the slopes are reciprocals, or multiplicative inverses, of each other. These are our two properties of slopes of perpendicular lines.
::除了斜坡有相反的标志之外,还要注意斜坡是相互对等的,或相互重复的反向,这是我们两个垂直线斜坡的特性。

Perpendicular Lines
::直直直线

Two lines are perpendicular if their slopes are negative reciprocals of each other. That is, if the of one line is $\begin{aligned} m_{1} \end{aligned}$ and the slope of the other line is $\begin{aligned} m_{2} \end{aligned}$ , then the slopes are related by
::如果两条线的斜坡是相互负对等的,则两条线是垂直的。也就是说,如果一条线是m1,另一条线的斜坡是m2,则两条斜坡是相关的。

$\begin{aligned} m_{1} = - \frac{1}{m_{2}} OR m_{1} \cdot m_{2} = - 1 \end{aligned}$

::m1,%1,m2ORM1,m2%1,%1,%1,%1,%1,%1,%1,%2,%1,%1,%1,%1,%1,%1,%2

by CK-12 demonstrates how to graph perpendicular lines.
::用 CK-12 演示如何绘制垂直直线图。

Example 1
::例1

Assuming that the boxes on the grid are 1 unit each, determine if any of the streets below are perpendicular to each other.
::假设网格上的盒子各是一个单元,确定下面的街道是否相互相连。

Solution: Using the graphed endpoints of each street, we find the slope of A Street by finding the rise over the run . Since we go down 5 boxes and to the right by 10 boxes, the slope is $\begin{aligned} - \frac{1}{2} \end{aligned}$ .
::解决方案:使用每条街道的图表端点,我们发现A街的斜坡是横越高处。自从我们从5个箱子下到右到10个箱子,斜坡就是 - 12。

For 7th Avenue, we go up 10 boxes and to the right 2 boxes, so the slope is 2.
::第七大道上,我们上10个箱子右2个箱子,斜坡是2个

Lastly, for Lincoln Boulevard, we go up 4 boxes and to the right 8 boxes, so the slope is $\begin{aligned} \frac{1}{2} \end{aligned}$ .
::最后,对于林肯大道,我们上4个箱子右8个箱子,所以斜坡是12个。

Because they are negative reciprocals of each other, A St. and 7th Ave. are perpendicular: $\begin{aligned} - \frac{1}{2} \cdot 2 = - 1 \end{aligned}$ . Lincoln Blvd. is not perpendicular to 7th Ave. since their slopes are both positive and it is not perpendicular to A St. since their slopes are not reciprocals of each other.
::A St. 和 7th Ave. 相互对等是负面的,因此,A St. 和 7th Ave. 是垂直的: - 1221. Lincoln Blvd. 与 7 ve. 并不垂直,因为它们的斜坡都是正的,与 A St. 并不垂直,因为它们的斜坡不是对等的。

by CK-12 demonstrates how to graph perpendicular lines in the coordinate plane .
::CK-12 演示如何在坐标平面上绘制垂直直线图。

Example 2
::例2

Are $\begin{aligned} x - 4 y = 8 \end{aligned}$ and $\begin{aligned} 2 x + 8 y = - 32 \end{aligned}$ parallel, perpendicular, or neither?
::x-4y=8和2x+8y32是平行的,垂直的,还是两者兼有?

Solution: To determine if the two lines are parallel or perpendicular, we need to change them both into slope-intercept form to find the slopes.
::解决办法:为了确定这两条线是平行的还是垂直的,我们需要将这两条线都改为斜坡间隔形式,以找到斜坡。

$\begin{aligned} x - 4 y & = 8 2 x + 8 y = - 32 \\ - 4 y & = - x + 8 8 y = - 2 x - 32 \\ y & = \frac{1}{4} x - 2 y = - \frac{1}{4} x - 4 \end{aligned}$

::x- 4y= 82x+8y_ 32- 4y_ x+8 8 y_ 2x- 32y= 14x- 2 y 14x- 4

Now, compare the slopes. One is $\begin{aligned} \frac{1}{4} \end{aligned}$ and the other is $\begin{aligned} - \frac{1}{4} \end{aligned}$ . They are not the same, so they are not parallel. To be perpendicular, the slopes need to be reciprocals, which they are not. Therefore , these two lines are neither parallel or perpendicular.
::现在,比较斜坡。一个是14, 另一个是14。它们不一样, 所以不平行。要垂直, 斜坡必须是对等的, 而不是对等的。因此, 这两条线既不是平行的, 也不是对等的。

Example 3
::例3

Find the equation of the line that is perpendicular to $\begin{aligned} 2 x - 3 y = 15 \end{aligned}$ and passes through (6, 5).
::查找直径为 2x-3y=15 的直线的方程,然后通过 (6, 5) 。

Solution: First, we need to change the line from general form to slope-intercept form.
::解决办法:首先,我们需要将线从一般形式改为斜坡干扰形式。

$\begin{aligned} 2 x - 3 y & = 15 \\ - 3 y & = - 2 x + 15 \\ y & = \frac{2}{3} x - 5 \end{aligned}$

::2x-3y=15-3y2x+15y=23x-5

The slope of the line that is perpendicular to this one will satisfy the equation $\begin{aligned} \frac{2}{3} \cdot m = - 1 \end{aligned}$ . We solve for m .
::直径于此线的斜坡将满足方程式23m1。我们解决 m。

$\begin{aligned} \frac{2}{3} \cdot m & = - 1 \\ \frac{3}{2} \cdot \frac{2}{3} \cdot m & = - 1 \cdot \frac{3}{2} \\ m & = - \frac{3}{2} \end{aligned}$

::23号 23号 31号 32号 32号 32号

Notice that the new slope is the negative reciprocal of the original slope. Using slope-intercept form, we use the given point to find the $\begin{aligned} y - \end{aligned}$ intercept .
::注意新斜坡是原斜坡的负对等值。我们使用斜坡截取形式, 使用给定点来查找 y- 截取。

$\begin{aligned} 5 & = - \frac{3}{2} (6) + b \\ 5 & = - 9 + b \\ 14 & = b \end{aligned}$

::532(6)+b59+b14=b

The equation of the line that is perpendicular to $\begin{aligned} y = \frac{2}{3} x - 5 \end{aligned}$ is $\begin{aligned} y = - \frac{3}{2} x + 14 \end{aligned}$ or in general form $\begin{aligned} 3 x + 2 y = 28 \end{aligned}$ .
::与 y=23x-5 垂直的直线的方程式是 y32x+14 或一般形式为 3x+2y=28。

by CK-12 demonstrates how to write equations of perpendicular lines.
::通过 CK-12 演示如何写出垂直线的方程式。

Example 4
::例4

Write the equation of the line that passes through (6, -10) and is perpendicular to the line that passes through (4, -6) and (3, -4).
::写出通过(6,10)的线的方程式,该线的方程式与通过(4,6)和(3,4)的线的方程式是垂直的。

Solution: First, we need to find the slope of the line that our line will be perpendicular to. Use the points (4, -6) and (3, -4) to find the slope.
::解答: 首先, 我们需要找到线的斜坡, 我们的线将垂直。用点数( 4, 6) 和点数( 3, 4) 来找到斜坡。

$\begin{aligned} m = \frac{- 4 - (- 6)}{3 - 4} = \frac{2}{- 1} = - 2 \end{aligned}$

::m*4 - (-6)3 - 4=2 - 1 *2

Therefore, the slope of the other line is the negative reciprocal of -2. That makes the slope $\begin{aligned} \frac{1}{2} \end{aligned}$ . Use the point (6, -10) to find the $\begin{aligned} y - \end{aligned}$ intercept.
::因此,另一条线的斜坡是-2.负对等。这就使斜坡12. 使用点(6,10)来找到y-interview。

$\begin{aligned} - 10 & = \frac{1}{2} (6) + b \\ - 10 & = 3 + b \\ - 7 & = b \end{aligned}$

::- 10=12(6)+b-10=3+b-7=b

The equation of the perpendicular line is $\begin{aligned} y = \frac{1}{2} x - 7 \end{aligned}$ .
::垂直直线的方程式是y=12x-7。

Example 5
::例5

Write the equation of the line that passes through (4, -7) and is perpendicular to $\begin{aligned} y = 2 \end{aligned}$ .
::写入通过(4,7)的直线的方程(4,7),该直线与y=2垂直。

Solution: The line $\begin{aligned} y = 2 \end{aligned}$ is a horizontal line and has zero slope . The reciprocal of 0 is undefined , so the line that is perpendicular to this is a vertical line. Therefore, to find the perpendicular line that passes through (4, -7), we only need the x- coordinate. The perpendicular line would be $\begin{aligned} x = 4 \end{aligned}$ .
::分辨率 : y= 2 线是水平线, 零斜度为零。 0 的对等值是未定义的, 因此与此相关的直线是一条垂直线。因此, 要找到通过( 4, - 7) 的直线, 我们只需要 x 坐标。直线是 x= 4 。

Summary
::摘要

Perpendicular lines meet at right angles and have slopes that are negative reciprocals of each other.
::垂直线在右角相交,斜坡是相互负对等的斜坡。

Review
::回顾

Are the pairs of lines parallel, perpendicular, or neither?
::这两条线是平行的,直直的,还是两者兼而有之?

1. $\begin{aligned} 4 x + 2 y = 5 \end{aligned}$ and $\begin{aligned} 5 x - 10 y = - 20 \end{aligned}$
::1. 4x+2y=5和5x-10y=20

2. $\begin{aligned} 9 x + 12 y = 8 \end{aligned}$ and $\begin{aligned} 6 x + 8 y = - 1 \end{aligned}$
::2. 9x+12y=8和6x+8y=1

3. $\begin{aligned} 5 x - 5 y = 20 \end{aligned}$ and $\begin{aligned} x + y = 7 \end{aligned}$
::3. 5x-5y=20和x+y=7

4. $\begin{aligned} 8 x - 4 y = 12 \end{aligned}$ and $\begin{aligned} 4 x - y = - 15 \end{aligned}$
::4. 8x-4y=12和4x-y15

Find the equation of the line, given the following information.
::查找线条的方程, 给出以下信息。

5. Passes through (-6, -2) and is perpendicular to $\begin{aligned} y = 4 \end{aligned}$ .
::5. 通过(-6,-2),与y=4垂直。

6. Passes through (4, 5) and is perpendicular to $\begin{aligned} y = - \frac{1}{3} x - 1 \end{aligned}$ .
::6. 通过(4,5),与y13x-1垂直。

7. Passes through (4, 7) and is perpendicular to $\begin{aligned} x - y = - 5 \end{aligned}$ .
::7. 通过(4、7)和直通X-y5。

8. Passes through (-12, 4) and is perpendicular to $\begin{aligned} y = - 3 x + 5 \end{aligned}$ .
::8. 通过(12、4)和与y3x+5垂直。

9. Passes through (1, -9) and is perpendicular to $\begin{aligned} x = 8 \end{aligned}$ .
::9. 通过(1,-9),直经x=8。

10. Passes through (3, -10) and is perpendicular to the line that passes through (-2, 2) and (-8, 1).
::10. 通过(3,-10),与通过(-2,2)和(-8,1)的线垂直。

Explore More
::探索更多

1. An interior deSourcesigner lays out some kitchen tiles in a design program below.
::1. 以下设计方案中,一个内地的DOFSigner将一些厨房瓷砖放出来。

Find the equations of the two grout lines and determine if the tiles are arranged perpendicular to each other.
::查找两个水泥槽线的方程,确定瓷砖是否是相互垂直排列的。

2. Show that the diagonals of the square below are perpendicular.
::2. 显示以下方形的对角是垂直的。

3. If the equation of a line in general form is $\begin{aligned} A x + B y = C \end{aligned}$ , how would you describe the relationship of the slope of a line perpendicular to this line using A and B?
::3. 如果一般形式的一条线的方程式是Ax+By=C,你如何用A和B描述与该线垂直的一条线的斜坡之间的关系?

4. Are (3,0), (-1,-4), and (1,-6) the vertices of a right triangle?
::4. (3,0),(1,4)和(1,6)是右三角的顶点吗?

5. Find the equation of the perpendicular bisector (the line that intersects a line segment at a right angle and passes through the line segment's midpoint) that goes through the line segment connecting (-2,3) and (2,-3).
::5. 查找垂直两侧区段的方程式(在右角交叉一条线段,穿过线段中点的直线段线段的平方程),横穿连接线段(-2、3和2、3)的直线段。

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

PLIX
::PLIX

Try this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

章节大纲

常规

直直直线

Perpendicular Lines ::直直直线

Perpendicular Lines ::直直直线

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

PLIX ::PLIX

Perpendicular Lines
::直直直线

Perpendicular Lines
::直直直线

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

PLIX
::PLIX