Course: 9.7 找出一条线的曲线

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查找线条的曲线
Jacques is at the top of a hill in the Whistler Ski Resort. As he looks around he sees a flag with the coordinates $\begin{align*}(98, 60)\end{align*}$ blowing in the wind. Jacques wonders if these coordinates have anything to do with the coordinates $\begin{align*}(82, 124) \end{align*}$ he saw on a flag on his way up on the lift. What could these coordinates be telling Jacques about the steepness of the hill?
::雅克(Jacques)站在Whistler Ski Ski 度假村的山顶上。当他环顾四周时,他看到一面有座标(98,60)的国旗在风中吹着。雅克想知道这些座标是否与他上电梯途中在旗上看到的座标(82,124)有关。这些座标能告诉Jacques山的陡峭之处是什么?

In this concept, you will learn to find the slope of a line.
::在这个概念中,你会学会找到一条线的斜坡。

Slopes
::缩筒

The graph of a linear function is a straight line. Although the graphs are all straight lines, the slant or steepness of the lines vary from graph to graph. The slant or steepness of a straight line is referred to as the slope of the line. The slope of a line indicates the rate at which a change in $\begin{align*}x\end{align*}$ (the horizontal coordinates) creates a change in the value of $\begin{align*}y\end{align*}$ (the vertical coordinates).
::线性函数的图形是一条直线。虽然图表都是一条直线,但直线的倾斜度或陡度因图而异。直线的斜度或陡度称为线的斜度。直线的斜度是指线的斜度。直线的斜度表示 x (水平坐标) 变化导致 y (垂直坐标) 值变化的速率。

In mathematics, the slope of a line is represented by the letter $\begin{align*}m\end{align*}$ as defined by:
::在数学方面,一条线的斜坡由字母m表示,字母m的定义是:

$\begin{aligned} m = \frac{Δ y}{Δ x} \to \frac{change in y}{change in x} \to \frac{rise}{run} \end{aligned}$
$\begin{align*}m = \frac{\Delta y}{\Delta x} \rightarrow \frac{\text{change in }y}{\text{change in }x} \rightarrow \frac{\text{rise}}{\text{run}}\end{align*}$
::_____________________________________________________________________________________________________________\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\在 x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

The slope of a graph can be found by counting as shown on the graph below.
::图表的斜坡可按下图所示的数值计算。

$\begin{aligned} m = \frac{rise}{run} \end{aligned}$
$\begin{align*}m=\frac{\text{rise}}{\text{run}}\end{align*}$
::m=中年

First, find two exact points (points on the line that are on the corner of a box on the grid) on the graph.
::首先,在图形中找到两个精确点(线上在网格上一个盒子角上的点)。

Next, begin with a point to the left and count horizontally to the right until you are either directly above or directly below the second point.
::下一步,从左侧的一个点开始,向右水平水平点数,直到您直接在第二点上方或直接在第二点下方。

Then, write the number that you have just moved horizontally in a positive direction as the denominator (run) of the slope.
::然后,写下刚刚水平向正方向移动的数值,作为斜坡的分母(运行)。

Next, count the number vertically to the second point. If you move vertically upward then the number is positive. If you move vertically downward then the number is negative.
::接下来,垂直数到第二点。如果垂直向上移动,则数字为正数。如果垂直向下移动,数字为负数。

Then, write this value as the numerator (rise) of the slope.
::然后将此值作为斜坡的分子( 立方) 写入。

For the following graph:
::下图:

$\begin{aligned} m = \frac{rise}{run} m = \frac{2}{1} \end{aligned}$
$\begin{align*}m = \frac{\text{rise}}{\text{run}} \quad \ m = \frac{2}{1}\end{align*}$
::m=月中 m=21

The slope of a line remains unchanged everywhere on the line. The slope of this line is positive as it moves upward from left to right. If the slope of the line were negative then the line would move downward from left to right.
::线的斜坡在线上的每一处保持不变。该线的斜坡是正的, 因为它从左向右移动。如果线的斜坡是负的, 那么线会从左向右移动。

If you are not given a graph, the slope of a line can be determined by using the coordinates of two points on the line.
::如果不给您提供一个图形,则使用线上两个点的坐标即可确定线的斜度。

Let’s look at another example.
::让我们再看看另一个例子。

The points $\begin{align*}A(15, 8)\end{align*}$ and $\begin{align*}B(10, 7)\end{align*}$ are on a straight line. What is the slope of the line?
::A(15,8)和B(10,7)点在一条直线上。线的坡度是多少?

When two points on the straight line are known, the slope can be calculated using the formula $\begin{align*}m=\frac{y_2-y_1}{x_2-x_1}\end{align*}$ where $\begin{align*}(x_1, y_1)\end{align*}$ are the coordinates of the point chosen to be the first point and $\begin{align*}(x_2, y_2)\end{align*}$ are the coordinates of the point chosen to be the second point. Of the two given points the one chosen as the first and the one chosen as the second will not affect the value of the slope of the line.
::当知道直线上的两点时,可以使用公式m=y2-y1x2-x1计算斜度,其中(x1,y1)是选定为第一个点的坐标,(x2,y2)是选定为第二个点的坐标。在两个给定点中,第一个选定为第一个点,第二个选定为第二个点,不影响线的斜度值。

First, choose and label the points as the first point and the second point.
::首先,选择点并标为第一点和第二点。

$\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ 15, & 8 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ 10, & 7 \end{matrix}) \end{aligned}$
$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 15, & 8 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 10, & 7 \end{pmatrix}\end{align*}$
:1,y115,8)和(x2,y210,7)

Next, write the formula for finding the slope.
::下一个,写出寻找斜坡的公式。

$\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$
$\begin{align*}m = \frac{y_2 - y_1}{x_2 - x_1}\end{align*}$
::m=y2 - y1x2 - x1

Next, fill in the values for the coordinates as they have been named.
::下一步,填写指定坐标的值。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{7 - 8}{10 - 15} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{7-8}{10-15} \end{array}\end{align*}$
::my2-y1x2-x1m=7-810-15

Next, simplify the numerator and the denominator of the formula.
::接下来,简化公式的分子和分母。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{- 1}{- 5} \\ m & = & \frac{1}{5} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{\text{-}1}{\text{-}5}\\ m &=& \frac{1}{5} \end{array}\end{align*}$
::m=-1-5m=15

The slope of the straight line is $\begin{align*}\frac{1}{5}\end{align*}$ .
::直线的斜坡是15。

Let’s look at the slopes of the two special straight lines.
::让我们看看两条特别直线的斜坡。

Find the slope of the line having $\begin{align*}x = \text{-}4\end{align*}$ as its equation .
::查找以 x=4 为其方程的线的斜度。

First, graph the straight line and determine the coordinates of two points that the line passes through.
::首先,绘制直线,并确定两点的坐标。

The vertical line passes through the point $\begin{align*}(-4, 0)\end{align*}$ and is parallel to the $\begin{align*}y\end{align*}$ -axis. The line also passes through the point $\begin{align*}(-4, -1)\end{align*}$ .
::垂直线穿过点(-4,0),与 Y 轴平行。该线也穿过点(-4,-1)。

First, name the points as being the first point and the second point.
::首先,将点命名为第一点和第二点。

$\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ - 4, & 0 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ - 4, & - 1 \end{matrix}) \end{aligned}$
$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ -4, & 0 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ -4, & -1 \end{pmatrix}\end{align*}$
:x1,y1-4,0)和(x2,y2-4,4-1)

Next, write the formula for finding the slope.
::下一个,写出寻找斜坡的公式。

$\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$
$\begin{align*}m = \frac{y_2-y_1}{x_2-x_1}\end{align*}$
::m=y2 - y1x2 - x1

Next, fill in the values for the coordinates as they have been named.
::下一步,填写指定坐标的值。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{- 1 - 0}{- 4 - - 4} \\ m & = & \frac{- 1 - 0}{- 4 + 4} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{\text{-}1-0}{\text{-}4-\text{-}4}\\ m &=& \frac{\text{-}1-0}{\text{-}4+4} \end{array}\end{align*}$
::my2-y1x2-x1m=-1-0-4-4m=-1-0-4-4+4)

Next, simplify the numerator and the denominator of the formula.
::接下来,简化公式的分子和分母。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{- 1}{0} \\ m & = & undefined \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{\text{-}1}{0}\\ m &=& \text{undefined} \end{array}\end{align*}$
::m=-10m=未定义

Remember division by zero is undefined .
::记住零除法未定义。

The slope of the vertical line having the equation $\begin{align*}x=\text{-}4\end{align*}$ is undefined.
::公式 x= 4 的垂直线的斜坡没有定义。

All vertical lines have a slope that is undefined.
::所有垂直线都有未定义的斜坡。

Find the slope of the line having $\begin{align*}y=5\end{align*}$ as its equation.
::查找以 y=5 为等式的线的斜坡。

First, graph the straight line and determine the coordinates of two points that the line passes through.
::首先,绘制直线,并确定两点的坐标。

The horizontal line passes through the point $\begin{align*}(0, 5)\end{align*}$ and is parallel to the $\begin{align*}x\end{align*}$ -axis. The line also passes through the point $\begin{align*}(1, 5)\end{align*}$ .
::水平线穿过点(0,5),与 X 轴平行。该线也穿过点(1,5)。

First, name the points as being the first point and the second point.
::首先,将点命名为第一点和第二点。

$\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ 0, & 5 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ 1, & 5 \end{matrix}) \end{aligned}$
$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 0, & 5 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 1, & 5 \end{pmatrix}\end{align*}$
:x1,y10,5)和(x2,y21,5)

Next, write the formula for finding the slope.
::下一个,写出寻找斜坡的公式。

$\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$
$\begin{align*}m = \frac{y_2-y_1}{x_2-x_1}\end{align*}$
::m=y2 - y1x2 - x1

Next, fill in the values for the coordinates as they have been named.
::下一步,填写指定坐标的值。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{5 - 5}{1 - 0} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{5-5}{1-0} \end{array}\end{align*}$
::my2-y1x2-x1m=5-51-51-0

Next, simplify the numerator and the denominator of the formula.
::接下来,简化公式的分子和分母。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{0}{1} \\ m & = & 0 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{0}{1}\\ m &=& 0 \end{array}\end{align*}$
::m= 01m=0

Remember any number divided by zero equals zero.
::记住任何数字除以零等于零。

The slope of the horizontal line having the equation $\begin{align*}y=5\end{align*}$ is zero.
::Y=5 等式的水平线斜度为零。

All horizontal lines have a slope of zero.
::所有水平线的斜坡为零。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Jacques and the coordinates on the flags. He needs to figure out if the coordinates have anything to do with the steepness of the ski hill. How can Jacques figure this out?
::早些时候,有人给了你Jacques和国旗坐标的问题。他需要弄清楚坐标是否与滑雪山的陡坡有关。Jacques怎么能够弄清楚呢?

He can use the formula to calculate the slope of the ski hill.
::他可以用公式来计算滑雪山的斜坡

First, name the points as being the first point and the second point.
::首先,将点命名为第一点和第二点。

$\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ 98, & 60 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ 82, & 124 \end{matrix}) \end{aligned}$
$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 98, & 60 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 82, & 124 \end{pmatrix}\end{align*}$
:x1,y198,60)和(x2,y282,124)

Next, write the formula for finding the slope.
::下一个,写出寻找斜坡的公式。

$\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$
$\begin{align*}m = \frac{y_2-y_1}{x_2-x_1}\end{align*}$
::m=y2 - y1x2 - x1

Next, fill in the values for the coordinates as they have been named.
::下一步,填写指定坐标的值。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{124 - 60}{82 - 98} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{124-60}{82-98} \end{array}\end{align*}$
::my2-y1x2-x1m=124-6082-98

Next, simplify the numerator and the denominator of the formula.
::接下来,简化公式的分子和分母。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{64}{- 16} \\ m & = & \frac{- 4}{1} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{64}{\text{-}16}\\ m &=& \frac{\text{-}4}{1} \end{array}\end{align*}$
::m=64-16m=-41

The slope of the line is $\begin{align*}\frac{\text{-}4}{1}\end{align*}$ .
::线的斜坡是 -41

Remember, the slope of a line indicates the rate at which a change in $\begin{align*}x\end{align*}$ creates a change in the value of $\begin{align*}y\end{align*}$ . For the ski hill, a horizontal change of positive 1 changes the vertical value 4 units downward. The ski hill is very steep.
::记住, 线的斜坡表示 x 变化导致y 值变化的速率。对于滑雪山, 正值1的水平变化将垂直值 4 单位向下改变。滑雪山非常陡峭。

Example 2
::例2

What is the slope of the straight line shown on the following graph?
::下图显示的直线坡度是多少?

First, select two exact points on the line.
::首先,在线上选择两个精确点。

Next, begin with a point to the left and count horizontally to the right until you are either directly above or directly below the second point.
::下一步,从左侧的一个点开始,向右水平水平点数,直到您直接在第二点上方或直接在第二点下方。

Then, write the number that you have just moved horizontally in a positive direction as the denominator (run) of the slope.
::然后,写下刚刚水平向正方向移动的数值,作为斜坡的分母(运行)。

Next, count the number vertically to the second point. If you move vertically upward then the number is positive. If you move vertically downward then the number is negative.
::接下来,垂直数到第二点。如果垂直向上移动,则数字为正数。如果垂直向下移动,数字为负数。

Then, write this value as the numerator (rise) of the slope.
::然后将此值作为斜坡的分子( 立方) 写入。

For the above graph:
::上图:

$\begin{aligned} m = \frac{rise}{run} m = \frac{2}{3} \end{aligned}$
$\begin{align*}m = \frac{\text{rise}}{\text{run}} \quad m = \frac{{\color{blue}2}}{{\color{red}3}}\end{align*}$
::m=月圆=23

Example 3
::例3

What is the slope of the line passing through the points $\begin{align*}A(-6, 10)\end{align*}$ and $\begin{align*}B(-2, 7)\end{align*}$ ?
::穿过A(6)、10和B(2、7)点的线的坡度是多少?

First, name the points as being the first point and the second point.
::首先,将点命名为第一点和第二点。

$\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ - 6, & 10 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ - 2, & 7 \end{matrix}) \end{aligned}$
$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ -6, & 10 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ -2, & 7 \end{pmatrix}\end{align*}$
:x1,y1-6,10)和(x2,y2-2,7)

Next, write the formula for finding the slope.
::下一个,写出寻找斜坡的公式。

$\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$
$\begin{align*}m = \frac{y_2-y_1}{x_2-x_1}\end{align*}$
::m=y2 - y1x2 - x1

Next, fill in the values for the coordinates as they have been named.
::下一步,填写指定坐标的值。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{7 - 10}{- 2 - - 6} \\ m & = & \frac{7 - 10}{- 2 + 6} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{7-10}{\text{-}2-\text{-}6}\\ m &=& \frac{7-10}{\text{-}2+6} \end{array}\end{align*}$
::my2-y1x2-x1m=7-10-2-6m=7-10-2+6

Next, simplify the numerator and the denominator of the formula.
::接下来,简化公式的分子和分母。

$\begin{aligned} m = \frac{- 3}{4} \end{aligned}$
$\begin{align*}m = \frac{\text{-}3}{4}\end{align*}$
::m=-34

The slope of the line is $\begin{align*}\frac{\text{-}3}{4}\end{align*}$ .
::线的斜坡是 -34

Example 4
::例4

Describe the graph of a line that has $\begin{align*}y=-3\end{align*}$ as its equation and state the slope of the line.
::描述以 y3 为其方程的线条图,并标出线的斜度。

First, draw a graph to model the equation $\begin{align*}y=-3\end{align*}$ .
::首先,绘制一个图形来模拟y3的方程。

The graph is a horizontal line parallel to the $\begin{align*}x\end{align*}$ -axis. The slope of a horizontal line is zero.
::图形是X轴平行的水平线。水平线的斜度为零。

$\begin{aligned} m = 0 \end{aligned}$
$\begin{align*}m=0\end{align*}$
::m=0

Example 5
::例5

Use the formula for slope to calculate the slope of the straight line passing through the points $\begin{align*}(-8, -5)\end{align*}$ and $\begin{align*}(12, -6)\end{align*}$ .
::使用坡度公式计算通过点(-8、5)和点(12、6)的直线的斜度。

First, name the points as being the first point and the second point.
::首先,将点命名为第一点和第二点。

$\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ - 8, & - 5 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ 12, & - 6 \end{matrix}) \end{aligned}$
$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ -8, & -5 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 12, & -6 \end{pmatrix}\end{align*}$
:x1,y1-8,-5)和(x2,y212,-6)

Next, write the formula for finding the slope.
::下一个,写出寻找斜坡的公式。

$\begin{aligned} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$
$\begin{align*}m = \frac{y_2-y_1}{x_2-x_1}\end{align*}$
::m=y2 - y1x2 - x1

Next, fill in the values for the coordinates as they have been named.
::下一步,填写指定坐标的值。

$\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{- 6 - - 5}{12 - - 8} \\ m & = & \frac{- 6 + 5}{12 + 8} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{\text{-}6-\text{-}5}{12-\text{-}8}\\ m &=& \frac{\text{-}6+5}{12+8} \end{array}\end{align*}$
::my2-y1x2-x1m=-6-512-8m=-6+512+8

Next, simplify the numerator and the denominator of the formula.
::接下来,简化公式的分子和分母。

$\begin{aligned} m = \frac{- 1}{20} \end{aligned}$
$\begin{align*}m = \frac{\text{-}1}{20}\end{align*}$
::m=-120

The slope of the line is $\begin{align*}\frac{\text{-}1}{20}\end{align*}$ .
::线的斜坡是 -120

Review
::回顾

Answer true or false for each of the following questions.
::回答下列每个问题是否真实或虚假。

1. True or false. The equation of a line is always linear.
::1. 正确或虚假,线的等式始终是线性的。

2. True or false. A linear equation will be shown as a straight line on a graph.
::2. 真实的或虚假的,线性方程式将作为直线线在图表中显示。

3. True or false. The $\begin{align*}x\end{align*}$ -intercept is where the line crosses the $\begin{align*}y\end{align*}$ axis.
::3. 真实的或虚假的。 X 界面是线穿过 y 轴的位置。

4. True or false. The $\begin{align*}y\end{align*}$ -intercept is where the line crosses the $\begin{align*}y\end{align*}$ axis.
::4. 真实的或虚假的。 Y 界面是线穿过 y 轴的位置。

5. True or false. A vertical line has an undefined slope.
::5. 真实的或虚假的,纵向线有一个未界定的斜坡。

6. True or false. A horizontal line has a slope of 0.
::6. 真实的或虚假的,水平线的斜坡为0。

7. True or false. Slope is the distance that the line travels on the coordinate graph.
::7. 准确或虚假。斜坡是坐标图上行行距离。

8. True or false. Slope is found using a ratio.
::8. 真实的或假的,斜坡是用比例表示的。

9. True or false. You can figure out the slope of a line if you have been given one set of points.
::9. 正确或虚假。如果给您一组点,您可以找出线的斜坡。

10. True or false. You will need two sets of points that a line passes through to figure out the slope.
::10. 正确或虚假:你需要两组分数,一行通过,以找出斜坡。

Figure out the slope of a line that passes through each of the following pairs of points.
::找出横穿以下每一对点的线条的斜坡。

11. $\begin{align*}(2, 3) (3, 4)\end{align*}$

12. $\begin{align*}(4, 5) (2, 3)\end{align*}$

13. $\begin{align*}(2, 1) (\text{-}1, 3)\end{align*}$

14. $\begin{align*}(3, 1) (4, 3)\end{align*}$

15. $\begin{align*}(5, 7) (3, 6)\end{align*}$

16. $\begin{align*}(3, 0) (4, 1)\end{align*}$

17. $\begin{align*}(6, 4) (2, 7)\end{align*}$

18. $\begin{align*}(2, 0) (0, 1)\end{align*}$

19. $\begin{align*}(6, 1) (1, 6)\end{align*}$

20. $\begin{align*}(4, 4) (5, 0)\end{align*}$

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

查找线条的曲线

Slopes ::缩筒

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Slopes
::缩筒

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)