Course: 9.11 书写线性平方

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书写线条式

Janet bought a flowering crab apple tree for her front lawn. The tree was 24 inches tall when she bought it and was told it would grow 6 inches in height every week. How can Janet figure out the height of the tree at the end of six weeks?
::珍妮特为她的前草坪买了一棵开花的螃蟹苹果树。当她买了这棵树时,树高24英寸,据说每星期会长6英寸高。珍妮特如何在6周结束时找到树的高度?

In this concept, you will learn to write linear equations.
::在这个概念中,你会学会写线性方程。

Linear Equation
::线性对数

A linear equation can be written in many different forms. To write a linear equation in slope-intercept form , the and the $\begin{align*}y\end{align*}$ - intercept of the line must be known. These values can be either given or calculated using the information given. Remember the slope-intercept of a linear equation is written as:
::线性方程式可以以多种不同的形式写成。要以斜度截面形式写出线性方程式, 必须知道线条的线性方程式和Y的截面。这些数值可以使用给定的信息给定或计算。记住线性方程式的斜度截面是:

$\begin{align*}y=mx+b\end{align*}$ such that ‘ $\begin{align*}m\end{align*}$ ’ is the slope of the line, ‘ $\begin{align*}b\end{align*}$ ’ is the $\begin{align*}y\end{align*}$ -intercept and ( $\begin{align*}x, y\end{align*}$ ) are the coordinates of a point on the line.
::y=mx+b 表示 'm' 是线的斜坡,`b' 表示 Y 界面,(x,y)表示线上点的坐标。

Let’s look at an example.
::让我们举个例子。

Write the equation of the line that has a slope of $\begin{align*}\frac{1}{2}\end{align*}$ and passes through the point $\begin{align*}(0, -4)\end{align*}$ .
::写入线的方程式,线的斜度为12,穿过点(0,-4)。

First, determine what information is given.
::首先,确定提供什么信息。

The value of the slope ( $\begin{align*}m\end{align*}$ ) is $\begin{align*}\frac{1}{2}\end{align*}$ and the point $\begin{align*}(0, -4)\end{align*}$ is the value of the $\begin{align*}y\end{align*}$ -intercept ( $\begin{align*}b\end{align*}$ ).
::斜度(m) 值为 12, 点(0) - 4) 值为 y 界面(b) 值。

Next, write the slope-intercept form for the equation of a line.
::下一步,为直线的方程写入斜坡界面表。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, to write the equation in slope intercept form , the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ must be known. Fill these values into the equation.
::其次,要以斜坡截击形式写入方程式,必须知道 ' m ' 和 'b ' 的值。将这些值填入方程式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & \frac{1}{2} x + (- 4) \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& \frac{1}{2}x+(-4) \end{array}\end{align*}$
::y=mx+by=12x+(- 4)

Then, simplify the equation.
::然后,简化方程。

\begin{aligned} \begin{array}{rcl} y & = & \frac{1}{2} x + (- 4) \\ y & = & \frac{1}{2} x - 4 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& \frac{1}{2}x+(-4)\\ y &=& \frac{1}{2}x-4 \end{array}\end{align*}$
::y=12x+(- 4)y=12x- 4

The answer is $\begin{align*}y = \frac{1}{2}x-4\end{align*}$ .
::答案是y=12x-4。

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

Write the equation of the line that has a slope of 3 and passes through the point $\begin{align*}(9, 6)\end{align*}$ .
::写下斜度为3的线的方程,然后通过点(9,6)。

First, determine what information is given.
::首先,确定提供什么信息。

The value of the slope ( $\begin{align*}m\end{align*}$ ) is 3 and the line passes through the point $\begin{align*}(9, 6)\end{align*}$ which are the coordinates of a point ( $\begin{align*}x, y\end{align*}$ ) on the line.
::斜度(m) 值为 3, 线通过点(9,6), 即线上点(x,y)的坐标。

Next, write the slope-intercept form for the equation of a line.
::下一步,为直线的方程写入斜坡界面表。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, to write the equation in slope intercept form, the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ must be known. Fill the value for ‘ $\begin{align*}m\end{align*}$ ’ into the equation.
::其次,要以斜坡拦截形式写入方程式,必须知道 ' m ' 和 'b ' 的值。在方程式中填入`m ' 的值。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & 3 x + b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& 3x+b \end{array}\end{align*}$
::y=mx+by=3x+b y=mx+by=3x+b

Next, since the value of the $\begin{align*}y\end{align*}$ -intercept ( $\begin{align*}b\end{align*}$ ) is not known, use the coordinates $\begin{align*}\begin{pmatrix} x, & y \\ 9, & 6 \end{pmatrix}\end{align*}$ of the point to calculate the $\begin{align*}y\end{align*}$ -intercept.
::其次,由于 Y 界面(b) 的值未知,使用点的坐标(x,y9,6) 来计算 y 界面。

\begin{aligned} \begin{array}{rcl} y & = & 3 x + b \\ 6 & = & 3 (9) + b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& 3x+b\\ 6 &=& 3(9)+b \end{array}\end{align*}$
::y=3x+b6=3(9)+b

Next, perform the multiplication on the right side of the equation to clear the parenthesis.
::下一步,在方程右侧执行乘法以清除括号。

\begin{aligned} \begin{array}{rcl} 6 & = & 3 (9) + b \\ 6 & = & 27 + b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} 6 &=& 3(9)+b\\ 6 &=& 27+b \end{array}\end{align*}$
::6=3(9)+b6=27+b

Next, subtract 27 from both sides of the equation and simplify to solve for ‘ $\begin{align*}b\end{align*}$ ’.
::接下来,从方程的两侧减去27个,并简化 'b' 的解答。

\begin{aligned} \begin{array}{rcl} 6 & = & 27 + b \\ 6 - 27 & = & 27 - 27 + b \\ - 21 & = & b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} 6 &=& 27+b\\ 6-27 &=& 27-27+b\\ -21 &=& b \end{array}\end{align*}$
::6=27+B6-27=27-27+B-21=b

Then, fill in the value for ‘ $\begin{align*}b\end{align*}$ ’ into the slope-intercept form of the equation and simplify.
::然后,将 " b " 的值填入方程式的斜度拦截形式并简化。

\begin{aligned} \begin{array}{rcl} y & = & 3 x + b \\ y & = & 3 x + (- 21) \\ y & = & 3 x - 21 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& 3x+b\\ y &=& 3x+(-21)\\ y &=& 3x-21 \end{array}\end{align*}$
::y=3x+by=3x+(- 21y=3x- 21)

The answer is $\begin{align*}y = 3x-21\end{align*}$ .
::答案是y=3x-21。

If neither the slope nor the $\begin{align*}y\end{align*}$ -intercept is given, then both must be calculated using the coordinates of two points on the line.
::如果未给出斜坡或 Y 界面,则两者都必须使用线上两个点的坐标计算。

Let’s look at an example.
::让我们举个例子。

Write the equation, in slope-intercept form, of the line that passes through the points $\begin{align*}(-2, 6)\end{align*}$ and $\begin{align*}(4, -6)\end{align*}$ .
::以斜坡间距形式写出通过点(-2,6)和点(4,6)的线的方程。

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} \\ - 2, & 6 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ 4, & - 6 \end{matrix}) \end{aligned}

$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ -2, & 6 \end{pmatrix} \ \text{and}\ \begin{pmatrix} x_2, & y_2 \\ 4, & -6 \end{pmatrix}\end{align*}$
:

x1,y1-2,6)和(x2,y24,-6)

Next, use the coordinates of these points to fill in the formula for calculating the slope of the line.
::其次,使用这些点的坐标来填写计算线坡度的公式。

\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{- 6 - 6}{4 - - 2} \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{-6-6}{4--2} \end{array}\end{align*}$
::my2 - y1x2 - x1m6 - 64%2

Next, simplify the right side of the equation and express the answer in simplest form.
::接下来,简化方程的右侧,以最简单的形式表达答案。

\begin{aligned} \begin{array}{rcl} m & = & \frac{- 6 - 6}{4 - - 2} \\ m & = & \frac{- 12}{6} \\ m & = & - 2 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} m &=& \frac{-6-6}{4--2}\\ m &=& \frac{-12}{6}\\ m &=& -2 \end{array}\end{align*}$
::m6 - 642m126m2

The answer is -2.
::答案是 -2

The slope of the line is -2.
::线的斜坡是 -2。

Next, use the slope of the line and the coordinates of one point to calculate the $\begin{align*}y\end{align*}$ -intercept of the line.
::下一步,使用线的斜坡和一个点的坐标来计算线的 Y 插口。

\begin{aligned} m = - 2 and (\begin{matrix} x_{1}, & y_{1} \\ - 2, & 6 \end{matrix}) \end{aligned}

$\begin{align*}m=-2 \ \text{and} \begin{pmatrix} x_1, & y_1 \\ -2, & 6 \end{pmatrix}\end{align*}$
::2 和 (x1,y1-2,6)

Next, substitute the values into the slope-intercept form of an equation.
::其次,将数值替换为方程式的斜坡间接形式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ 6 & = & - 2 (- 2) + b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ 6 &=& -2(-2)+b \end{array}\end{align*}$
::y=mx+b6=2(-2)+b

Next, perform the multiplication to clear the parenthesis.
::下一步,执行乘法清除括号。

\begin{aligned} \begin{array}{rcl} 6 & = & - 2 (- 2) + b \\ 6 & = & 4 + b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} 6 &=& -2(-2)+b\\ 6 &=& 4+b \end{array}\end{align*}$
::6(-2)+b6=4+b

Next, subtract 4 from both sides of the equation to solve for ‘ $\begin{align*}b\end{align*}$ ’.
::接下来,从方程的两侧减去4, 以解决 'b' 。

\begin{aligned} \begin{array}{rcl} 6 & = & 4 + b \\ 6 - 4 & = & 4 - 4 + b \\ 2 & = & b \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} 6 &=& 4+b\\ 6-4 &=& 4-4+b\\ 2 &=& b \end{array}\end{align*}$
::6=4+B6-4=4-4-4-4+b2=b

Then, substitute the values for ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ into the slope-intercept form of the equation of a line.
::然后,将 " m " 和 " b " 的值换成线形方程式的斜度间距形式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & - 2 x + 2 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& -2x+2 \end{array}\end{align*}$
::y=mx+by=%2x+2

The answer is $\begin{align*}y=-2x+2\end{align*}$ .
::答案是y2x+2。

The equation of a line can also be written using the values in a given t-table. Remember the values in the table are ordered pairs ( $\begin{align*}x, y\end{align*}$ ). If none of the values are given in the form $\begin{align*}\begin{pmatrix} x, & y \\ 0, & \# \end{pmatrix}\end{align*}$ then both the slope and the $\begin{align*}y\end{align*}$ -intercept would have to be calculate as shown in the previous example.
::线条的方程也可以使用给定的table 中的值来写入。记住表格中的值是按顺序排列的对( x, y) 。如果窗体( x, y0, #) 没有给出这些值, 那么则必须按上一个示例所示计算斜度和 y- intercept 。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Janet and the flowering crab apple tree. She needs to figure out the height of the tree at the end of six weeks, given that it grows 6 inches a week. How can Janet figure this out?
::早些时候,有人给了你关于珍妮特和开花的螃蟹苹果树的问题。她需要6周后找出树的高度,因为树每星期生长6英寸。珍妮特怎么能知道呢?

She can use an equation of a line written in slope intercept form.
::她可以使用斜坡截击形式书写线的方程式。

First, write down the given information.
::首先,写下给定的信息。

The height of the tree increases 6 inches in one week. This means $\begin{align*}m=\frac{\text{change in }y}{\text{change in }x}=\frac{6}{1}=6\end{align*}$ . The original height of the tree, was 24 inches. This means $\begin{align*}b=(0, 24)=24\end{align*}$ .
::树的高度在一周内增加 6 英寸。这意味着 ychange 的 m= change in x= 61=6. 树的原始高度为 24 英寸。这意味着 b= (0) 24= 24 。

Next, write the equation of a line in slope-intercept form.
::下一步,以斜坡界面的形式写出线的方程式。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, substitute the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ into the equation.
::其次,将`m ' 和`b ' 的值换成等式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & 6 x + 24 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& 6x+24 \end{array}\end{align*}$
::y=mx+by=6x+24 y=mx+by=6x+24

Next, substitute ‘6’ into the equation for ‘ $\begin{align*}x\end{align*}$ ’ to calculate the height of the tree, ‘ $\begin{align*}y\end{align*}$ ’ in six weeks.
::接下来,将“6”改为“x”,在六周内计算树的高度,“y”。

\begin{aligned} \begin{array}{rcl} y & = & 6 x + 24 \\ y & = & 6 (6) + 24 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& 6x+24\\ y &=& 6(6)+24 \end{array}\end{align*}$
::y=6x+24y=6(6)+24

Then, simplify the equation.
::然后,简化方程。

\begin{aligned} \begin{array}{rcl} y & = & 36 + 24 \\ y & = & 60 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& 36+24\\ y &=& 60 \end{array}\end{align*}$
::y=36+24y=60

The answer is 60.
::答案是60岁

The height of the tree at the end of 6 weeks will be 60 inches or 5 feet.
::6周结束时树的高度为60英寸或5英尺。

Example 2
::例2

Write the equation of the line that has a slope of 4 and crosses the $\begin{align*}y\end{align*}$ -axis at the point $\begin{align*}(0, -3)\end{align*}$ .
::写出斜度为 4 的线的方程, 并在点( 0, - 3) 上横跨 Y 轴。

First, write down the given information.
::首先,写下给定的信息。

$\begin{align*}m=4\end{align*}$ and the given point is the $\begin{align*}y\end{align*}$ -intercept of the line. Therefore , $\begin{align*}b = -3\end{align*}$ .
::m=4, 给定点是线条的 Y 界面。因此, b3 。

Next, write the equation of a line in slope-intercept form.
::下一步,以斜坡界面的形式写出线的方程式。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, substitute the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ into the equation.
::其次,将`m ' 和`b ' 的值换成等式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & 4 x + (- 3) \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& 4x+(-3) \end{array}\end{align*}$
::y=mx+by=4x+(-3)

Then, simplify the equation.
::然后,简化方程。

\begin{aligned} y = 4 x - 3 \end{aligned}

$\begin{align*}y=4x-3\end{align*}$
::y=4x-3 y=4x-3

The answer is $\begin{align*}y=4x-3\end{align*}$ .
::答案是y=4x-3。

Example 3
::例3

Write an equation in slope-intercept form to model the values given in the following t-table.
::以斜坡界面形式写入一个方程式,以模拟下表给出的值。

$\begin{align}x\end{align}$	$\begin{align}y\end{align}$
0	6
1	1
2	-4
3	-9
4	-14

The value of the $\begin{align*}y\end{align*}$ -intercept ( $\begin{align*}b\end{align*}$ ) is 6 and the line passes through the points $\begin{align*}(1, 1), (2, -4), (3, -9)\end{align*}$ and $\begin{align*}(4, -14)\end{align*}$ which are the coordinates of points ( $\begin{align*}x, y\end{align*}$ ) on the line. Choose one of the points to represent ( $\begin{align*}x, y\end{align*}$ ).
::y- interfits (b) 的值为 6, 横线通过点(1, 1, 2, (2) - 4, (3, 9) 和 4, -14) 的坐标。选择要代表的点之一 (x, y) 。

Next, write the slope-intercept form for the equation of a line.
::下一步,为直线的方程写入斜坡界面表。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, to write the equation of a line in slope intercept form, the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ must be known. Fill the value for ‘ $\begin{align*}b\end{align*}$ ’ into the equation.
::其次,要以斜坡截击形式写入线的方程式,必须知道 'm ' 和 'b ' 的值。将`b ' 的值填入方程式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & m x + 6 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& mx+6 \end{array}\end{align*}$
::y=mx+by=mx+6 y=mx+6 y=mx+by=mx+6

Next, since the value of the $\begin{align*}y\end{align*}$ -intercept ( $\begin{align*}b\end{align*}$ ) is known, use the coordinates $\begin{align*}\begin{pmatrix} x, & y \\ 1, & 1 \end{pmatrix}\end{align*}$ of the point to calculate the slope.
::下一步,由于 Y 界面(b) 的值已知,使用点的坐标(x,y1,1) 来计算斜度。

\begin{aligned} \begin{array}{rcl} y & = & m x + 6 \\ 1 & = & m (1) + 6 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+6\\ 1 &=& m(1)+ 6 \end{array}\end{align*}$
::y=mx+61=m(1)+6

Next, perform the multiplication on the right side of the equation to clear the parenthesis.
::下一步,在方程右侧执行乘法以清除括号。

\begin{aligned} \begin{array}{rcl} 1 & = & m (1) + 6 \\ 1 & = & m + 6 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} 1 &=& m(1)+6\\ 1 &=& m+6 \end{array}\end{align*}$
::1=m(1)+61=m+6

Next, subtract 6 from both sides of the equation and simplify to solve for ‘ $\begin{align*}m\end{align*}$ ’.
::接下来,从方程两侧减去6, 并简化解答 'm ' 。

\begin{aligned} \begin{array}{rcl} 1 & = & m + 6 \\ 1 - 6 & = & m + 6 - 6 \\ - 5 & = & m \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} 1 &=& m+6\\ 1-6 &=& m+6-6\\ -5 &=& m \end{array}\end{align*}$
::1=m+61-6=m+6-6-6-6-6-5=m

Then, fill in the value for ‘ $\begin{align*}m\end{align*}$ ’ into the slope-intercept form of the equation and simplify.
::然后,将`m ' 的值填入方程式的斜度拦截形式并简化。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & m x + 6 \\ y & = & - 5 x + 6 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& mx+6\\ y &=& -5x+6 \end{array}\end{align*}$
::y=mx+by=mx+6y @%5x+6

The answer is $\begin{align*}y = -5x+6\end{align*}$ .
::答案是y5x+6。

Example 4
::例4

Write the equation of the line in slope-intercept form that passes through the points $\begin{align*}(7, 0)\end{align*}$ and $\begin{align*}(0, 4)\end{align*}$ .
::以斜坡截取形式写出横穿点(7,0)和点(0,4)的线的方程式。

First, write down the given information.
::首先,写下给定的信息。

The given points $\begin{align*}(7, 0)\end{align*}$ and $\begin{align*}(0, 4)\end{align*}$ are the $\begin{align*}x\end{align*}$ - and $\begin{align*}y\end{align*}$ - intercepts respectively of the line.
::给定点(7,0)和(0,4)分别为线条的 x 和 y 界面。

Next, write the slope-intercept form for the equation of a line.
::下一步,为直线的方程写入斜坡界面表。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, to write the equation in slope intercept form, the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ must be known. Fill the value for ‘ $\begin{align*}b\end{align*}$ ’ into the equation.
::其次,要以斜坡拦截形式写入方程式,必须知道 ' m ' 和 'b ' 的值。将`b ' 的值填入方程式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & m x + 4 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& mx+4 \end{array}\end{align*}$
::y=mx+by=mx+4 y=mx+4 y=mx+by=mx+4

Next, name the points as being the first point and the second point,
::下个点命名为第一点和第二点,

\begin{aligned} (\begin{matrix} x_{1}, & y_{1} \\ 7, & 0 \end{matrix}) and (\begin{matrix} x_{2}, & y_{2} \\ 0, & 4 \end{matrix}) \end{aligned}

$\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 7, & 0 \end{pmatrix}\ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 0, & 4 \end{pmatrix}\end{align*}$
:

x1,y17,0)和(x2,y20,4)

Next, use the coordinates of these points to fill in the formula for calculating the slope of the line.
::其次,使用这些点的坐标来填写计算线坡度的公式。

\begin{aligned} \begin{array}{rcl} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{4 - 0}{0 - 7} \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} m &=& \frac{y_2-y_1}{x_2-x_1}\\ m &=& \frac{4-0}{0-7} \end{array}\end{align*}$
::my2-y1x2-x1m=4-00-7

Next, simplify the right side of the equation and express the answer in simplest form.
::接下来,简化方程的右侧,以最简单的形式表达答案。

\begin{aligned} \begin{array}{rcl} m & = & \frac{4 - 0}{0 - 7} \\ m & = & \frac{4}{- 7} \\ m & = & - \frac{4}{7} \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} m &=& \frac{4-0}{0-7}\\ m &=& \frac{4}{-7}\\ m &=& - \frac{4}{7}\\ \end{array}\end{align*}$
::m=4-00-7m=4-7m47

Then, fill in the value for ‘ $\begin{align*}m\end{align*}$ ’ into the slope-intercept form of the equation and simplify.
::然后,将`m ' 的值填入方程式的斜度拦截形式并简化。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & m x + 4 \\ y & = & - \frac{4}{7} x + 4 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& mx+4\\ y &=& -\frac{4}{7}x+4 \end{array}\end{align*}$
::y=mx+by=mx+4y47x+4

The answer is $\begin{align*}y = -\frac{4}{7}x+4\end{align*}$ .
::答案是y47x+4。

Example 5
::例5

Write the equation of a line that has a slope of zero and a $\begin{align*}y\end{align*}$ -intercept of -5.
::写出一条线的方程式,线面的斜度为零,Y- interviews of - 5。

First, write down the given information.
::首先,写下给定的信息。

$\begin{align*}m=0\end{align*}$ and the $\begin{align*}y\end{align*}$ -intercept of the line is $\begin{align*}b=-5\end{align*}$ .
::m=0 线的 Y 界面是 b5 。

Next, write the equation of a line in slope-intercept form.
::下一步,以斜坡界面的形式写出线的方程式。

\begin{aligned} y = m x + b \end{aligned}

$\begin{align*}y=mx+b\end{align*}$
::y=mx+b y=mx+b

Next, substitute the values of ‘ $\begin{align*}m\end{align*}$ ’ and ‘ $\begin{align*}b\end{align*}$ ’ into the equation.
::其次,将`m ' 和`b ' 的值换成等式。

\begin{aligned} \begin{array}{rcl} y & = & m x + b \\ y & = & 0 x + (- 5) \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& mx+b\\ y &=& 0x+(-5) \end{array}\end{align*}$
::y=mx+by=0x+(-5)

Then, simplify the equation.
::然后,简化方程。

\begin{aligned} \begin{array}{rcl} y & = & 0 - 5 \\ y & = & - 5 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} y &=& 0-5\\ y &=& -5 \end{array}\end{align*}$
::y=0 - 5y=5

The answer is $\begin{align*}y = -5\end{align*}$ .
::答案是你们5个

Review
::回顾

Write the equation of a line with the following slopes and $\begin{align*}y\end{align*}$ -intercepts.
::以下列斜坡和 Y 截面写入线条的方程式。

1. $\begin{align*}\text{slope} = 2, y \text{ int} = 4\end{align*}$
::1. 斜度=2,y intt=4

2. $\begin{align*}\text{slope} = -3, y \text{ int} = 2\end{align*}$
::2. 斜度3,3,y int=2

3. $\begin{align*}\text{slope} = -4, y \text{ int} = 4\end{align*}$
::3. 斜度4,4,y int=4

4. $\begin{align*}\text{slope} = 3, y \text{ int} = -5\end{align*}$
::4. 斜坡=3,y int5

5. $\begin{align*}\text{slope} = \frac{1}{2} , y \text{ int} = -2\end{align*}$
::5. 斜坡=12,y int2

6. $\begin{align*}\text{slope} = \frac{-1}{3} , y \text{ int} = 2\end{align*}$
::6. 斜度=13,y int=2

7. $\begin{align*}\text{slope} = 1, y \text{ int} = 8\end{align*}$
::7. 斜坡=1,y intt=8

8. $\begin{align*}\text{slope} = -2, y \text{ int} = 4\end{align*}$
::8. 斜度=2,y intt=4

9. $\begin{align*}\text{slope} = -1, y \text{ int} = -1\end{align*}$
::9. 斜坡1,1,y英寸1

10. $\begin{align*}\text{slope} = 5, y \text{ int} = -2\end{align*}$
::10. 斜坡=5,y int2

Write the following horizontal or vertical line equations.
::写入以下水平或垂直线方程式。

11. A horizontal line with a $\begin{align*}b\end{align*}$ value of 7.
::11. 水平线,B值为7。

12. A horizontal line with a $\begin{align*}b\end{align*}$ value of -4.
::12. b 值为-4的横向线条。

13. A vertical line with an $\begin{align*}x\end{align*}$ value of 2.
::13. 垂直线,x值为2。

14. A vertical line with an $\begin{align*}x\end{align*}$ value of -5.
::14. 垂直线,x值为-5。

Write the equation of a line that passes through the following points and graph them.
::写出横穿以下各点的线条的方程并绘制图表。

15. $\begin{align*}(3, -3)\end{align*}$ and $\begin{align*}(-3, 1)\end{align*}$
::15. (3,-3)和(-3,3,1)

16. $\begin{align*}(2, 3)\end{align*}$ and $\begin{align*}(0, -3)\end{align*}$
::16. (2,3)和(0,3)

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

书写线条式

Linear Equation ::线性对数

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)

Linear Equation
::线性对数

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)