7.4 The slope-intercept form of a line

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter the student is shown how graphs provide information that is not always evident from the equation alone. The chapter begins by establishing the relationship between the variables in an equation, the number of coordinate axes necessary to construct its graph, and the spatial dimension of both the coordinate system and the graph. Interpretation of graphs is also emphasized throughout the chapter, beginning with the plotting of points. The slope formula is fully developed, progressing from verbal phrases to mathematical expressions. The expressions are then formed into an equation by explicitly stating that a ratio is a comparison of two quantities of the same type (e.g., distance, weight, or money). This approach benefits students who take future courses that use graphs to display information.The student is shown how to graph lines using the intercept method, the table method, and the slope-intercept method, as well as how to distinguish, by inspection, oblique and horizontal/vertical lines. Objectives of this module: be more familiar with the general form of a line, be able to recognize the slope-intercept form of a line, be able to interpret the slope and intercept of a line, be able to use the slope formula to find the slope of a line.

Overview

• The General Form of a Line
• The Slope-Intercept Form of a Line
• Slope and Intercept
• The Formula for the Slope of a Line

The general form of a line

We have seen that the general form of a linear equation in two variables is $ax+by=c$ (Section [link] ). When this equation is solved for $y$ , the resulting form is called the slope-intercept form. Let's generate this new form.

$\begin{array}{rrrr}\hfill ax+by& \hfill =& c\hfill & \hfill \text{Subtract}\text{\hspace{0.17em}}ax\text{\hspace{0.17em}}\text{from both sides}\text{.}\\ \hfill by& \hfill =& -ax+c\hfill & \hfill \text{Divide}\text{\hspace{0.17em}}both\text{\hspace{0.17em}}\text{sides by}\text{\hspace{0.17em}}b\\ \hfill \frac{by}{b}& \hfill =& \frac{-ax}{b}+\frac{c}{b}\hfill & \hfill \\ \hfill \frac{\overline{)b}y}{\overline{)b}}& \hfill =& \frac{-ax}{b}+\frac{c}{b}\hfill & \hfill \\ \hfill y& \hfill =& \frac{-ax}{b}+\frac{c}{b}\hfill & \hfill \\ \hfill y& \hfill =& \frac{-ax}{b}+\frac{c}{b}\hfill & \hfill \end{array}$

This equation is of the form $y=mx+b$ if we replace $\frac{-a}{b}$ with $m$ and constant $\frac{c}{b}$ with $b$ . ( Note: The fact that we let $b=\frac{c}{b}$ is unfortunate and occurs beacuse of the letters we have chosen to use in the general form. The letter $b$ occurs on both sides of the equal sign and may not represent the same value at all. This problem is one of the historical convention and, fortunately, does not occur very often.)

The following examples illustrate this procedure.

Solve $3x+2y=6$ for $y$ .

$\begin{array}{rrrr}\hfill 3x+2y& \hfill =& 6\hfill & \hfill \text{Subtract 3}x\text{\hspace{0.17em}}\text{from both sides}\text{.}\\ \hfill 2y& \hfill =& -3x+6\hfill & \text{Divide both sides by 2}\text{.}\hfill \\ \hfill y& \hfill =& -\frac{3}{2}x+3\hfill & \hfill \end{array}$

This equation is of the form $y=mx+b$ . In this case, $m=-\frac{3}{2}$ and $b=3$ .

Solve $-15x+5y=20$ for $y$ .

$\begin{array}{rrr}\hfill -15x+5y& \hfill =& 20\hfill \\ \hfill 5y& \hfill =& \hfill 15x+20\\ \hfill y& \hfill =& 3x+4\hfill \end{array}$

This equation is of the form $y=mx+b$ . In this case, $m=3$ and $b=4$ .

Solve $4x-y=0$ for $y$ .

$\begin{array}{rrr}\hfill 4x-y& \hfill =& 0\hfill \\ \hfill -y& \hfill =& -4x\hfill \\ \hfill y& \hfill =& 4x\hfill \end{array}$

This equation is of the form $y=mx+b$ . In this case, $m=4$ and $b=0$ . Notice that we can write $y=4x$ as $y=4x+0$ .

The slope-intercept form of a line $y=mx+b$

A linear equation in two variables written in the form $y=mx+b$ is said to be in slope-intercept form.

Sample set a

The following equations are in slope-intercept form:

$\begin{array}{cc}y=6x-7.& \text{In}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{case}\text{\hspace{0.17em}}m=6\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}b=-7.\end{array}$

$\begin{array}{cc}y=-2x+9.& \text{In}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{case}\text{\hspace{0.17em}}m=-2\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}b=9.\end{array}$

$\begin{array}{cc}y=\frac{1}{5}x+4.8& \text{In}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{case}\text{\hspace{0.17em}}m=\frac{1}{5}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}b=4.8.\end{array}$

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