 Last updated
 Save as PDF
 Page ID
 66274
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vectorC}[1]{\textbf{#1}}\)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)
\( \newcommand{\vectE}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
Learning Objectives
By the end of this section, you will be able to:
 Find the slope of a line
 Graph a line given a point and the slope
 Graph a line using its slope and intercept
 Choose the most convenient method to graph a line
 Graph and interpret applications of slopeintercept
 Use slopes to identify parallel and perpendicular lines
Be Prepared
Before you get started, take this readiness quiz.
 Simplify \(\dfrac{1–4}{8−2}\).
 Divide \(\dfrac{0}{4}\) and \(\dfrac{4}{0}\).
 Simplify \(\dfrac{15}{3}\), \(\dfrac{15}{3}\), and \(\dfrac{15}{3}\).
Find the Slope of a Line
When we graph linear equations, we may notice that some lines tilt up and some lines tilt down as they go from left to right. Some lines are very steep and some lines are flatter.
In mathematics, the measure of the steepness of a line is called the slope of the line.
The concept of slope has many applications in the real world. In construction, the pitch of a roof, the slant of the plumbing pipes, and the steepness of the stairs are all applications of slope.
We can assign a numerical value to the inclination of a line by finding the ratio of the rise and run. This is the slope.
Defintion \(\PageIndex{1}\)
Given two points \(P_1\) and \(P_2\) on a line, the rise is the vertical "distance" and the run is the horizontal "distance" traveled and moving from \(P_1\) to \(P_2\), as shown in this illustration.
The "distance" is positive when we are moving up or to the right, and negative when we are moving down or to the left.
The slope of a line is \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\).
To find the slope of a line, we locate any two points on the line, preferably whose coordinates are integers. Then we sketch a right triangle where the two points are vertices and one side is horizontal and one side is vertical.
To find the slope of the line, we measure the "distance" along the vertical and horizontal sides of the triangle, that is, the rise and the run, respectively. The rise and the run can be positive, negative or zero.
Find the slope of a line from its graph using \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\)
 Locate two points on the line whose coordinates are integers.
 Starting with one point, sketch a right triangle, going from the first point to the second point.
 Determine the rise and the run on the legs of the triangle. This can be done by counting jumps vertically and horizontally when moving from the first point to the second.
 Take the ratio of rise to run to find the slope: \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\).
Example \(\PageIndex{2}\)
Find the slope of the line shown.
 Solution

Find the slope of a line from its graph. Locate two points on the graph whose
coordinates are integers.\(P_1=(0,5)\) and \(P_2=(3,3)\) Starting at \((0,5)\), sketch a right triangle to
\((3,3)\) as shown in this graph.Determine the rise. Since it goes down, it is negative. The rise is \(−2\). Determine the run. The run is 3. Write the slope formula. \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\) Substitute the values of the rise and run. \(m=\dfrac{−2}{3}\) Simplify. \(m=−\dfrac{2}{3}\) Answer the question. The slope of the line is \(−\dfrac{2}{3}\). Interpret this as a rate of change. As \(x\) increases by 3 units, \(y\) decreases by 2 units.
Try It \(\PageIndex{3}\)
Find the slope of the line shown.
 Answer

The slope of the line is \(\dfrac{4}{3}\).
Try It \(\PageIndex{4}\)
Find the slope of the line shown.
 Answer

The slope of the line is \(\dfrac{3}{5}\).
How do we find the slope of horizontal and vertical lines? To find the slope of the horizontal line, \(y=4\), we could graph the line, find two points on it, and determine the rise and the run. Let’s see what happens when we do this, as shown in the graph below.
Find the slope of a line from its graph.  

Locate two points on the graph whose coordinates are integers.  \(P_1=(0,4)\) and \(P_2=(3,4)\) 
Determine the rise.  The rise is \(0\). 
Determine the run.  The run is \(3\). 
Write the slope formula.  \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\) 
Substitute the values of the rise and run.  \(m=\dfrac{0}{3}\) 
Simplify.  \(m=0\) 
Answer the question.  The slope of the horizontal line \(y=4\) is \(0\). 
Let’s also consider a vertical line, the line \(x=3\), as shown in the graph.
Find the slope of a line from its graph.  

Locate two points on the graph whose coordinates are integers.  \(P_1=(3,0)\) and \(P_2=(3,2)\) 
Determine the rise.  The rise is \(2\). 
Determine the run.  The run is \(0\). 
Write the slope formula.  \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\) 
Substitute the values of the rise and run.  \(m=\dfrac{2}{0}\) 
Simplify.  \(m\) is undefined 
Answer the question.  The slope of the vertical line \(x=3\) is undefined. 
The slope is undefined since division by zero is undefined. So we say that the slope of the vertical line \(x=3\) is undefined.
All horizontal lines have slope \(0\). When the \(y\)coordinates are the same, the rise is \(0\).
The slope of any vertical line is undefined. When the \(x\)coordinates of a line are all the same, the run is \(0\).
Slope of a horizontal and vertical line
The slope of a horizontal line, \(y=b\), is 0.
The slope of a vertical line, \(x=a\), is undefined.
Example \(\PageIndex{5}\)
Find the slope of each line:
a. \(x=8\)
b. \(y=−5\)
 Solution

a. \(x=8\)
This is a vertical line. Its slope is undefined.
b. \(y=−5\)
This is a horizontal line. It has slope 0.
Try It \(\PageIndex{6}\)
Find the slope of the line \(x=−4\).
 Answer

The slope of the line is undefined.
Try It \(\PageIndex{7}\)
Find the slope of the line \(y=7\).
 Answer

The slope of the line is \(0\).
How does the sign of the slope manifest itself in the line?
 When both the rise and the run are positive, the slope is positive. In this case, the line is "going up" from left to right.
 When both the rise and the run are negative, the slope is also positive. In this case, the line is "going up" from left to right.
 If the rise is positive and the run is negative, the slope is negative. In this case, the line is "going down" from left to right.
 If the rise is negative and the run is positive, the slope is negative. In this case, the line is "going down" from left to right.
 If the rise is zero, the slope is zero. In this case, the line is horizontal.
 If the run is zero, the slope is undefined. In this case, the line is vertical.
Quick guide to the slope of lines
Sometimes we’ll need to find the slope of a line between two points when we don’t have a graph to help determine the rise and the run. We could plot the points on grid paper, then count out the jumps to determine the rise and the run, but as we’ll see, there is a way to find the slope without graphing. Before we get to it, we need to introduce some algebraic notation.
We have seen that an ordered pair \((x,y)\) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol \((x,y)\) be used to represent two different points? Mathematicians use subscripts to distinguish the points.
\((x_1, y_1)\) read “\(x\) sub \(1\), \(y\) sub \(1\).”
\((x_2, y_2)\) read “\(x\) sub \(2\), \(y\) sub \(2\).”
We will use \((x_1,y_1)\) to identify the first point and \((x_2,y_2)\) to identify the second point.
If we had more than two points, we could use \((x_3,y_3)\), \((x_4,y_4)\), and so on.
Let’s see how the rise and run relate to the coordinates of the two points by taking another look at the slope of the line between the points \((2,3)\) and \((7,6)\), as shown in this graph.
Find the slope of the line given two points.  

Since we have two points, we will use subscript notation.  \((x_1,y_1)=(2,3)\) and \((x_2, y_2 )=(7, 6)\) 
Write the slope formula.  \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\) 
Determine the rise and the run by counting jumps on the grid.  On the graph, we counted jumps and found a rise of \(3\) and a run of \(5\). 
Substitute the values of the rise and the run.  \(m=\dfrac{3}{5}\) Notice that the rise of \(3\) can be found by subtracting the \(y\)coordinates, \(6\) and \(3\), and the run of \(5\) can be found by subtracting the \(x\)coordinates \(7\) and \(2\). 
We rewrite the rise and run by putting in the coordinates.  \(m=\dfrac{63}{72}\) But \(6\) is \(y_2\) the \(y\)coordinate of the second point and \(3\) is \(y_1\), the \(y\)coordinate of the first point. 
We can rewrite the slope using subscript notation.  \(m=\dfrac{y_2y_1}{72}\) Also \(7\) is the \(x\)coordinate of the second point and \(2\) is the \(x\)coordinate of the first point. 
So again we rewrite the slope using subscript notation.  \(m=\dfrac{y_2y_1}{x_2x_1}\) 
Conclusion.  We’ve shown that \(m=\dfrac{y_2−y_1}{x_2−x_1}\) is \(m=\dfrac{\text{rise}}{\text{run}}\). We can use this formula to find the slope of a line when we have two points on the line. 
Definition \(\PageIndex{8}\)
The slope of the line between two points \(P_1=(x_1,y_1)\) and \(P_2=(x_2,y_2)\) is
\[m=\dfrac{y_2−y_1}{x_2−x_1}.\nonumber\]
The slope is
\[y\text{ of the second point minus }y\text{ of the first point} \nonumber\] \[\text{over} \nonumber\] \[x\text{ of the second point minus }x\text{ of the first point}. \nonumber\]
Example \(\PageIndex{9}\)
Use the slope formula to find the slope of the line through the points \((−2,−3)\) and \((7,4)\).
 Solution

Find the slope of the line given two points. Write the two points. \(P_1=(−2,−3)\) and \(P_2=(−7,4)\) Identify \(x_1\) and \(y_1\). \(P_1=\underbrace{(2,3)}_{(x_1, y_1)}\)
\(x_1=2\), \(y_1=3\)
Identify \(x_2\) and \(y_2\). \(P_2=\underbrace{(7 , 4)}_{(x_2, y_2)}\)
\(x_2=7\), \(y_2=4\)
Write the slope formula. \(m=\dfrac{y_2y_1}{x_2x_1}\)
Substitute the values. \(m=\dfrac{4(3)}{7(2)}\) Simplify. \(m=\dfrac{7}{5}\) Simplify. \(m=\dfrac{7}{5}\) Answer the question. The slope of the line is \(m=\dfrac{7}{5}\). Let’s verify this slope on the graph shown.
\(\begin{aligned}m&=\dfrac{\mathrm{rise}}{\mathrm{run}}\\ &=\dfrac{7}{−5}\\ &=\dfrac{7}{5}\end{aligned}\)
Try It \(\PageIndex{10}\)
Use the slope formula to find the slope of the line through the points \((−3,4)\) and \((2,−1)\).
 Answer

The slope of the line is \(1\).
Try It \(\PageIndex{11}\)
Use the slope formula to find the slope of the line through the points \((−2,6)\) and \((−3,−4)\).
 Answer

The slope of the line is \(10\).
Graph a Line Given a Point and the Slope
Up to now, we have graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.
We can also graph a line when we know one point and the slope of the line. We will start by plotting the point and then use the definition of slope to draw the graph of the line.
Example \(\PageIndex{12}\)
Graph the line passing through the point \((1,−1)\) whose slope is \(m=\dfrac{3}{4}\).
 Solution

We can check our work by finding a third point. Since the slope is \(m=\dfrac{3}{4}\), it can also be written as \(m=\dfrac{−3}{−4}\) (negative divided by negative is positive!). Go back to \((1,−1)\) and count out the rise, \(−3\), and the run, \(−4\).
Try It \(\PageIndex{13}\)
Graph the line passing through the point \((2,−2)\) with the slope \(m=\dfrac{4}{3}\).
 Answer
Try It \(\PageIndex{14}\)
Graph the line passing through the point \((−2,3)\) with the slope \(m=\dfrac{1}{4}\).
 Answer
Graph a line given a point and the slope
 Plot the given point.
 Use the slope formula \(m=\dfrac{\mathrm{rise}}{\mathrm{run}}\) to identify the rise and the run.
 Starting at the given point, count jumps for the rise and run to mark the second point.
 Draw a line passing through the points.
Graph a Line Using its Slope and Intercept
We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using one point and the slope of the line. Once we see how an equation in slopeintercept form and its graph are related, we will have one more method we can use to graph lines.
Let’s look at the graph of the equation \(y=\dfrac{1}{2}x+3\) and find its slope and \(y\)intercept.
The red lines in the graph show us the rise is 1 and the run is 2. Substituting into the slope formula:
\[\begin{aligned} m &=\dfrac{\text{rise}}{\text{run}} \\ &=\dfrac{1}{2} \end{aligned}\nonumber\]
The \(y\)intercept is \((0,3)\).
Look at the equation of this line.
\(y=\dfrac{1}{2}x+3\)
Look at the slope and \(y\)intercept.
\(\begin{aligned}\text{slope:} & \; m=\dfrac{1}{2} \\ y\text{intercept:} & \; (0,3)\end{aligned}\)
When a linear equation is solved for \(y\), the coefficient of the \(x\) term is the slope \(m\) and the constant term is the \(y\)coordinate of the \(y\)intercept, say \(b\). We say that the equation \(y=\dfrac{1}{2}x+3\) is in slopeintercept form.
\(y=\underbrace{\dfrac{1}{2}}_{m}x+\underbrace{3}_{b}\)
\(y=mx+b\)
Slopeintercept form of an equation of a line
The slopeintercept form of an equation of a line with slope \(m\) and \(y\)intercept, \((0,b)\), is \(y=mx+b\).
Let’s practice finding the values of the slope and \(y\)intercept from the equation of a line.
Example \(\PageIndex{15}\)
Identify the slope and \(y\)intercept of the line from the equation:
a. \(y=−\dfrac{4}{7}x−2\)
b. \(x+3y=9\)
 Solution

a. We compare our equation to the slopeintercept form of the equation.
\(y=−\dfrac{4}{7}x−2\) Write the slopeintercept form of the equation of the line. \(y=mx+b\) Write the equation of the line. Note that it is in slopeintercept form. \(y=\dfrac{4}{7}x2\) Identify the slope. The slope is \(m=\dfrac{4}{7}\). Identify the \(y\)intercept. The \(y\)intercept is \((0,2)\). b. When an equation of a line is not given in slopeintercept form, our first step will be to solve the equation for \(y\).
\(x+3y=9\) Solve for \(y\). \(x+3y=9\) Subtract \(x\) from each side. \(3y=x+9\) Divide both sides by 3. \(\dfrac{3y}{3} = \dfrac{x+9}{3}\) Simplify. \(y=\dfrac{1}{3}x+3\) Write the slopeintercept form of the equation of the line. \(y=mx+b\) Write the equation of the line in slopeintercept form. \(y=\dfrac{1}{3}x+3\) Identify the slope. The slope is \(m=\dfrac{1}{3}\). Identify the \(y\)intercept. The \(y\)intercept is \((0,3)\).
Try It \(\PageIndex{16}\)
Identify the slope and \(y\)intercept from the equation of the line.
a. \(y=\dfrac{2}{5}x−1\)
b. \(x+4y=8\)
 Answer

a. The slope is \(m=\dfrac{2}{5}\), and the \(y\)intercept is \((0,−1)\).
b. The slope is \(m=−\dfrac{1}{4}\), and the \(y\)intercept is \((0,2)\).
Try It \(\PageIndex{17}\)
Identify the slope and \(y\)intercept from the equation of the line.
a. \(y=−\dfrac{4}{3} x+1\)
b. \(3x+2y=12\)
 Answer

a. The slope is \(m=−\dfrac{4}{3}\), and the \(y\)intercept is \((0,1)\).
b. The slope is \(m=−\dfrac{3}{2}\), and the \(y\)intercept is \((0,6)\).
We have graphed a line using the slope and a point. Now that we know how to find the slope and \(y\)intercept of a line from its equation, we can use the \(y\)intercept as the point, and then count jumps determined by the slope from there to find a second point.
Example \(\PageIndex{18}\)
Graph the line of the equation \(y=−x+4\) using its slope and \(y\)intercept.
 Solution

\(y=x+4\) The equation is in slope–intercept form, \(y=mx+b\). \(y=−x+4\) Identify the slope and \(y\)intercept. The slope is \(m=−1\).
The \(y\)intercept is \((0,4)\).Plot the \(y\)intercept. See the graph. Rewrite the slope in the fraction form. \(m=\dfrac{1}{1}\) Identify the rise and the run. rise = \(1\)
run =\(1\)
Count jumps using the rise and run to mark the second point. Draw the line as shown in the graph.
Try It \(\PageIndex{19}\)
Graph the line of the equation \(y=−x−3\) using its slope and \(y\)intercept.
 Answer
Try It \(\PageIndex{20}\)
Graph the line of the equation \(y=−x−1\) using its slope and \(y\)intercept.
 Answer
Now that we have graphed lines by using the slope and \(y\)intercept, let’s summarize all the methods we have used to graph lines.
Choose the Most Convenient Method to Graph a Line
Now that we have seen several methods we can use to graph lines, how do we know which method to use for a given equation?
While we could plot points, use the slopeintercept form, or find the intercepts for any equation, if we recognize the most convenient way to graph a certain type of equation, our work will be easier.
Generally, plotting points is not the most efficient way to graph a line. Let’s look for some patterns to help determine the most convenient method to graph a line.
Here are five equations we graphed in this section, and the method we used to graph each of them.
\[ \begin{array} {lll} {} &{\textbf{Equation}} &{\textbf{Method}} \\ {\text{#1}} &{x=2} &{\text{Vertical line}} \\ {\text{#2}} &{y=−1} &{\text{Horizontal line}} \\ {\text{#3}} &{−x+2y=6} &{\text{Intercepts}} \\ {\text{#4}} &{4x−3y=12} &{\text{Intercepts}} \\ {\text{#5}} &{y=−x+4} &{\text{Slopeintercept}} \\ \end{array} \nonumber\]
Equations #1 and #2 each have just one variable. Remember, in equations of this form the value of that one variable is the same for every solution; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.
In Equations #3 and #4, both \(x\) and \(y\) are on the same side of the equation. These two equations are of the form \(Ax+By=C\). We substituted \(y=0\) to find the \(x\)intercept and \(x=0\) to find the \(y\)intercept, and then found a third point by choosing another value for \(x\) or \(y\).
Equation #5 is written in slopeintercept form. After identifying the slope and \(y\)intercept from the equation we used them to graph the line.
This leads to the following strategy.
Strategy for choosing the most convenient method to graph a line
Consider the form of the equation.
 If it only has one variable, it is a vertical or horizontal line.
 \(x=a\) is a vertical line passing through the \(x\)axis at \(a\).
 \(y=b\) is a horizontal line passing through the \(y\)axis at \(b\).
 If \(y\) is isolated on one side of the equation, in the form \(y=mx+b\), graph by using the slope and \(y\)intercept.
 Identify the slope and \(y\)intercept and then graph.
 If the equation is of the form \(Ax+By=C\), find the intercepts.
 Find the \(x\) and \(y\)intercepts, a third point, and then graph.
Example \(\PageIndex{21}\)
Determine the most convenient method to graph each line:
a. \(y=5\)
b. \(4x−5y=20\)
c. \(x=−3\)
d. \(y=−\dfrac{5}{9}x+8\)
 Solution

a. \(y=5\)
This equation has only one variable, \(y\). Its graph is a horizontal line crossing the \(y\)axis at \(5\).
b. \(4x−5y=20\)
This equation is of the form \(Ax+By=C\). The easiest way to graph it will be to find the intercepts and one more point.c. \(x=−3\)
There is only one variable, \(x\). The graph is a vertical line crossing the \(x\)axis at \(−3\).d. \(y=−\dfrac{5}{9}x+8\)
Since this equation is in \(y=mx+b\) form, it will be easiest to graph this line by using the slope and \(y\)intercepts.
Try It \(\PageIndex{22}\)
Determine the most convenient method to graph each line:
a. \(3x+2y=12\)
b. \(y=4\)
c. \(y=\dfrac{1}{5}x−4\)
d. \(x=−7\)
 Answer

a. intercepts
b. horizontal line
c. slopeintercept
d. vertical line
Try It \(\PageIndex{23}\)
Determine the most convenient method to graph each line:
a. \(x=6\)
b. \(y=−\dfrac{3}{4}x+1\)
c. \(y=−8\)
d. \(4x−3y=−1\)
 Answer

a. vertical line
b. slopeintercept
c. horizontal line
d. intercepts
Graph and Interpret Applications of SlopeIntercept
Many realworld applications are modeled by linear equations. We will take a look at a few applications here so we can see how equations written in slopeintercept form relate to realworld situations.
Usually, when a linear equation models uses realworld data, different letters are used for the variables, instead of using only \(x\) and \(y\). The variable names remind us of what quantities are being measured.
Also, we often will need to extend the axes in our rectangular coordinate system to bigger positive and negative numbers to accommodate the data in the application.
Example \(\PageIndex{24}\)
The equation \(F=\dfrac{9}{5}C+32\) is used to convert temperatures, \(C\), on the Celsius scale to temperatures, \(F\), on the Fahrenheit scale.
a. Find the Fahrenheit temperature for a Celsius temperature of \(0\).
b. Find the Fahrenheit temperature for a Celsius temperature of \(20\).
c. Interpret the slope and \(F\)intercept of the equation.
d. Graph the equation.
 Solution

a.
Find the Fahrenheit temperature for a Celsius temperature of \(0\). Write the conversion equation. \(F=\dfrac{9}{5}C+32\) Find \(F\) when \(C=0\). \(F=\dfrac{9}{5}(0)+32\) Simplify. \(F=32\) Answer the question. The Fahrenheit temperature for a Celsius temperature of \(0\) is \(20\). b.
Find the Fahrenheit temperature for a Celsius temperature of \(20\). Write the conversion equation. \(F=\dfrac{9}{5}C+32\) Find \(F\) when \(C=20\). \(F=\dfrac{9}{5}(20)+32\) Simplify. \(F=36+32\) Simplify. \(F=68\) Answer the question. The Fahrenheit temperature for a Celsius temperature of \(20\) is \(68\). c.
Interpret the slope and \(F\)intercept of the equation.
Even though this equation uses \(F\) and \(C\), it is still in slopeintercept form.\(\begin{aligned} y &=mx+b \\ F &= mC+b \\ F &= \dfrac{9}{5}C+32\end{aligned}\)
The slope, \(\dfrac{9}{5}\), means that the temperature Fahrenheit (\(F\)) increases \(9\) degrees when the temperature Celsius (\(F\)) increases \(5\) degrees.
The \(F\)intercept means that when the temperature is \(0°\) on the Celsius scale, it is \(32°\) on the Fahrenheit scale.
d. Graph the equation. We will need to use a larger scale than our usual. Start at the \(F\)intercept \((0,32)\), and then count out the rise of \(9\) and the run of \(5\) to get a second point as shown in the graph.
Try It \(\PageIndex{25}\)
The equation \(h=2s+50\) is used to estimate a woman’s height in inches, \(h\), based on her shoe size, \(s\).
a. Estimate the height of a child who wears women’s shoe size \(0\).
b. Estimate the height of a woman with shoe size \(8\).
c. Interpret the slope and \(h\)intercept of the equation.
d. Graph the equation.
 Answer

a. \(50\) inches
b. \(66\) inches
c. The slope, \(2\), means that the height, \(h\), increases by \(2\) inches when the shoe size, \(s\), increases by \(1\). The \(h\)intercept means that when the shoe size is \(0\), the height is \(50\) inches.d.
Try It \(\PageIndex{26}\)
The equation \(T=\dfrac{1}{4}n+40\) is used to estimate the temperature in degrees Fahrenheit, \(T\), based on the number of cricket chirps, \(n\), in one minute.
a. Estimate the temperature when there are no chirps.
b. Estimate the temperature when the number of chirps in one minute is \(100\).
c. Interpret the slope and \(T\)intercept of the equation.
d. Graph the equation.
 Answer

a. \(40\) degrees
b. \(65\) degrees
c. The slope, \(\dfrac{1}{4}\), means that the temperature Fahrenheit (\(T\)) increases \(1\) degree when the number of chirps, \(n\), increases by \(4\). The \(T\)intercept means that when the number of chirps is \(0\), the temperature is \(40°\).
d.
The cost of running some types business have two components—a fixed cost and a variable cost. The fixed cost is always the same regardless of how many units are produced. This is the cost of rent, insurance, equipment, advertising, and other items that must be paid regularly. The variable cost depends on the number of units produced. It is for the material and labor needed to produce each item.
Example \(\PageIndex{27}\)
Sam drives a delivery van. The equation \(C=0.5m+60\) models the relation between his weekly cost, \(C\), in dollars and the number of miles, \(n\), that he drives.
a. Find Sam’s cost for a week when he drives \(0\) miles.
b. Find the cost for a week when he drives \(250\) miles.
c. Interpret the slope and \(C\)intercept of the equation.
d. Graph the equation.
 Solution

a.
Find Sam’s cost for a week when he drives \(0\) miles. Write the equation that relates Sam’s cost (\(C\)) per week when he drives \(n\) miles. \(C=0.5n+60\) Find \(C\) when \(n=0\). \(C=0.5(0)+60\) Simplify. \(C=60\) Answer the question. Sam’s costs are \(\$60\) when he drives \(0\) miles. b.
Find Sam’s cost for a week when he drives \(250\) miles. Write the equation that relates Sam’s cost (\(C\)) per week when he drives \(n\) miles. \(C=0.5n+60\) Find \(C\) when \(n=250\). \(C=0.5(250)+60\) Simplify. \(C=185\) Answer the question. Sam’s costs are \(\$185\) when he drives \(250\) miles. c. Interpret the slope and \(C\)intercept of the equation.
\(\begin{aligned} y &=mx+b \\ C &= 0.5n+60\end{aligned}\)
The slope, \(0.5\), means that the weekly cost, \(C\), increases by \(\$0.50\) when the number of miles driven, \(n\), increases by \(1\).
The \(C\)intercept means that when the number of miles driven is \(0\), the weekly cost is \(\$60\).d. Graph the equation. We will need to use a larger scale than our usual. Start at the \(C\)intercept \((0,60)\).
To count out the slope \(m= 0.5\), we rewrite it as an equivalent fraction that will make our graphing easier.
\(m=0.5\) Rewrite as a fraction. \(m=\dfrac{0.5}{1}\) Multiply numerator and denominator by \(100\). \(m=\dfrac{0.5(100)}{1(100)}\) Simplify. \(m=\dfrac{50}{100}\) So to graph the next point go up \(50\) from the intercept of \(60\) and then to the right \(100\). The second point will be \((100, 110)\).
Try It \(\PageIndex{28}\)
Stella has a home business selling gourmet pizzas. The equation \(C=4p+25\) models the relation between her weekly cost, \(C\), in dollars and the number of pizzas, \(p\), that she sells.
a. Find Stella’s cost for a week when she sells no pizzas.
b. Find the cost for a week when she sells \(15\) pizzas.
c. Interpret the slope and \(C\)intercept of the equation.
d. Graph the equation.
 Answer

a. \(\$25\)
b. \(\$85\)
c. The slope, \(4\), means that the weekly cost, \(C\), increases by \(\$4\) when the number of pizzas sold, \(p\), increases by \(1\). The \(C\)intercept means that when the number of pizzas sold is \(0\), the weekly cost is \(\$25\).
d.
Try It \(\PageIndex{29}\)
Loreen has a calligraphy business. The equation \(C=1.8n+35\) models the relation between her weekly cost, \(C\), in dollars and the number of wedding invitations, \(n\), that she writes.
a. Find Loreen’s cost for a week when she writes no invitations.
b. Find the cost for a week when she writes \(75\) invitations.
c. Interpret the slope and \(C\)intercept of the equation.
d. Graph the equation.
 Answer

a. \(\$35\)
b. \(\$170\)
c. The slope, \(1.8\), means that the weekly cost, \(C\), increases by \($1.80\) when the number of invitations, \(n\), increases by \(1\). The \(C\)intercept means that when the number of invitations is \(0\), the weekly cost is \(\$35\).
d.
Use Slopes to Identify Parallel and Perpendicular Lines
Two lines that don't intersect are called parallel. Parallel lines have the same steepness and never intersect.
We say this more formally in terms of the rectangular coordinate system. Two lines that have the same slope and different \(y\)intercepts are called parallel lines.
Verify that both lines have the same slope, \(m=\dfrac{2}{5}\), and different \(y\)intercepts.
What about vertical lines? The slope of a vertical line is undefined but we see that vertical lines that have different \(x\)intercepts are parallel, like the lines shown in this graph.
Definition \(\PageIndex{30}\)
Parallel lines are lines in the same plane that do not intersect.
Parallel lines
 Parallel lines have the same slope and different \(y\)intercepts.
 If \(m_1\) and \(m_2\) are the slopes of two parallel lines then \(m_1=m_2\).
 Parallel vertical lines have different \(x\)intercepts.
Since nonvertical parallel lines have the same slope and different \(y\)intercepts, we can now just look at the slopeintercept form of the equations of lines and decide if the lines are parallel.
Example \(\PageIndex{31}\)
Use slopes and \(y\)intercepts to determine if the lines are parallel:
a. \(3x−2y=6\) and \(y=\dfrac{3}{2}x+1\)
b. \(y=2x−3\) and \(−6x+3y=−9\)
 Solution

a.
Verify that the lines are parallel. \(3x−2y=6\) \(y=\dfrac{3}{2}x+1\) Is the equation in slopeintercept form, \(y=mx+b\)? No Yes Solve the first equation for \(y\) and write it in slopeintercept form. \(\begin{aligned}−2y&=−3x+6 \\ \dfrac{2y}{2}&=\dfrac{3x+6}{2}\\y&=\dfrac{3}{2}x−3 \end{aligned}\)
Now both equations are in slopeintercept form. \(y=\dfrac{3}{2}x−3\) \(y=\dfrac{3}{2}x+1\) Identify the slope and \(y\)intercept of both lines. The slope is \(m=\dfrac{3}{2}\).
The \(y\)intercept is \((0,−3)\).
The slope is \(m=\dfrac{3}{2}\).
The \(y\)intercept is \((0,1)\).
Conclusion. The lines have the same slope and different \(y\)intercepts and so they are parallel. You may want to graph the lines to confirm whether they are parallel.
b.Verify that the lines are parallel. \(y=2x−3\) \(−6x+3y=−9\) Is the equation in slopeintercept form, \(y=mx+b\)? Yes No Solve the second equation for \(y\) and write it in slopeintercept form. \(\begin{aligned}3y&=6x−9\\ \dfrac{3y}{3}&=\dfrac{6x9}{3}\\y&=2x−3 \end{aligned}\)
Now both equations are in slopeintercept form. \(y=2x−3\) \(y=2x−3\) Identify the slope and \(y\)intercept of both lines. The slope is \(m=2\).
The \(y\)intercept is \((0,−3)\).
The slope is \(m=2\).
The \(y\)intercept is \((0,3)\).
Conclusion. The lines have the same slope, but they also have the same \(y\)intercepts. Their equations represent the same line and we say the lines are coincident. They are not parallel; they are the same line.
Try It \(\PageIndex{32}\)
Use slopes and \(y\)intercepts to determine if the lines are parallel:
a. \(2x+5y=5\) and \(y=−\dfrac{2}{5}x−4\)
b. \(y=−\dfrac{1}{2}x−1\) and \(x+2y=−2\)
 Answer

a. The lines are parallel.
b. The lines are not parallel as the equations represent the same line.
Try It \(\PageIndex{33}\)
Use slopes and \(y\)intercepts to determine if the lines are parallel:
a. \(4x−3y=6\) and \(y=\dfrac{4}{3}x−1\)
b. \(y=\dfrac{3}{4}x−3\) and \(3x−4y=12\)
 Answer

a. The lines are parallel.
b. The lines are not parallel as the equations represent the same line.
Example \(\PageIndex{34}\)
Use slopes and \(y\)intercepts to determine if the lines are parallel:
a. \(y=−4\) and \(y=3\)
b. \(x=−2\) and \(x=−5\)
 Solution

a. \(y=−4\) and \(y=3\)
We recognize right away from the equations that these are horizontal lines, and so we know their slopes are both \(0\).
Since the horizontal lines cross the \(y\)axis at \(y=−4\) and at \(y=3\), we know the \(y\)intercepts are \((0,−4)\) and \((0,3)\).
The lines have the same slope and different \(y\)intercepts and so they are parallel.b. \(x=−2\) and \(x=−5\)
We recognize right away from the equations that these are vertical lines, and so we know their slopes are undefined.
Since the vertical lines cross the \(x\)axis at \(x=−2\) and \(x=−5\), we know the \(y\)intercepts are \((−2,0)\) and \((−5,0)\).
The lines are vertical and have different \(x\)intercepts and so they are parallel.
Try It \(\PageIndex{35}\)
Use slopes and \(y\)intercepts to determine if the lines are parallel:
a. \(y=8\) and \(y=−6\)
b. \(x=1\) and \(x=−5\)
 Answer

a. The lines are parallel.
b. The lines are parallel.
Try It \(\PageIndex{36}\)
Use slopes and \(y\)intercepts to determine if the lines are parallel:
a. \(y=1\) and \(y=−5\)
b. \(x=8\) and \(x=−6\)
 Answer

a. The lines are parallel.
b. The lines are parallel.
Let’s look at the lines whose equations are \(y=\dfrac{1}{4}x−1\) and \(y=−4x+2\).
These lines lie in the same plane and intersect in right angles. We call these lines perpendicular.
If we look at the slope of the first line, \(m_1=\dfrac{1}{4}\), and the slope of the second line, \(m_2=−4\), we can see that they are negative reciprocals of each other. If we multiply them, their product is \(−1\).
\[\begin{aligned} &m_1m_2 \\ = & \dfrac{1}{4}(4) \\ = & 1 \end{aligned} \nonumber\]
This is always true for perpendicular lines and leads us to the following definition.
Definition \(\PageIndex{37}\)
Perpendicular lines are lines in the same plane that intersect at a right angle.
Perpendicular lines
 If \(m_1\) and \(m_2\) are the slopes of two perpendicular lines, then their slopes are negative reciprocals of each other, \(m_1=\dfrac{1}{m_2}\). In other words, the product of their slopes is \(1\), that is, \(m_1m_2=1\).
 A vertical line and a horizontal line are always perpendicular to each other.
We were able to look at the slopeintercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.
We find the slopeintercept form of the equation, and then see if the slopes are opposite reciprocals. If the product of the slopes is \(−1\), the lines are perpendicular.
Example \(\PageIndex{38}\)
Use slopes to determine if the lines are perpendicular:
a. \(y=−5x−4\) and \(x−5y=5\)
b. \(7x+2y=3\) and \(2x+7y=5\)
 Solution

a.
Verify that the lines are perpendicular. \(y=−5x−4\) \(x−5y=5\) Is the equation in slopeintercept form, \(y=mx+b\)? Yes No Solve the second equation for \(y\) and write it in slopeintercept form. \(\begin{aligned} 5y &= x+5 \\ y&=\dfrac{1}{5}x1\end{aligned}\)
Now both equations are in slopeintercept form. \(y=−5x−4\) \(y=\dfrac{1}{5}x1\)
Identify the slope of each line. \(m_1=−5\) \(m_2=\dfrac{1}{5}\) Conclusion. The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes. Since
\[m_1m_2=−5\cdot \dfrac{1}{5}=−1,\nonumber\]
it checks.
b.Verify that the lines are perpendicular. \(7x+2y=3\) \(2x+7y=5\) Is the equation in slopeintercept form, \(y=mx+b\)? No No Solve both equations for \(y\) and write it in slopeintercept form. \(\begin{aligned} 2y&=7x+3\\ y &=\dfrac{7}{2}x+\dfrac{3}{2}\end{aligned}\)
\(\begin{aligned} 7y &= 2x+5 \\ y &= \dfrac{2}{7}x+\dfrac{5}{7}\end{aligned}\)
Now both equations are in slopeintercept form. \(y =\dfrac{7}{2}x+\dfrac{3}{2}\) \(y=\dfrac{1}{5}x1\)
Identify the slope of each line. \(m_1=\dfrac{7}{2}\) \(m_2=\dfrac{2}{7}\) Conclusion. The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.
Try It \(\PageIndex{39}\)
Use slopes to determine if the lines are perpendicular:
a. \(y=−3x+2\) and \(x−3y=4\)
b. \(5x+4y=1\) and \(4x+5y=3\)
 Answer

a. The lines are perpendicular.
b. The lines are not perpendicular.
Try It \(\PageIndex{40}\)
Use slopes to determine if the lines are perpendicular:
a. \(y=2x−5\) and \(x+2y=−6\)
b. \(2x−9y=3\) and \(9x−2y=1\)
 Answer

a. The lines are perpendicular.
b. The lines are not perpendicular.
Key Concepts
 Slope of a Line
 The slope of a line is \(m=\dfrac{\text{rise}}{\text{run}}\).
 The rise measures the vertical change and the run measures the horizontal change when moving from one point on the line to another on the line.
 How to find the slope of a line from its graph using \(m=\dfrac{\text{rise}}{\text{run}}\).
 Locate two points on the line whose coordinates are integers.
 Starting with one point, sketch a right triangle, going from the first point to the second point.
 Count vertical and horizontal jumps needed when moving along the legs of the triangle to find the rise and the run.
 Take the ratio of rise to run to find the slope: \(m=\dfrac{\text{rise}}{\text{run}}\).
 Slope of a line between two points.
The slope of the line between two points \((x_1,y_1)\) and \((x_2,y_2)\) is
\[m=\dfrac{y_2−y_1}{x_2−x_1}. \nonumber\]
 How to graph a line given a point and the slope.
 Plot the given point.
 Use the slope formula \(m=\dfrac{\text{rise}}{\text{run}}\) to identify the rise and the run.
 Starting at the given point, count jumps corresponding to the rise and run to mark the second point.
 Draw the line passing through the points.
 Slope Intercept Form of an Equation of a Line
 The slope–intercept form of an equation of a line with slope \(m\) and \(y\)intercept, \((0,b)\) is \(y=mx+b\).
 Parallel Lines
 Parallel lines are lines in the same plane that do not intersect.
 Parallel lines have the same slope and different \(y\)intercepts.
 If \(m_1\) and \(m_2\) are the slopes of two parallel lines then \(m_1=m_2\).
 Parallel vertical lines have different \(x\)intercepts.
 Perpendicular Lines
 Perpendicular lines are lines in the same plane that intersect at a right angle.
 If \(m_1\) and \(m_2\) are the slopes of two perpendicular lines, then their slopes are negative reciprocals of each other, \(m_1=−\dfrac{1}{m_2}\). Equivalently, the product of their slopes is \(−1\), that is, \(m_1m_2=−1\).
 A vertical line and a horizontal line are always perpendicular to each other.
Glossary
 parallel lines
 Parallel lines are lines in the same plane that do not intersect.
 perpendicular lines
 Perpendicular lines are lines in the same plane that form a right angle.
Practice Makes Perfect
Find the Slope of a Line
In the following exercises, find the slope of each line shown.
1.
 Answer

\(m=\dfrac{2}{5}\)
2.
3.
 Answer

\(m=\dfrac{5}{4}\)
4.
5.
 Answer

\(m = \dfrac{1}{3}\)
6.
7.
 Answer

\(m = \dfrac{5}{2}\)
8.
In the following exercises, find the slope of each line.
9. \(y=3\)
 Answer

\(m = 0\)
10. \(y=−2\)
11. \(x=−5\)
 Answer

undefined
12. \(x=4\)
In the following exercises, use the slope formula to find the slope of the line between each pair of points.
13. \((2,5),\;(4,0)\)
 Answer

\(m = \dfrac{5}{2}\)
14. \((3,6),\;(8,0)\)
15. \((−3,3),\;(4,−5)\)
 Answer

\(m = \dfrac{8}{7}\)
16. \((−2,4),\;(3,−1)\)
17. \((−1,−2),\;(2,5)\)
 Answer

\(m = \dfrac{7}{3}\)
18. \((−2,−1),\;(6,5)\)
19. \((4,−5),\;(1,−2)\)
 Answer

\(m = 1\)
20. \((3,−6),\;(2,−2)\)
Graph a Line Given a Point and the Slope
In the following exercises, graph each line with the given point and slope.
21. \((2,5)\); \(m=−\dfrac{1}{3}\)
 Answer
22. \((1,4)\); \(m=−\dfrac{1}{2}\)
23. \((−1,−4)\); \(m=\dfrac{4}{3}\)
 Answer
24. \((−3,−5)\); \(m=\dfrac{3}{2}\)
25. \(y\)intercept: \((0, 3)\); \(m=−\dfrac{2}{5}\)
 Answer
26. \(x\)intercept: \((−2,0)\); \(m=\dfrac{3}{4}\)
27. \((−4,2)\); \(m=4\)
 Answer
28. \((1,5)\); \(m=−3\)
Graph a Line Using Its Slope and Intercept
In the following exercises, identify the slope and yintercept of each line.
29. \(y=−7x+3\)
 Answer

\(m=−7\); \((0,3)\)
30. \(y=4x−10\)
31. \(3x+y=5\)
 Answer

\(m=−3\); \((0,5)\)
32. \(4x+y=8\)
33. \(6x+4y=12\)
 Answer

\(m=−\dfrac{3}{2}\); \((0,3)\)
34. \(8x+3y=12\)
35. \(5x−2y=6\)
 Answer

\(m=\dfrac{5}{2}\); \((0,−3)\)
36. \(7x−3y=9\)
In the following exercises, graph the line of each equation using its slope and yintercept.
37. \(y=3x−1\)
 Answer
38. \(y=2x−3\)
39. \(y=−x+3\)
 Answer
40. \(y=−x−4\)
41. \(y=−\dfrac{2}{5}x−3\)
 Answer
42. \(y=−\dfrac{3}{5}x+2\)
43. \(3x−2y=4\)
 Answer
44. \(3x−4y=8\)
Choose the Most Convenient Method to Graph a Line
In the following exercises, determine the most convenient method to graph each line.
45. \(x=2\)
 Answer

vertical line
46. \(y=5\)
47. \(y=−3x+4\)
 Answer

slopeintercept
48. \(x−y=5\)
49. \(x−y=1\)
 Answer

intercepts
50. \(y=\dfrac{2}{3}x−1\)
51. \(3x−2y=−12\)
 Answer

intercepts
52. \(2x−5y=−10\)
Graph and Interpret Applications of Slope–Intercept
53. The equation \(P=31+1.75w\) models the relation between the amount of Tuyet’s monthly water bill payment, \(P\), in dollars, and the number of units of water, \(w\), used.
a. Find Tuyet’s payment for a month when \(0\) units of water are used.
b. Find Tuyet’s payment for a month when \(12\) units of water are used.
c. Interpret the slope and \(P\)intercept of the equation.
d. Graph the equation.
 Answer

a. \($31\)
b. \($52\)
c. The slope, \(1.75\), means that the payment, \(P\), increases by \($1.75\) when the number of units of water used, \(w\), increases by \(1\). The \(P\)intercept means that when the number units of water Tuyet used is \(0\), the payment is \($31\).
d.
54. The equation \(P=28+2.54w\) models the relation between the amount of Randy’s monthly water bill payment, \(P\), in dollars, and the number of units of water, \(w\), used.
a. Find the payment for a month when Randy used \(0\) units of water.
b. Find the payment for a month when Randy used \(15\) units of water.
c. Interpret the slope and \(P\)intercept of the equation.
d. Graph the equation.
55. Bruce drives his car for his job. The equation \(R=0.575m+42\) models the relation between the amount in dollars, \(R\), that he is reimbursed and the number of miles, \(m\), he drives in one day.
a. Find the amount Bruce is reimbursed on a day when he drives \(0\) miles.
b. Find the amount Bruce is reimbursed on a day when he drives \(220\) miles.
c. Interpret the slope and \(R\)intercept of the equation.
d. Graph the equation.
 Answer

a. \($42\)
b. \($168.50\)
c. The slope, \(0.575\) means that the amount he is reimbursed, \(R\), increases by \($0.575\) when the number of miles driven, \(m\), increases by \(1\). The \(R\)intercept means that when the number miles driven is \(0\), the amount reimbursed is \($42\).
d.
56. Janelle is planning to rent a car while on vacation. The equation \(C=0.32m+15\) models the relation between the cost in dollars, \(C\), per day and the number of miles, \(m\), she drives in one day.
a. Find the cost if Janelle drives the car \(0\) miles one day.
b. Find the cost on a day when Janelle drives the car \(400\) miles.
c. Interpret the slope and \(C\)intercept of the equation.
d. Graph the equation.
57. Cherie works in retail and her weekly salary includes commission for the amount she sells. The equation \(S=400+0.15c\) models the relation between her weekly salary, \(S\), in dollars and the amount of her sales, \(c\), in dollars.
a. Find Cherie’s salary for a week when her sales were \($0\).
b. Find Cherie’s salary for a week when her sales were \($3,600\).
c. Interpret the slope and \(S\)intercept of the equation.
d. Graph the equation.
 Answer

a. \($400\)
b. \($940\)
c. The slope, \(0.15\), means that Cherie’s salary, S, increases by \($0.15\) for every \($1\) increase in her sales. The \(S\)intercept means that when her sales are \($0\), her salary is \($400\).
d.
58. Patel’s weekly salary includes a base pay plus commission on his sales. The equation \(S=750+0.09c\) models the relation between his weekly salary, \(S\), in dollars and the amount of his sales, \(c\), in dollars.
a. Find Patel’s salary for a week when his sales were \(0\).
b. Find Patel’s salary for a week when his sales were \(18,540\).
c. Interpret the slope and \(S\)intercept of the equation.
d. Graph the equation.
59. Costa is planning a lunch banquet. The equation \(C=450+28g\) models the relation between the cost in dollars, \(C\), of the banquet and the number of guests, \(g\).
a. Find the cost if the number of guests is \(40\).
b. Find the cost if the number of guests is \(80\).
c. Interpret the slope and \(C\)intercept of the equation.
d. Graph the equation.
 Answer

a. \($1570\)
b. \($5690\)
c. The slope gives the cost per guest. The slope, \(28\), means that the cost, \(C\), increases by \($28\) when the number of guests increases by \(1\). The \(C\)intercept means that if the number of guests was \(0\), the cost would be \($450\).
d.
60. Margie is planning a dinner banquet. The equation \(C=750+42g\) models the relation between the cost in dollars, \(C\), of the banquet and the number of guests, \(g\).
a. Find the cost if the number of guests is \(50\).
b. Find the cost if the number of guests is \(100\).
c. Interpret the slope and \(C\)intercept of the equation.
d. Graph the equation.
Use Slopes to Identify Parallel and Perpendicular Lines
In the following exercises, use slopes and \(y\)intercepts to determine if the lines are parallel, perpendicular, or neither.
61. \(y=\dfrac{3}{4}x−3\); \(3x−4y=−2\)
 Answer

parallel
62. \(3x−4y=−2\); \(y=\dfrac{3}{4}x−3\)
63. \(2x−4y=6\); \(x−2y=3\)
 Answer

neither
64. \(8x+6y=6\); \(12x+9y=12\)
65. \(x=5\); \(x=−6\)
 Answer

parallel
66. \(x=−3\); \(x=−2\)
67. \(4x−2y=5\); \(3x+6y=8\)
 Answer

perpendicular
68. \(8x−2y=7\); \(3x+12y=9\)
69. \(3x−6y=12\); \(6x−3y=3\)
 Answer

neither
70. \(9x−5y=4\); \(5x+9y=−1\)
71. \(7x−4y=8\); \(4x+7y=14\)
 Answer

perpendicular
72. \(5x−2y=11\); \(5x−y=7\)
73. \(3x−2y=8\); \(2x+3y=6\)
 Answer

perpendicular
74. \(2x+3y=5\); \(3x−2y=7\)
75. \(3x−2y=1\); \(2x−3y=2\)
 Answer

neither
76. \(2x+4y=3\); \(6x+3y=2\)
77. \(y=2\); \(y=6\)
 Answer

parallel
78. \(y=−1\); \(y=2\)
Writing Exercises
79. How does the graph of a line with slope \(m=12\) differ from the graph of a line with slope \(m=2\)?
 Answer

Answers will vary.
80. Why is the slope of a vertical line “undefined”?
81. Explain how you can graph a line given a point and its slope.
 Answer

Answers will vary.
82. Explain in your own words how to decide which method to use to graph a line.
Self Check
a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
b. After reviewing this checklist, what will you do to become confident for all objectives?