Determining Whether Lines are Parallel or Perpendicular

The two lines in Figure 1 are parallel lines: they will never intersect. Notice that they have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become the same line.

Figure 1 Parallel lines.

Figure 2

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 3 are perpendicular.

Figure 3. Perpendicular lines.

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if [latex]{m}_{1}\text{ and }{m}_{2}[/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[/latex].

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\frac{1}{8}[/latex], and the reciprocal of [latex]\frac{1}{8}[/latex] is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

The product of the slopes is –1.

Example 1: Identifying Parallel and Perpendicular Lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

[latex]\begin{align}&f\left(x\right)=2x+3 &&h\left(x\right)=-2x+2 \\ &g\left(x\right)=\frac{1}{2}x - 4 && j\left(x\right)=2x - 6 \end{align}[/latex]

Show Solution

Parallel lines have the same slope. Because the functions [latex]f\left(x\right)=2x+3[/latex] and [latex]j\left(x\right)=2x - 6[/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and [latex]\frac{1}{2}[/latex] are negative reciprocals, the equations, [latex]g\left(x\right)=\frac{1}{2}x - 4[/latex] and [latex]h\left(x\right)=-2x+2[/latex] represent perpendicular lines.

Analysis of the Solution

A graph of the lines is shown in Figure 21.

Figure 4. The graph shows that the lines [latex]f\left(x\right)=2x+3[/latex] and [latex]j\left(x\right)=2x - 6[/latex] are parallel, and the lines [latex]g\left(x\right)=\frac{1}{2}x - 4[/latex] and [latex]h\left(x\right)=-2x+2[/latex] are perpendicular.

Writing the Equation of a Line Parallel or Perpendicular to a Given Line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Writing Equations of Parallel Lines

Suppose for example, we are given the following equation.

[latex]f\left(x\right)=3x+1[/latex]

We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f(x). So the lines formed by all of the following functions will be parallel to f(x).

[latex]\begin{align}&g\left(x\right)=3x+6 \\ &h\left(x\right)=3x+1 \\ &p\left(x\right)=3x+\frac{2}{3} \end{align}[/latex]

Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

[latex]\begin{gathered}y-{y}_{1}=m\left(x-{x}_{1}\right) \\ y - 7=3\left(x - 1\right) \\ y - 7=3x - 3 \\ y=3x+4 \end{gathered}[/latex]

So [latex]g(x)=3x+4[/latex] is parallel to [latex]f(x)=3x+1[/latex] and passes through the point (1,7).

How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

1. Find the slope of the function.
2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.
3. Simplify.

Example 2: Finding a Line Parallel to a Given Line

Find a line parallel to the graph of [latex]f\left(x\right)=3x+6[/latex] that passes through the point (3,0).

Show Solution

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3, x = 3, and f(x) = 0 into the slope-intercept form to find the y-intercept.

[latex]\begin{gathered}g\left(x\right)=3x+b \\ 0=3\left(3\right)+b \\ b=-9 \end{gathered}[/latex]

The line parallel to f(x) that passes through (3, 0) is [latex]g\left(x\right)=3x - 9[/latex].

Analysis of the Solution

We can confirm that the two lines are parallel by graphing them. Figure 5 shows that the two lines will never intersect.

Figure 5

Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:

[latex]f\left(x\right)=2x+4[/latex]

The slope of the line is 2, and its negative reciprocal is [latex]-\frac{1}{2}[/latex]. Any function with a slope of [latex]-\frac{1}{2}[/latex] will be perpendicular to f(x). So the lines formed by all of the following functions will be perpendicular to f(x).

[latex]\begin{align}&g\left(x\right)=-\frac{1}{2}x+4 \\ &h\left(x\right)=-\frac{1}{2}x+2 \\ &p\left(x\right)=-\frac{1}{2}x-\frac{1}{2} \end{align}[/latex]

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f(x) and passes through the point (4, 0). We already know that the slope is [latex]-\frac{1}{2}[/latex]. Now we can use the point to find the y-intercept by substituting the given values into the slope-intercept form of a line and solving for b.

[latex]\begin{gathered}g\left(x\right)=mx+b \\ 0=-\frac{1}{2}\left(4\right)+b \\ 0=-2+b \\ 2=b \\ b=2 \end{gathered}[/latex]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a y-intercept of 2 is

[latex]g\left(x\right)=-\frac{1}{2}x+2[/latex].

So [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]f\left(x\right)=2x+4[/latex] and passes through the point (4,0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

Q & A

A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?

No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.

How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

1. Find the slope of the function.
2. Determine the negative reciprocal of the slope.
3. Substitute the new slope and the values for x and y from the coordinate pair provided into [latex]g\left(x\right)=mx+b[/latex].
4. Solve for b.
5. Write the equation for the line.

Example 3: Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to [latex]f\left(x\right)=3x+3[/latex] that passes through the point (3,0).

Show Solution

The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\frac{1}{3}[/latex]. Using this slope and the given point, we can find the equation for the line.

[latex]\begin{gathered}g\left(x\right)=-\frac{1}{3}x+b \\ 0=-\frac{1}{3}\left(3\right)+b \\ 1=b \\ b=1 \end{gathered}[/latex]

The line perpendicular to f(x) that passes through (3, 0) is [latex]g\left(x\right)=-\frac{1}{3}x+1[/latex].

Analysis of the Solution

A graph of the two lines is shown in Figure 6.

Figure 6

Try It

Given the function [latex]h\left(x\right)=2x - 4[/latex], write an equation for the line passing through (0, 0) that is

a. parallel to h(x)
b. perpendicular to h(x)

Show Solution

a. [latex]f\left(x\right)=2x[/latex]
b. [latex]g\left(x\right)=-\frac{1}{2}x[/latex]

Try It

How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

1. Determine the slope of the line passing through the points.
2. Find the negative reciprocal of the slope.
3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values.
4. Simplify.

Example 4: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point

A line passes through the points (–2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).

Show Solution

From the two points of the given line, we can calculate the slope of that line.

[latex]\begin{align}{m}_{1}&=\frac{5 - 6}{4-\left(-2\right)} \\ &=\frac{-1}{6} \\ &=-\frac{1}{6} \end{align}[/latex]

Find the negative reciprocal of the slope.

[latex]\begin{align}{m}_{2}&=\frac{-1}{-\frac{1}{6}} \\ &=-1\left(-\frac{6}{1}\right) \\ &=6 \end{align}[/latex]

We can then solve for the y-intercept of the line passing through the point (4, 5).

[latex]\begin{gathered}g\left(x\right)=6x+b \\ 5=6\left(4\right)+b \\ 5=24+b \\ -19=b \\ b=-19 \end{gathered}[/latex]

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is

[latex]y=6x - 19[/latex]

Try It

A line passes through the points, (–2, –15) and (2, –3). Find the equation of a perpendicular line that passes through the point, (6, 4).

Show Solution

[latex]y=-\frac{1}{3}x+6[/latex]

Modeling Real-World Problems with Linear Functions

In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

How To: Given a linear function f and the initial value and rate of change, evaluate f(c).

1. Determine the initial value and the rate of change (slope).
2. Substitute the values into [latex]f\left(x\right)=mx+b[/latex].
3. Evaluate the function at [latex]x=c[/latex].

Example 5: Using a Linear Function to Determine the Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year?

Show Solution

The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200.

Figure 7

The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.

We can write the formula [latex]N\left(t\right)=15t+200[/latex].

With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12.

[latex]\begin{align}N\left(12\right)&=15\left(12\right)+200 \\ &=180+200 \\ &=380 \end{align}[/latex]

Marcus will have 380 songs in 12 months.

Analysis of the Solution

Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Example 6: Using a Linear Function to Calculate Salary Plus Commission

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation.

Show Solution

The given information gives us two input-output pairs: (3, 760) and (5, 920). We start by finding the rate of change.

[latex]\begin{align}m&=\frac{920 - 760}{5 - 3} \\ &=\frac{$160}{\text{2 policies}} \\ &=$80\text{ per policy} \end{align}[/latex]

Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.

We can then solve for the initial value.

[latex]\begin{align}I\left(n\right)=&80n+b \\ 760=&80\left(3\right)+b && \text{when }n=3, I\left(3\right)=760 \\ 760 - &80\left(3\right)=b \\ 520=&b \end{align}[/latex]

The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.

We can now write the final equation.

[latex]I\left(n\right)=80n+520[/latex]

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.

Example 7: Using Tabular Form to Write an Equation for a Linear Function

The table below relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

w, number of weeks	0	2	4	6
P(w), number of rats	1000	1080	1160	1240

Show Solution

We can see from the table that the initial value for the number of rats is 1000, so b = 1000.

Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

[latex]P\left(w\right)=40w+1000[/latex]

If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240)

[latex]\begin{align}m&=\frac{1240 - 1080}{6 - 2} \\ &=\frac{160}{4} \\ &=40 \end{align}[/latex]

Q & A

Is the initial value always provided in a table of values like the table in Example 12?

No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into [latex]f\left(x\right)=mx+b[/latex], and solve for b.

Try It

A new plant food was introduced to a young tree to test its effect on the height of the tree. The table below shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment.

x	0	2	4	8	12
H(x)	12.5	13.5	14.5	16.5	18.5

Show Solution

[latex]H\left(x\right)=0.5x+12.5[/latex]

Identifying Steps to Model and Solve Problems

When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
5. When needed, write a formula for the function.
6. Solve or evaluate the function using the formula.
7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
8. Clearly convey your result using appropriate units, and answer in full sentences when necessary.

Building Linear Models

Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.

• Output: M, money remaining, in dollars
• Input: t, time, in weeks

So, the amount of money remaining depends on the number of weeks: M(t)

We can also identify the initial value and the rate of change.

• Initial Value: She saved $3,500, so $3,500 is the initial value for M.
• Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

Figure 8

The rate of change is constant, so we can start with the linear model [latex]M\left(t\right)=mt+b[/latex]. Then we can substitute the intercept and slope provided.

To find the x-intercept, we set the output to zero, and solve for the input.

[latex]\begin{align}0&=-400t+3500 \\ t&=\frac{3500}{400} \\ &=8.75 \end{align}[/latex]

The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is [latex]0\le t\le 8.75[/latex].

In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

Using a Given Intercept to Build a Model

Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is–$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.

[latex]\begin{align}f\left(x\right)&=mx+b \\ &=-250x+1000 \end{align}[/latex]

Now we can set the function equal to 0, and solve for x to find the x-intercept.

[latex]\begin{gathered}0=-250x+1000 \\ 1000=250x \\ 4=x \\ x=4 \end{gathered}[/latex]

The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan.

Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

How To: Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

1. Identify the input and output values.
2. Convert the data to two coordinate pairs.
3. Find the slope.
4. Write the linear model.
5. Use the model to make a prediction by evaluating the function at a given x value.
6. Use the model to identify an x value that results in a given y value.
7. Answer the question posed.

Example 8: Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.

1. Predict the population in 2013.
2. Identify the year in which the population will reach 15,000.

Show Solution

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: t, years since 2004
• Output: P(t), the town’s population

To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

[latex]m=\frac{\text{change in output}}{\text{change in input}}[/latex]

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to [latex]t=0[/latex], giving the point [latex]\left(0,\text{6200}\right)[/latex]. Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to [latex]t=\text{5}[/latex], giving the point [latex]\left(5,\text{8100}\right)[/latex].

The two coordinate pairs are [latex]\left(0,\text{6200}\right)[/latex] and [latex]\left(5,\text{8100}\right)[/latex]. Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

[latex]\begin{align} m&=\frac{8100 - 6200}{5 - 0} \\ &=\frac{1900}{5} \\ &=380\text{ people per year} \end{align}[/latex]

We already know the y-intercept of the line, so we can immediately write the equation:

[latex]P\left(t\right)=380t+6200[/latex]

To predict the population in 2013, we evaluate our function at t = 9.

[latex]\begin{align}P\left(9\right)&=380\left(9\right)+6,200 \\ &=9,620\hfill \end{align}[/latex]

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set [latex]P\left(t\right)=15000[/latex] and solve for t.

[latex]\begin{gathered}15000=380t+6200 \\ 8800=380t \\ t\approx 23.158 \end{gathered}[/latex]

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

Try It

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.

1. Write a linear model to represent the cost C of the company as a function of x, the number of doughnuts produced.
2. Find and interpret the y-intercept.

Show Solution

[latex]C\left(x\right)=0.25x+25,000[/latex] The y-intercept is (0, 25,000). If the company does not produce a single doughnut, they still incur a cost of $25,000.

Try It

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

1. Predict the population in 2014.
2. Identify the year in which the population will reach 54,000.

Show Solution

41,100; 2020

Try It

Using a Diagram to Model a Problem

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Example 9: Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

Show Solution

In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: “How long will it take them to be 2 miles apart?”

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

• Input: t, time in hours.
• Output: A(t), distance in miles, and E(t), distance in miles.

Because it is not obvious how to define our output variable, we’ll start by drawing a picture.

Figure 9

Initial Value: They both start at the same intersection so when [latex]t=0[/latex], the distance traveled by each person should also be 0. Thus the initial value for each is 0.

Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3.

Using those values, we can write formulas for the distance each person has walked.

[latex]\begin{align}&A\left(t\right)=4t\\ &E\left(t\right)=3t\end{align}[/latex]

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful.

Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances.

Figure 4

This picture shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

[latex]\begin{align}D{\left(t\right)}^{2}&=A{\left(t\right)}^{2}+E{\left(t\right)}^{2} \\ &={\left(4t\right)}^{2}+{\left(3t\right)}^{2} \\ &=16{t}^{2}+9{t}^{2} \\ &=25{t}^{2} \\ D\left(t\right)&=\pm \sqrt{25{t}^{2}} && \text{Solve for }D\left(t\right)\text{ using the square root} \\ &=\pm 5|t| \end{align}[/latex]

In this scenario we are considering only positive values of [latex]t[/latex], so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t.

[latex]\begin{align}D\left(t\right)&=2 \\ 5t&=2 \\ t&=\frac{2}{5}=0.4 \end{align}[/latex]

They will fall out of radio contact in 0.4 hours, or 24 minutes.

Q & A

Should I draw diagrams when given information based on a geometric shape?

Yes. Sketch the figure and label the quantities and unknowns on the sketch.

Example 10: Using a Diagram to Model Distance between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

Show Solution

It might help here to draw a picture of the situation. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0).

Figure 10

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:

[latex]m=\frac{10 - 0}{30 - 0}=\frac{1}{3}[/latex]

The equation of the road from Westborough to Agritown would be

[latex]W\left(x\right)=\frac{1}{3}x[/latex]

From this, we can determine the perpendicular road to Eastborough will have slope [latex]m=-3[/latex]. Because the town of Eastborough is at the point (20, 0), we can find the equation:

[latex]\begin{align}E\left(x\right)&=-3x+b \\ 0&=-3\left(20\right)+b && \text{Substitute in (20, 0)} \\ b&=60 \\ E\left(x\right)&=-3x+60 \end{align}[/latex]

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

[latex]\begin{align}\frac{1}{3}x&=-3x+60 \\ \frac{10}{3}x&=60 \\ 10x&=180 \\ x&=18 && \text{Substituting this back into }W\left(x\right) \\ y&=W\left(18\right) \\ &=\frac{1}{3}\left(18\right) \\ &=6 \end{align}[/latex]

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

[latex]\begin{align}\text{distance}&=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}} \\ &=\sqrt{{\left(18 - 0\right)}^{2}+{\left(6 - 0\right)}^{2}} \\ &\approx 18.974\text{ miles} \end{align}[/latex]

Analysis of the Solution

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

Try It

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

Show Solution

21.15 miles

Building Systems of Linear Models

Figure 11

Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible.

How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

1. Identify the input and output of each linear model.
2. Identify the slope and y-intercept of each linear model.
3. Find the solution by setting the two linear functions equal to another and solving for x, or find the point of intersection on a graph.

Example 11: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile¹. When will Keep on Trucking, Inc. be the better choice for Jamal?

Show Solution

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

Input	d, distance driven in miles
Outputs	K(d): cost, in dollars, for renting from Keep on TruckingM(d) cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: K(0) = 20 and M(0) = 16
Rate of Change	K(d) = $0.59/mile and P(d) = $0.63/mile

A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{gathered}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{gathered}[/latex]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)

Figure 12

These graphs are sketched in Figure 12, with K(d) in blue.

To find the intersection, we set the equations equal and solve:

[latex]\begin{align}K\left(d\right)&=M\left(d\right) \\ 0.59d+20&=0.63d+16 \\ 4&=0.04d \\ 100&=d \\ d&=100 \end{align}[/latex]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].

Try it 5

Example 12: Finding a Break-Even Point

A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140.

1. Find the cost function, C, to produce x helmets, in dollars.
2. Find the revenue function, R, from the sales of x helmets, in dollars.
3. Find the break-even point, the point of intersection of the two graphs C and R.

Show Solution

1. he cost function in the sum of the fixed cost, $125,000, and the variable cost, $120 per helmet.
   [latex]C\left(x\right)=120x+250,000[/latex]
2. The revenue function is the total revenue from the sale of [latex]x[/latex] helmets, [latex]R\left(x\right)=140x[/latex].
3. The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the x-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve for x.

[latex]\begin{align}C\left(x\right)&=R\left(x\right) \\ 250,000+120x&=140x \\ 250,000&=20x \\ 12,500&=x \\ x&=12,500 \end{align}[/latex]

To find [latex]y[/latex], evaluate either the revenue or the cost function at 12,500.

[latex]\begin{align}R\left(20\right)&=140\left(12,500\right) \\ &=$1,750,000 \end{align}[/latex]

The break-even point is (12,500, 1,750,000).

Analysis of the Solution

This means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75 million dollars.

Figure 25

Key Concepts

• Parallel lines have the same slope.
• Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x– and y-values of the given point into the equation, [latex]f(x)=mx+b[/latex], and using the b that results. Similarly, the point-slope form of an equation can also be used.
• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.

Glossary

parallel lines

two or more lines with the same slope

perpendicular lines

two lines that intersect at right angles and have slopes that are negative reciprocals of each other

Section 1.5 Homework Exercises

1. If the graphs of two linear functions are parallel, describe the relationship between the slopes and the y-intercepts.

2. If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts.

3. Explain how to find the input variable in a word problem that uses a linear function.

4. Explain how to find the output variable in a word problem that uses a linear function.

5. Explain how to interpret the initial value in a word problem that uses a linear function.

6. Explain how to determine the slope in a word problem that uses a linear function.

7. Explain how to find a line perpendicular to a linear function that passes through a given point.

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular:

8. [latex]\begin{cases}4x - 7y=10\hfill \\ 7x+4y=1\hfill \end{cases}[/latex]

9. [latex]\begin{cases}3y+x=12\\ -y=8x+1\end{cases}[/latex]

10. [latex]\begin{cases}3y+4x=12\\ -6y=8x+1\end{cases}[/latex]

11. [latex]\begin{cases}6x - 9y=10\\ 3x+2y=1\end{cases}[/latex]

12. [latex]\begin{cases}y=\frac{2}{3}x+1\\ 3x+2y=1\end{cases}[/latex]

13. [latex]\begin{cases}y=\frac{3}{4}x+1\\ -3x+4y=1\end{cases}[/latex]

For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither?

14. Line 1: Passes through [latex]\left(0,6\right)[/latex] and [latex]\left(3,-24\right)[/latex]
Line 2: Passes through [latex]\left(-1,19\right)[/latex] and [latex]\left(8,-71\right)[/latex]

15. Line 1: Passes through [latex]\left(-8,-55\right)[/latex] and [latex]\left(10,89\right)[/latex]
Line 2: Passes through [latex]\left(9,-44\right)[/latex] and [latex]\left(4,-14\right)[/latex]

16. Line 1: Passes through [latex]\left(2,3\right)[/latex] and [latex]\left(4,-1\right)[/latex]
Line 2: Passes through [latex]\left(6,3\right)[/latex] and [latex]\left(8,5\right)[/latex]

17. Line 1: Passes through [latex]\left(1,7\right)[/latex] and [latex]\left(5,5\right)[/latex]
Line 2: Passes through [latex]\left(-1,-3\right)[/latex] and [latex]\left(1,1\right)[/latex]

18. Line 1: Passes through [latex]\left(0,5\right)[/latex] and [latex]\left(3,3\right)[/latex]
Line 2: Passes through [latex]\left(1,-5\right)[/latex] and [latex]\left(3,-2\right)[/latex]

19. Write an equation for a line parallel to [latex]g\left(x\right)=3x - 1[/latex] and passing through the point [latex]\left(4,9\right)[/latex].

20. Write an equation for a line parallel to [latex]f\left(x\right)=-5x - 3[/latex] and passing through the point [latex]\left(2,\text{ -}12\right)[/latex].

21. Write an equation for a line perpendicular to [latex]p\left(t\right)=3t+4[/latex] and passing through the point [latex]\left(3,1\right)[/latex].

22. Write an equation for a line perpendicular to [latex]h\left(t\right)=-2t+4[/latex] and passing through the point [latex]\left(\text{-}4,\text{ -}1\right)[/latex].

23. A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?

24. A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form [latex]p\left(n\right)=mn+b[/latex] that gives the price p they can charge for n shirts.

25. A phone company charges for service according to the formula: [latex]C\left(n\right)=24+0.1n[/latex], where n is the number of minutes talked, and [latex]C\left(n\right)[/latex] is the monthly charge, in dollars. Find and interpret the rate of change and initial value.

26. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form [latex]y=\mathrm{mn}+b[/latex] that gives the yield when n stalks are planted.

27. A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.

28. A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, [latex]P\left(t\right)[/latex], for the population t years after 2003.

29. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: [latex]I\left(x\right)=1054x+23,286[/latex], where x is the number of years after 1990. Which of the following interprets the slope in the context of the problem?

a. As of 1990, average annual income was $23,286.
b. In the ten-year period from 1990–1999, average annual income increased by a total of $1,054.
c. Each year in the decade of the 1990s, average annual income increased by $1,054.
d. Average annual income rose to a level of $23,286 by the end of 1999.

30. When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, the Celsius temperature, [latex]F\left(C\right)[/latex].

a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.
b. Find and interpret [latex]F\left(28\right)[/latex].
c. Find and interpret [latex]F\left(-40\right)[/latex].

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues.

31. Predict the population in 2016.

32. Identify the year in which the population will reach 0.

For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.

33. Predict the population in 2016.

34. Identify the year in which the population will reach 75,000.

For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.

35. Find the linear function that models the town’s population P as a function of the year, t, where t is the number of years since the model began.

36. Find a reasonable domain and range for the function P.

37. If the function P is graphed, find and interpret the x- and y-intercepts.

38. If the function P is graphed, find and interpret the slope of the function.

39. When will the output reached 100,000?

40. What is the output in the year 12 years from the onset of the model?

For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.

41. Find the linear function that models the baby’s weight W as a function of the age of the baby, in months, t.

42. Find a reasonable domain and range for the function W.

43. If the function W is graphed, find and interpret the x– and y-intercepts.

44. If the function W is graphed, find and interpret the slope of the function.

45. When did the baby weight 10.4 pounds?

46. What is the output when the input is 6.2? Interpret your answer.

For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were afflicted.

47. Find the linear function that models the number of people inflicted with the common cold C as a function of the year, t.

48. Find a reasonable domain and range for the function C.

49. If the function C is graphed, find and interpret the x– and y-intercepts.

50. If the function C is graphed, find and interpret the slope of the function.

51. When will the output reach 0?

52. In what year will the number of people be 9,700?

For the following exercises, use the graph below, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980.

53. Find the linear function y, where y depends on t, the number of years since 1980.

54. Find and interpret the y-intercept.

55. Find and interpret the x-intercept.

56. Find and interpret the slope.

For the following exercises, use the graph in below, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980.

57. Find the linear function y, where y depends on t, the number of years since 1980.

58. Find and interpret the y-intercept.

59. Find and interpret the x-intercept.

60. Find and interpret the slope.

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown below. Assume that the house values are changing linearly.

61. In which state have home values increased at a higher rate?

62. If these trends were to continue, what would be the median home value in Mississippi in 2010?

63. If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown below. Assume that the house values are changing linearly.

64. In which state have home values increased at a higher rate?

65. If these trends were to continue, what would be the median home value in Indiana in 2010?

66. If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

67. In 2004, a school population was 1001. By 2008 the population had grown to 1697. Assume the population is changing linearly.

a. How much did the population grow between the year 2004 and 2008?
b. How long did it take the population to grow from 1001 students to 1697 students?
c. What is the average population growth per year?
d. What was the population in the year 2000?
e. Find an equation for the population, P, of the school t years after 2000.
f. Using your equation, predict the population of the school in 2011.

68. In 2003, a town’s population was 1,431. By 2007 the population had grown to 2,134. Assume the population is changing linearly.

a. How much did the population grow between the year 2003 and 2007?
b. How long did it take the population to grow from 1,431 people to 2,134 people?
c. What is the average population growth per year?
d. What was the population in the year 2000?
e. Find an equation for the population, P of the town t years after 2000.
f. Using your equation, predict the population of the town in 2014.

69. A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118.

a. Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used.
b. Interpret the slope and y-intercept of the equation.
c. Use your equation to find the total monthly cost if 687 minutes are used.

70. A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80.

a. Find a linear equation for the monthly cost of the data plan as a function of x, the number of MB used.
b. Interpret the slope and y-intercept of the equation.
c. Use your equation to find the total monthly cost if 250 MB are used.

71. In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly.

a. Find a formula for the moose population, P since 1990.
b. What does your model predict the moose population to be in 2003?

72. In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990.

a. Find a formula for the owl population, P. Let the input be years since 2003.
b. What does your model predict the owl population to be in 2012?

73. The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year.

a. Give a linear equation for the remaining federal helium reserves, R, in terms of t, the number of years since 2010.
b. In 2015, what will the helium reserves be?
c. If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?

74. Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year:

a. Give a linear equation for the remaining oil reserves, R, in terms of t the number of years since now.
b. Seven years from now, what will the oil reserves be?
c. If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?

75. You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

76. You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable?

77. When hired at a new job selling jewelry, you are given two pay options:

Option A: Base salary of $17,000 a year with a commission of 12% of your sales
Option B: Base salary of $20,000 a year with a commission of 5% of your sales

How much jewelry would you need to sell for option A to produce a larger income?

78. When hired at a new job selling electronics, you are given two pay options:

Option A: Base salary of $14,000 a year with a commission of 10% of your sales
Option B: Base salary of $19,000 a year with a commission of 4% of your sales

How much electronics would you need to sell for option A to produce a larger income?

79. When hired at a new job selling electronics, you are given two pay options:

Option A: Base salary of $20,000 a year with a commission of 12% of your sales
Option B: Base salary of $26,000 a year with a commission of 3% of your sales

How much electronics would you need to sell for option A to produce a larger income?

80. When hired at a new job selling electronics, you are given two pay options:

Option A: Base salary of $10,000 a year with a commission of 9% of your sales
Option B: Base salary of $20,000 a year with a commission of 4% of your sales

How much electronics would you need to sell for option A to produce a larger income?