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## SectionComparing Linear Functions

###### Supplemental Videos

The main topics of this section are also presented in the following videos:

### SubsectionUsing Ratios for Comparison

Which is more expensive, a 64-ounce bottle of Velvolux dish soap that costs $3.52, or a 60-ounce bottle of Rainfresh dish soap that costs$3.36?

You are probably familiar with the notion of comparison shopping. To decide which dish soap is the better buy, we compute the unit price, or price per ounce, for each bottle. The unit price for Velvolux is

\begin{equation*} \frac{352 \text{ cents}}{64 \text{ ounces}}= 5.5 \text{ cents per ounce} \end{equation*}

and the unit price for Rainfresh is

\begin{equation*} \frac{336 \text{ cents}}{60 \text{ ounces}}= 5.6 \text{ cents per ounce} \end{equation*}

The Velvolux costs less per ounce, so it is the better buy. By computing the price of each brand for the same amount of soap, it is easy to compare them.

In many situations, a ratio, similar to a unit price, can provide a basis for comparison. Example142 uses a ratio to measure a rate of growth.

###### Example142

Which grow faster, Hybrid A wheat seedlings, which grow 11.2 centimeters in 14 days, or Hybrid B seedlings, which grow 13.5 centimeters in 18 days?

Solution

We compute the growth rate for each strain of wheat. Growth rate is expressed as a ratio, $\dfrac{\text{centimeters}}{\text{days}}\text{,}$ or centimeters per day. The growth rate for Hybrid A is

\begin{equation*} \frac{11.2 \text{ centimeters}}{14 \text{ days}}= 0.8 \text{ centimeters per day} \end{equation*}

and the growth rate for Hybrid B is

\begin{equation*} \frac{13.5 \text{ centimeters}}{18 \text{ days}}= 0.75 \text{ centimeters per day} \end{equation*}

Because their rate of growth is larger, the Hybrid A seedlings grow faster.

By computing the growth of each strain of wheat seedling over the same unit of time, a single day, we have a basis for comparison. In this case, the ratio $\dfrac{\text{centimeters}}{\text{day}}$ measures the rate of growth of the wheat seedlings.

###### Example143

Delbert traveled $258$ miles on $12$ gallons of gas, and Francine traveled $182$ miles on $8$ gallons of gas. Compute the ratio $\dfrac{\text{miles}}{\text{gallon}}$ for each car. Whose car gets the better gas mileage?

Solution

Delbert's gas mileage was

\begin{equation*} \dfrac{258\text{ miles}}{12\text{ gallons}}=21.5\text{ mpg} \end{equation*}

and Francine's was

\begin{equation*} \dfrac{182\text{ miles}}{8\text{ gallons}}=22.75\text{ mpg} \end{equation*}

Delbert gets better mileage.

In Example143, the ratio $\dfrac{\text{miles}}{\text{gallon}}$ measures the rate at which each car uses gasoline. By computing the mileage for each car for the same amount of gas, we have a basis for comparison. We can use this same idea, finding a common basis for comparison, to measure the steepness of an incline.

### SubsectionSome Properties of Lines

For any constant $k\text{:}$

• The graph of the equation $y=k$ is a horizontal line through $(0,k)$ and its slope is zero.

• The graph of the equation $x=k$ is a vertical line through $(k,0)$ and its slope is undefined.

You may have a hard time remembering which is horizontal and which is vertical. One thing that can help is to remember that if $y=3\text{,}$ then $x$ can be whatever value it wants, but the value for $y$ is fixed at 3. So, though we can range over all possible $x$ values, $y$ is always three, which is why we have a horizontal line. You can come up with a similar statement for the case when $x=3$ to explain why this must be a vertical line.

Another important property that lines can have is how two lines intersect. We have two special cases that we address first lines that are perpendicular and lines that are parallel. Recall, from your last algebra class that:

Let $l_1: y = m_1x + b_1$ and $l_2: y = m_2x + b_2$ be two lines. Then, we say that

• $l_1$ and $l_2$ are parallel if $m_1 = m_2$ and $b_1\neq b_2.$

• $l_1$ and $l_2$ are perpendicular if $m_1 = \frac{-1}{m_2}$ (negative reciprocal).

Note that for two lines to parallel, they must have different $y$-intercepts, otherwise they are the same line.

###### Example144

Find the equation of a line passing through the point $(4,-1)$ and parallel to the line $y=3x+2$

Solution

Since the two lines are parallel we know that the slope of the new line must be the same as the slope of the original line. That is $m_1=3=m_2\text{.}$ We can now use point slope form $y=y_1+m(x-x_1)$ to get $y=(-1)+3(x-4)\text{.}$ At this point we have the equation of a line. However, most people are more comfortable writing lines in slope-intercept form so we will do that here (by simplifying and distributing): $y=3x-13\text{.}$

###### Example145

Find the equation of a line passing through the point $(4,-1)$ and perpendicular to the line $y=3x+2$

Solution

Since the two lines are perpendicular we know that the slope of the new line must be the opposite reciprocal of the slope of the original line. Since $m_1=3\text{,}$ $m_2=-\frac{1}{3}\text{.}$ We can now use point slope form $y=y_1+m(x-x_1)$ to get $y=(-1)-\frac{1}{3}(x-4)\text{.}$ At this point we have the equation of a line. In slope-interept form, we have $y=-\frac{1}{3}x-\frac{7}{3}$ after distributing and simplifying.

In addition to finding the equation for parallel and perpendicular lines, it is often useful to be able to describe the point at which two lines intersect. One way to find where two lines intersect is to first put the lines in slope intercept form and then set the two equations equal to each other.

###### Example146

Find the point where the lines $y=2x+4$ and $y=-x+3$ intersect.

Solution

Since both of these equations are solved for $y\text{,}$ we can set them equal to each other and then solve for $x\text{:}$

\begin{equation*} \begin{aligned} 2x+4 \amp= -x+3 \\ 3x+4 \amp= 3 \\ 3x \amp= -1 \\ x\amp= -\frac{1}{3} \end{aligned} \end{equation*}

To find the $y$-coordinate, we can substitute $x$ into either of the original equations: $y=2\left(-\frac{1}{3}\right)+4=\frac{10}{3}\text{.}$ Therefore the two lines intersect at the point $\left(-\frac{1}{3},\frac{10}{3}\right)\text{.}$

###### Example147

Renting a truck through Quick Rentals will cost you a fixed rate of 15 dollars plus 25 cents per mile. Renting a truck through Handy Trucks runs 25 dollars plus 15 cents per mile. If you were planning to rent a truck, how many miles would you need to drive in order for Handy Trucks to be a better deal?

Solution

We first need to recognize that we have two linear equations $y=0.25x+15$ and $y=0.15x+25\text{.}$ If we plug one mile into both equations we can see that Quick Rentals is much cheaper if you are only driving a mile. In order to find where Handy Trucks is cheaper it makes sense to try finding the point where both companies cost the same amount. In order to do this we find the intersection between the two lines.

\begin{gather} 0.25x+15=0.15+25\tag{3}\\ 0.10x=10\tag{4}\\ x=100\tag{5} \end{gather}

Therefore, as long as you are driving more than 100 miles, ($x >100$) we know that Handy Trucks is the better deal.

If you would like to review different strategies for finding where two lines intersect, refer to the Systems of Linear Equations Section from Intermediate Algebra.