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Chapter 1: Linear Equations
1.8: Slope-Intercept, Point-Slope, Standard form of lines, Vertical, Horizontal, Parallel, Perpendicular Lines and Applications

1.8: Slope-Intercept, Point-Slope, Standard form of lines, Vertical, Horizontal, Parallel, Perpendicular Lines and Applications

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Learning Objectives

Write linear equations
Use linear equations to solve real-world problems.
Apply slope to rates of change
Write the equation of a line parallel or perpendicular to a given line

Linear equations are everywhere in the real world. When you try to figure how long it will take you to get home while driving a certain speed, or determining if you have enough money to pay your bills, you are solving a linear equation. In this section, we will practice solving some linear equations, create linear equations from given information, and examine applications of linear equations and slope intercept form. Usually, when a linear equation model uses real-world data, different letters are used for the variables, instead of using only x and y but you will still use the concept of slope and $y$-intercept (starting value or initial condition).

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as \[y=mx+b\] where $m=\text{slope}$ and $b=\text{y−intercept.}$ Let us begin with the slope.

The slope of a line refers to the ratio of the vertical change in $y$ over the horizontal change in $x$ between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

\[m=\dfrac{y_2-y_1}{x_2-x_1}\]

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure $\PageIndex{1}$. The lines indicate the following slopes: $m=−3$, $m=2$, and $m=\dfrac{1}{3}$.

Coordinate plane with the x and y axes ranging from negative 10 to 10. Three linear functions are plotted: y = negative 3 times x minus 2; y = 2 times x plus 1; and y = x over 3 plus 2. — Figure $\PageIndex{1}$

Let's take a minute and practice writing lines in slope-intercept form.

Exercise $\PageIndex{1}$

a. Given a slope of $-\frac{2}{3}$ and a $y$-intercept of $6$, write the equation of the line in slope-intercept form. Then graph the equation.

Answer a

It will be in your best interest to memorize that slope-intercept form of a line is $y=mx+b$ where $m$ is the slope of the line and $b$ is the $y$-intercept, the initial condition, or the point of the form $(0,y)$.

This line will have the equation $y=-\frac{2}{3}x+6$. Notice on the graph below, the $y$-intercept is at $(0,6)$. Starting at the $y$-intercept, you can count down two and right three to find the next point on the line.

$graph lesson 1.5.png$

b. Given a slope of $-\frac{2}{3}$ and the point $(3,6)$, write the equation of the line in slope-intercept form. Then graph the equation.

Answer b

Starting with the equation $y=mx+b$ we will substitute in with the given information. We know that $y=6$ and $x=3$ from the ordered pair and we know that $m=-\frac{2}{3}$ since $m$=slope.

We now have the equation $6=-\frac{2}{3}(3)+b$

$6=-2+b$

$8=b$

The equation of the line is $y=-\frac{2}{3}x+8$.

Notice this line has the same slope as in part a, but it has a different $y$-intercept. Since they have the same slope, these two lines are parallel.

$graph lesson 1.5b.png$

c. Given the point $(-2,5)$ and the point $(2,1)$, write the equation of the line in slope-intercept form.

Answer c

We find the slope between the two points by using $m=\frac{y_2-y_1}{x_2-x_1}$ = $m=\frac{1-5}{2-(-2)}$ $=m=\frac{-4}{4}$

Notice it does NOT matter which point you choose first as long as you are consistent. $m=\frac{5-1}{-2-2}$ $=m=\frac{4}{-4}$

Now we may choose either point, I prefer not to deal with the negative signs so I will choose $(2,1)$ to go with $m=-1$

$1=-(1)(2)+b$

$1=-2+b$

$-3=b$

The equation of the line is $y=1x-3$.

$graph lesson 1.5b.png$

d. Given the point $(6,2)$ and the point $(8,5)$, write the equation of the line in slope-intercept form.

Answer d

We find the slope between the two points by using $m=\frac{y_2-y_1}{x_2-x_1}$ = $m=\frac{5-2}{8-6}$ $=m=\frac{3}{2}$

Now we may choose either point, let's choose $(6,2)$ to go with $m=\frac{3}{2}$

$2=\frac{3}{2}(6)+b$

$2=9+b$

$-7=b$

The equation of the line is $y=\frac{3}{2}x-7$

What do you notice about the slope of this line compared to the line in a.

Graph and see what you notice about the relationship between the two lines.

Exercise $\PageIndex{2}$

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $$34$ plus $$.05/min$ talk-time. Company B charges a monthly service fee of $$40$ plus $$.04/min$ talk-time.

Write a linear equation that models the packages offered by both companies.
If the average number of minutes used each month is $1,160$, which company offers the better plan?
If the average number of minutes used each month is $420$, which company offers the better plan?
How many minutes of talk-time would yield equal monthly statements from both companies?

Answer

The model for Company A can be written as $ A =0.05x+34$. This includes the variable cost of $ 0.05x$ plus the monthly service charge of $$34$. Company B’s package charges a higher monthly fee of $$40$, but a lower variable cost of $ 0.04x$. Company B’s model can be written as $ B =0.04x+$40$.

If the average number of minutes used each month is $1,160$, we have the following:

\[\begin{align*} \text{Company A}&= 0.05(1.160)+34\\ &= 58+34\\ &= 92 \end{align*}\]

\[\begin{align*} \text{Company B}&= 0.04(1,1600)+40\\ &= 46.4+40\\ &= 86.4 \end{align*}\]

So, Company B offers the lower monthly cost of $$86.40$ as compared with the $$92$ monthly cost offered by Company A when the average number of minutes used each month is $1,160$.

If the average number of minutes used each month is $420$, we have the following:

\[\begin{align*} \text{Company A}&= 0.05(420)+34\\ &= 21+34\\ &= 55 \end{align*}\]

\[\begin{align*} \text{Company B}&= 0.04(420)+40\\ &= 16.8+40\\ &= 56.8 \end{align*}\]

If the average number of minutes used each month is $420$, then Company A offers a lower monthly cost of $$55$ compared to Company B’s monthly cost of $$56.80$.

To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of $(x,y)$ coordinates: At what point are both the $x$-value and the $y$-value equal? We can find this point by setting the equations equal to each other and solving for $x$.

\[\begin{align*} 0.05x+34&= 0.04x+40\\ 0.01x&= 6\\ x&= 600 \end{align*}\]

Check the $x$-value in each equation.

$0.05(600)+34=64$

$0.04(600)+40=64$

Therefore, a monthly average of $600$ talk-time minutes renders the plans equal. See Figure $\PageIndex{2}$.

Coordinate plane with the x-axis ranging from 0 to 1200 in intervals of 100 and the y-axis ranging from 0 to 90 in intervals of 10. The functions A = 0.05x + 34 and B = 0.04x + 40 are graphed on the same plot — Figure $\PageIndex{2}$

Interpret Applications of Linear Equations

We will examine a few applications here so you can see how equations written in slope-intercept form relate to real-world situations.

Usually, when a linear equation model uses real-world 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.

Additionally, we often need to extend the axes in our rectangular coordinate system to accommodate larger positive and negative numbers to represent the data in the application.

Example $\PageIndex{1}$

The equation $F=\frac{9}{5}C+32$ 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.

Answer

a. $ \begin{array} {ll} {\text{Find the Fahrenheit temperature for a Celsius temperature of 0.}} &{F=\frac{9}{5}C+32} \\ {\text{Find F when C=0.}} &{F=\frac{9}{5}(0)+32} \\ {\text{Simplify.}} &{F=32} \\ \end{array} \nonumber$

b. $ \begin{array} {ll} {\text{Find the Fahrenheit temperature for a Celsius temperature of 20.}} &{F=\frac{9}{5}C+32} \\ {\text{Find F when C=20.}} &{F=\frac{9}{5}(20)+32} \\ {\text{Simplify.}} &{F=36+32} \\ {\text{Simplify.}} &{F=68} \\ \end{array} \nonumber$

c. Interpret the slope and F-intercept of the equation.
Even though this equation uses F and C, it is still in slope-intercept form.

$y equals m x plus b. F equals m C plus b. The y and F are emphasized in red. The x and C are emphasized in blue. F equals 9 divided by 5 C plus 32.$

The slope, $frac{9}{5}$, means that the temperature Fahrenheit (F) increases 9 degrees when the temperature Celsius (C) 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’ll 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.

$This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 40 to 80. The y-axis runs from negative 40 to 80. The line goes through the points (0, 32) and (5, 41).$

Example $\PageIndex{2}$

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.

$This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 14. The y-axis runs from negative 1 to 80. The line goes through the points (0, 50) and (10, 70).$

Example $\PageIndex{3}$

The equation $T=\frac{1}{4}n+40$ 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, $\frac{1}{4}$, means that the temperature Fahrenheit (F) 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.

$This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 140. The y-axis runs from negative 1 to 80. The line goes through the points (0, 40) and (40, 50).$

Setting up a Linear Equation to Solve a Real-World Application

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $$15.00/hr$, and she opened a savings account with an initial deposit of $$400$ on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $$2,500$, how many hours will she have to work to earn enough to pay for her vacation? If she can only work $4$ hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure $\PageIndex{3}$.

Coordinate plane where the x-axis ranges from 0 to 200 in intervals of 20 and the y-axis ranges from 0 to 3,000 in intervals of 500. The x-axis is labeled Hours Worked and the y-axis is labeled Savings Account Balance. A linear function is plotted with a y-intercept of 400 with a slope of 15. A dotted horizontal line extends from the point (0,2500). — Figure $\PageIndex{3}$

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

To set up or model a linear equation to fit a real-world application, we must first determine the independent and dependent variables. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $$0.10/mi$, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write $0.10x$. This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually add or subtract it, according to the problem. As these amounts do not change, we refer to them as fixed costs. Consider a car rental agency that charges $$0.10/mi$ plus a daily fee of $$50$. We can use these quantities to model an equation that can be used to find the daily car rental cost $C$.

$C=0.10x+50 \tag{2.4.1}$

Example $\PageIndex{4}$: Setting Up a Equation to Solve a Real-World Application

Write a linear equation that models the packages offered by both companies.
If the average number of minutes used each month is $1,160$, which company offers the better plan?
If the average number of minutes used each month is $420$, which company offers the better plan?
How many minutes of talk-time would yield equal monthly statements from both companies?

Solution