## Section4.5Slope-Intercept Form

###### ObjectivesPCC Course Content and Outcome Guide

In this section, we will explore one of the “standard” ways to write the equation of a line. It's known as slope-intercept form.

### Subsection4.5.1Slope-Intercept Definition

Recall Example 4.4.5, where Yara had $\50$ in her savings account when the year began, and decided to deposit $\20$ each week without withdrawing any money. In that example, we model using $x$ to represent how many weeks have passed. After $x$ weeks, Yara has added $20x$ dollars. And since she started with $\50\text{,}$ she has

\begin{equation*} y=20x+50 \end{equation*}

in her account after $x$ weeks. In this example, there is a constant rate of change of $20$ dollars per week, so we call that the slope as discussed in Section 4.4. We also saw in Figure 4.4.7 that plotting Yara's balance over time gives us a straight-line graph.

The graph of Yara's savings has some things in common with almost every straight-line graph. There is a slope, and there is a place where the line crosses the $y$-axis. Figure 4.5.2 illustrates this in the abstract.

We already have an accepted symbol, $m\text{,}$ for the slope of a line. The $y$-intercept is a point on the $y$-axis where the line crosses. Since it's on the $y$-axis, the $x$-coordinate of this point is $0\text{.}$ It is standard to call the $y$-intercept $(0,b)$ where $b$ represents the position of the $y$-intercept on the $y$-axis.

###### Checkpoint4.5.3

Use Figure 4.4.7 to answer this question.

One way to write the equation for Yara's savings was

\begin{equation*} y=20x+50\text{,} \end{equation*}

where both $m=20$ and $b=50$ are immediately visible in the equation. Now we are ready to generalize this.

###### Definition4.5.4Slope-Intercept Form

When $x$ and $y$ have a linear relationship where $m$ is the slope and $(0,b)$ is the $y$-intercept, one equation for this relationship is

$$y=mx+b\label{equation-slope-intercept-form}\tag{4.5.1}$$

and this equation is called the slope-intercept form of the line. It is called this because the slope and $y$-intercept are immediately discernible from the numbers in the equation.

###### Remark4.5.6

The number $b$ is the $y$-value when $x=0\text{.}$ Therefore it is common to refer to $b$ as the initial value or starting value of a linear relationship.

###### Example4.5.7

With a simple equation like $y=2x+3\text{,}$ we can see that this is a line whose slope is $2$ and which has initial value $3\text{.}$ So starting at $y=3$ when $x=0$ (that is, on the $y$-axis), each time we increase the $x$-value by $1\text{,}$ the $y$-value increases by $2\text{.}$ With these basic observations, we can quickly produce a table and/or a graph.

 $x$ $y$ start on $y$-axis $\longrightarrow$ $0$ $3$ initial $\longleftarrow$ value increase by $1\longrightarrow$ $1$ $5$ increase $\longleftarrow$ by $2$ increase by $1\longrightarrow$ $2$ $7$ increase $\longleftarrow$ by $2$ increase by $1\longrightarrow$ $3$ $9$ increase $\longleftarrow$ by $2$ increase by $1\longrightarrow$ $4$ $11$ increase $\longleftarrow$ by $2$
###### Example4.5.8

Decide whether data in the table has a linear relationship. If so, write the linear equation in slope-intercept form (4.5.1).

 $x$-values $y$-values $0$ $-4$ $2$ $2$ $5$ $11$ $9$ $23$
Explanation

To assess whether the relationship is linear, we have to recall from Section 4.3 that we should examine rates of change between data points. Note that the changes in $y$-values are not consistent. However, the rates of change are calculated as follows:

• When $x$ increases by $2\text{,}$ $y$ increases by $6\text{.}$ The first rate of change is $\frac{6}{2}=3\text{.}$

• When $x$ increases by $3\text{,}$ $y$ increases by $9\text{.}$ The second rate of change is $\frac{9}{3}=3\text{.}$

• When $x$ increases by $4\text{,}$ $y$ increases by $12\text{.}$ The third rate of change is $\frac{12}{4}=3\text{.}$

Since the rates of change are all the same, $3\text{,}$ the relationship is linear and the slope $m$ is $3\text{.}$

According to the table, when $x=0\text{,}$ $y=-4\text{.}$ So the starting value, $b\text{,}$ is $-4\text{.}$

So in slope-intercept form, the line's equation is $y=3x-4\text{.}$

### Subsection4.5.2Graphing Slope-Intercept Equations

###### Example4.5.10

The conversion formula for a Celsius temperature into Fahrenheit is $F=\frac{9}{5}C+32\text{.}$ This appears to be in slope-intercept form, except that $x$ and $y$ are replaced with $C$ and $F\text{.}$ Suppose you are asked to graph this equation. How will you proceed? You could make a table of values as we do in Section 4.2 but that takes time and effort. Since the equation here is in slope-intercept form, there is a nicer way.

Since this equation is for starting with a Celsius temperature and obtaining a Fahrenheit temperature, it makes sense to let $C$ be the horizontal axis variable and $F$ be the vertical axis variable. Note the slope is $\frac{9}{5}$ and the vertical intercept (here, the $F$-intercept) is $(0,32)\text{.}$

1. Set up the axes using an appropriate window and labels. Considering the freezing and boiling temperatures of water, it's reasonable to let $C$ run through at least $0$ to $100\text{.}$ Similarly it's reasonable to let $F$ run through at least $32$ to $212\text{.}$

2. Plot the $F$-intercept, which is at $(0,32)\text{.}$

3. Starting at the $F$-intercept, use slope triangles to reach the next point. Since our slope is $\frac{9}{5}\text{,}$ that suggests a “run” of $5$ and a “rise” of $9$ might work. But as Figure 4.5.11 indicates, such slope triangles are too tiny. Since $\frac{9}{5}=\frac{90}{50}\text{,}$ we can try a “run” of $50$ and a rise of $90\text{.}$

4. Connect your points with a straight line, use arrowheads, and label the equation.

###### Example4.5.12

Graph $y=-\frac{2}{3}x+10\text{.}$

###### Example4.5.14

Graph $y=3x+5\text{.}$

### Subsection4.5.3Writing a Slope-Intercept Equation Given a Graph

We can write a linear equation in slope-intercept form based on its graph. We need to be able to calculate the line's slope and see its $y$-intercept.

### Subsection4.5.4Writing a Slope-Intercept Equation Given Two Points

The idea that any two points uniquely determine a line has been understood for thousands of years in many cultures around the world. Once you have two specific points, there is a straightforward process to find the slope-intercept form of the equation of the line that connects them.

###### Example4.5.18

Find the slope-intercept form of the equation of the line that passes through the points $(0,5)$ and $(8,-5)\text{.}$

Explanation

We are trying to write down $y=mx+b\text{,}$ but with specific numbers for $m$ and $b\text{.}$ So the first step is to find the slope, $m\text{.}$ To do this, recall the slope formula (4.4.3) from Section 4.4. It says that if a line passes through the points $(x_1,y_1)$ and $(x_2,y_2)\text{,}$ then the slope is found by the formula $m=\frac{y_2-y_1}{x_2-x_1}\text{.}$

Applying this to our two points $(\overset{x_1}{0},\overset{y_1}{5})$ and $(\overset{x_2}{8},\overset{y_2}{-5})\text{,}$ we see that the slope is:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{-5}-\substitute{5}}{\substitute{8}-\substitute{0}}\\ \amp=\frac{-10}{8}\\ \amp=-\frac{5}{4} \end{align*}

We are trying to write $y=mx+b\text{.}$ Since we already found the slope, we know that we want to write $y=-\frac{5}{4}x+b$ but we need a specific number for $b\text{.}$ We happen to know that one point on this line is $(0,5)\text{,}$ which is on the $y$-axis because its $x$-value is $0\text{.}$ So $(0,5)$ is this line's $y$-intercept, and therefore $b=5\text{.}$ (We're only able to make this conclusion because this point has $0$ for its $x$-coordinate.) So, our equation is

\begin{equation*} y=-\frac{5}{4}x+5\text{.} \end{equation*}
###### Example4.5.19

Find the slope-intercept form of the equation of the line that passes through the points $(3,-8)$ and $(-6,1)\text{.}$

Explanation

The first step is always to find the slope between our two points: $(\overset{x_1}{3},\overset{y_1}{-8})$ and $(\overset{x_2}{-6},\overset{y_2}{1})\text{.}$ Using the slope formula (4.4.3) again, we have:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{1}-\substitute{(-8)}}{\substitute{{-}6}-\substitute{3}}\\ \amp=\frac{9}{-9}\\ \amp=-1 \end{align*}

Now that we have the slope, we can write $y=-1x+b\text{,}$ which simplifies to $y=-x+b\text{.}$ Unlike in Example 4.5.18, we are not given the value of $b$ because neither of our two given points have an $x$-value of $0\text{.}$ The trick to finding $b$ is to remember that we have two points that we know make the equation true! This means all we have to do is substitute either point into the equation for $x$ and $y$ and solve for $b\text{.}$ Let's arbitrarily choose $(3,-8)$ to plug in.

In conclusion, the equation for which we were searching is $y=-x-5\text{.}$

Don't be tempted to plug in values for $x$ and $y$ at this point. The general equation of a line in any form should have (at least one, and in this case two) variables in the final answer.

### Subsection4.5.5Modeling with Slope-Intercept Form

We can model many relatively simple relationships using slope-intercept form, and then solve related questions using algebra. Here are a few examples.

###### Example4.5.22

Uber is a ride-sharing company. Its pricing in Portland factors in how much time and how many miles a trip takes. But if you assume that rides average out at a speed of 30 mph, then their pricing scheme boils down to a base of $\7.35$ for the trip, plus $\3.85$ per mile. Use a slope-intercept equation and algebra to answer these questions.

1. How much is the fare if a trip is $5.3$ miles long?

2. With $\100$ available to you, how long of a trip can you afford?

Explanation

The rate of change (slope) is $\3.85$ per mile, and the starting value is $\7.35\text{.}$ So the slope-intercept equation is

\begin{equation*} y=3.85x+7.35\text{.} \end{equation*}

In this equation, $x$ stands for the number of miles in a trip, and $y$ stands for the amount of money to be charged.

If a trip is $5$ miles long, we substitute $x=5$ into the equation and we have:

\begin{align*} y\amp=3.85x+7.35\\ \amp=3.85(\substitute{5})+7.35\\ \amp=19.25+7.35\\ \amp=26.60 \end{align*}

And the $5$-mile ride will cost you about $\26.60\text{.}$ (We say “about,” because this was all assuming you average 30 mph.)

Next, to find how long of a trip would cost $\100\text{,}$ we substitute $y=100$ into the equation and solve for $x\text{:}$

\begin{align*} y\amp=3.85x+7.35\\ \substitute{100}\amp=3.85x+7.35\\ 100\subtractright{7.35}\amp=3.85x\\ 92.65\amp=3.85x\\ \frac{92.65}{3.85}\amp=x\\ 24.06\amp\approx x \end{align*}

So with $\100$ you could afford a little more than a $24$-mile trip.

### Subsection4.5.6Exercises

###### 1

Evaluate ${10B+2c}$ for $B = 7$ and $c = -4\text{.}$

###### 2

Evaluate ${-9C-a}$ for $C = -5$ and $a = -10\text{.}$

###### 3

Evaluate

\begin{equation*} \displaystyle\frac{y_2 - y_1}{x_2 - x_1} \end{equation*}

for $x_1 = 19\text{,}$ $x_2 = 9\text{,}$ $y_1 = 8\text{,}$ and $y_2 = 11\text{:}$

###### 4

Evaluate

\begin{equation*} \displaystyle\frac{y_2 - y_1}{x_2 - x_1} \end{equation*}

for $x_1 = -18\text{,}$ $x_2 = -5\text{,}$ $y_1 = -16\text{,}$ and $y_2 = -1\text{:}$

###### 5

Find the line’s slope and $y$-intercept.

A line has equation $y={3}x+1\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 6

Find the line’s slope and $y$-intercept.

A line has equation $y={4}x+7\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 7

Find the line’s slope and $y$-intercept.

A line has equation $y={-7}x - 7\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 8

Find the line’s slope and $y$-intercept.

A line has equation $y={-6}x - 1\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 9

Find the line’s slope and $y$-intercept.

A line has equation $y=x+3\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 10

Find the line’s slope and $y$-intercept.

A line has equation $y=x+5\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 11

Find the line’s slope and $y$-intercept.

A line has equation $y=-x+7\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 12

Find the line’s slope and $y$-intercept.

A line has equation $y=-x+9\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 13

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= -\frac{2}{3}x +8 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 14

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= -\frac{2}{9}x - 5 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 15

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= \frac{1}{2}x +8 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 16

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= \frac{1}{4}x - 7 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 17

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 7 +{6}x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 18

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 9 +{7}x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 19

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 8 -x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 20

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 9 -x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 21

Graph the equation $y=4x\text{.}$

###### 22

Graph the equation $y=5x\text{.}$

###### 23

Graph the equation $y=-3x\text{.}$

###### 24

Graph the equation $y=-2x\text{.}$

###### 25

Graph the equation $y=\frac{5}{2}x\text{.}$

###### 26

Graph the equation $y=\frac{1}{4}x\text{.}$

###### 27

Graph the equation $y=-\frac{1}{3}x\text{.}$

###### 28

Graph the equation $y=-\frac{5}{4}x\text{.}$

###### 29

Graph the equation $y=5x+2\text{.}$

###### 30

Graph the equation $y=3x+6\text{.}$

###### 31

Graph the equation $y=-4x+3\text{.}$

###### 32

Graph the equation $y=-2x+5\text{.}$

###### 33

Graph the equation $y=x-4\text{.}$

###### 34

Graph the equation $y=x+2\text{.}$

###### 35

Graph the equation $y=-x+3\text{.}$

###### 36

Graph the equation $y=-x-5\text{.}$

###### 37

Graph the equation $y=\frac{2}{3}x+4\text{.}$

###### 38

Graph the equation $y=\frac{3}{2}x-5\text{.}$

###### 39

Graph the equation $y=-\frac{3}{5}x-1\text{.}$

###### 40

Graph the equation $y=-\frac{1}{5}x+1\text{.}$

###### 41

A line’s graph is given.

This line’s slope-intercept equation is

###### 42

A line’s graph is given.

This line’s slope-intercept equation is

###### 43

A line’s graph is given.

This line’s slope-intercept equation is

###### 44

A line’s graph is given.

This line’s slope-intercept equation is

###### 45

A line’s graph is given.

This line’s slope-intercept equation is

###### 46

A line’s graph is given.

This line’s slope-intercept equation is

###### 47

A line’s graph is given.

This line’s slope-intercept equation is

###### 48

A line’s graph is given.

This line’s slope-intercept equation is

###### 49

A line passes through the points $(3,22)$ and $(4,27)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 50

A line passes through the points $(5,29)$ and $(2,14)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 51

A line passes through the points $(-2,20)$ and $(-5,35)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 52

A line passes through the points $(3,-12)$ and $(2,-7)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 53

A line passes through the points $(-2,-3)$ and $(-3,-2)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 54

A line passes through the points $(5,-7)$ and $(1,-3)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 55

A line passes through the points $(18,{16})$ and $(0,{1})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 56

A line passes through the points $(0,{7})$ and $(-15,{-11})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 57

A line passes through the points $(-9,{16})$ and $(0,{9})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 58

A line passes through the points $(-5,{9})$ and $(-15,{25})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 59

A gym charges members ${\40}$ for a registration fee, and then ${\24}$ per month. You became a member some time ago, and now you have paid a total of ${\448}$ to the gym. How many months have passed since you joined the gym?

months have passed since you joined the gym.

###### 60

Your cell phone company charges a ${\11}$ monthly fee, plus ${\0.18}$ per minute of talk time. One month your cell phone bill was ${\68.60}\text{.}$ How many minutes did you spend talking on the phone that month?

You spent talking on the phone that month.

###### 61

A school purchased a batch of T-shirts from a company. The company charged ${\4}$ per T-shirt, and gave the school a ${\75}$ rebate. If the school had a net expense of ${\1{,}565}$ from the purchase, how many T-shirts did the school buy?

The school purchased T-shirts.

###### 62

Izabelle hired a face-painter for a birthday party. The painter charged a flat fee of ${\65}\text{,}$ and then charged ${\2.50}$ per person. In the end, Izabelle paid a total of ${\137.50}\text{.}$ How many people used the face-painter’s service?

people used the face-painter’s service.

###### 63

A certain country has $406.56$ million acres of forest. Every year, the country loses $4.84$ million acres of forest mainly due to deforestation for farming purposes. If this situation continues at this pace, how many years later will the country have only $227.48$ million acres of forest left? (Use an equation to solve this problem.)

After years, this country would have $227.48$ million acres of forest left.

###### 64

Anthony has ${\80}$ in his piggy bank. He plans to purchase some Pokemon cards, which costs ${\1.75}$ each. He plans to save ${\60.75}$ to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Anthony can purchase at most Pokemon cards.

###### 65

By your cell phone contract, you pay a monthly fee plus ${\0.06}$ for each minute you spend on the phone. In one month, you spent $220$ minutes over the phone, and had a bill totaling ${\25.20}\text{.}$

Let $x$ be the number of minutes you spend on the phone in a month, and let $y$ be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone.

1. This line’s slope-intercept equation is .

2. If you spend $140$ minutes on the phone in a month, you would be billed .

3. If your bill was ${\39.60}$ one month, you must have spent minutes on the phone in that month.

###### 66

A company set aside a certain amount of money in the year 2000. The company spent exactly ${\42{,}000}$ from that fund each year on perks for its employees. In $2003\text{,}$ there was still ${\782{,}000}$ left in the fund.

Let $x$ be the number of years since 2000, and let $y$ be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.

1. The linear model’s slope-intercept equation is .

2. In the year $2009\text{,}$ there was left in the fund.

3. In the year , the fund will be empty.

###### 67

A biologist has been observing a tree’s height. This type of tree typically grows by $0.27$ feet each month. Ten months into the observation, the tree was $17.1$ feet tall.

Let $x$ be the number of months passed since the observations started, and let $y$ be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.

1. This line’s slope-intercept equation is .

2. $26$ months after the observations started, the tree would be feet in height.

3. months after the observation started, the tree would be $29.52$ feet tall.

###### 68

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose $1.7$ grams. Seven minutes since the experiment started, the remaining gas had a mass of $73.1$ grams.

Let $x$ be the number of minutes that have passed since the experiment started, and let $y$ be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.

1. This line’s slope-intercept equation is .

2. $33$ minutes after the experiment started, there would be grams of gas left.

3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

###### 69

A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In $2004\text{,}$ there was still ${\807{,}000}$ left in the fund. In $2005\text{,}$ there was ${\786{,}000}$ left.

Let $x$ be the number of years since 2000, and let $y$ be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.

1. The linear model’s slope-intercept equation is .

2. In the year $2009\text{,}$ there was left in the fund.

3. In the year , the fund will be empty.

###### 70

By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent $230$ minutes on the phone, and paid ${\17.45}\text{.}$ In another month, you spent $380$ minutes on the phone, and paid ${\19.70}\text{.}$

Let $x$ be the number of minutes you talk over the phone in a month, and let $y$ be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone.

1. This linear model’s slope-intercept equation is .

2. If you spent $130$ minutes over the phone in a month, you would pay .

3. If in a month, you paid ${\20.15}$ of cell phone bill, you must have spent minutes on the phone in that month.

###### 71

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way.

Nine minutes since the experiment started, the gas had a mass of $42.9$ grams.

Fifteen minutes since the experiment started, the gas had a mass of $35.1$ grams.

Let $x$ be the number of minutes that have passed since the experiment started, and let $y$ be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.

1. This line’s slope-intercept equation is .

2. $32$ minutes after the experiment started, there would be grams of gas left.

3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

###### 72

A biologist has been observing a tree’s height. $10$ months into the observation, the tree was $18.2$ feet tall. $16$ months into the observation, the tree was $19.22$ feet tall.

Let $x$ be the number of months passed since the observations started, and let $y$ be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.

1. This line’s slope-intercept equation is .

2. $26$ months after the observations started, the tree would be feet in height.

3. months after the observation started, the tree would be $25$ feet tall.

###### 73

Line $S$ has the equation $y = ax + b$ and Line $T$ has the equation $y =cx + d\text{.}$ Suppose $a \gt b \gt c \gt d \gt 0 \text{.}$

1. What can you say about Line $S$ and Line $T\text{,}$ given that $a \gt c\text{?}$ Give as much information about Line $S$ and Line $T$ as possible.

2. What can you say about Line $S$ and Line $T\text{,}$ given that $b \gt d\text{?}$ Give as much information about Line $S$ and Line $T$ as possible.