Overview of Lines

Subsection 2.1 What is a Line?

Mathematicians have interested in lines for thousands of years. One of the first "definitions" given for a line was by Euclid c. 300 BC. This description of a line was described in his postulated as,

1. A straight line segment can be drawn joining any two points.
2. Any straight line segment can be extended indefinitely in a straight line.

And while this is interesting, it is hard to work with. Since then many different ways to define a line have been introduced. But all of them fall a bit short of a mathematical definition. This is partly due to the fact that a line is a foundational mathematical object, and we generally build definitions off of these foundations.

The reason we bring this up is that lines are challenging and puzzling. They have been a point of interest in mathematics for many years and are still something that we as mathematicians are puzzling with today. Keep this in mind as we work through the next section.

Let's start by talking a look at the slope. The slope is a way that we can encode how to quantities vary along a line. This is like the DNA for the line, and tells us how to move from point to point along the line. The equation, \(\displaystyle m = \frac{y_2 - y_2}{x_2 - x_1} \) is a ratio of the change in y, (\(\Delta y = y_2 - y_1\)) over the change in x, (\(\Delta x = x_2 - x_1\)). We sometimes call this "rise over run".

Next let's explore how we can use the slope. Consider the following,

So in the last figure, we are starting at the point, \((2,2)\) and then we move up 1 and over 4 to get to another point, \((5,3)\text{.}\) Now if we were to connect the two points, we would have a, "line segment between \((2,2)\) and \((5,3)\text{.}\) And if we continue the line segment along this slope beyond the points, we would have a line.

Now one important note here is that, any point on the above line is related to the other any other point on the line by the same slope. We will see in the next sub section how this can be noted, but we can also see this in the above discription of a line when we said, "set of all points...". What we want to do now is connect the idea of "set of all points..." to the slope that we just explored.

Before we jump into the line equations, let's run a few problems.

For the next few problems, the slope can be negative, so take care as you subtract the values to find the changes. Feel free to use the following example as a guide for the next three problems.

Find the slope of the line between the points, \((-2,3) \) and \((4,-2)\text{.}\)

Solution

To find the slope we will use the equation \(\displaystyle m = \frac{y_2 - y_2}{x_2 - x_1} \) with \((x_1,y_1) = (-2,3) \) and \((x_2,y_2) = (4,-2)\)

So,

\begin{equation*} m = \frac{-2 - 3}{4 - (-2)} = \frac{-5}{6} \end{equation*}

giving us a negative slope.

When we consider a negative slope, we can think about it as "downhill" moving from left to right.

And one more possibility is a slope of zero. This will happen when the change in \(y\) is zero.

Find the slope of the line between the points, \((2,2) \) and \((4,2)\text{.}\) Note that the \(y \) value for both points is the same. So we have no change in \(y \) here.

Solution

To find the slope we will use the equation \(\displaystyle m = \frac{y_2 - y_2}{x_2 - x_1} \) with \((x_1,y_1) = (2,2) \) and \((x_2,y_2) = (4,2)\)

So,

\begin{equation*} m = \frac{2 - 2}{4 - 2} = \frac{0}{2} = 0 \end{equation*}

and a line with a zero change in slope would look like,

We will refer to this as a horizontal line. And in calculus, a slope of zero is very important to the study of maximization.

Now that we have had a chance to find the slope given two points, let's take a look at the line equation again.

Subsection 2.2 Equation of a Line

Working with the above equation is going to start with the "collection" of the things that we will need to build it. Namely, we need the slope and a point on the line. This is "collection" of these two mathematical elements, is vital to being able to write down a lines equation, and will often be what we are looking for in application problems. In the future (in calculus) the process of finding the slope plays a very important roll in the story.

The above step by step is going to be our go to when asked to find the equation of a line. From here we can find other forms of the equation of a line, as well as graph the line. Let's take a look at an example.

Now before we move on we should note that there are a few different equations for lines. Depending on the application, some of the equations are a better place to start then others. The one that we have been working with here is called, point-slope form. Another equation that is often used is called, slope-intercept form.

The beauty of the equation that we worked with first is that we can manipulate it to get a line into slope-intercept form.

Suppose that we have a line, \(y - 3 = \frac{5}{2} (x - 2) \) and we want to write this line in slope intercept form. To do that, we would solve for \(y \) and distribute the slope. So

Giving us the other form and we now know the \(y \) intercept as well.

Subsection 2.3 Graphing a Line

Putting together the work the we did with plotting points, the idea of slope and the equation of a line, we are now ready to break into graphing a line. Let's start with an example.

Graph the line, \(y - 2 = \frac{2}{3}(x-3).\)

Point-Slope Form of a Line.

The equation of the line in point slope form is,

\begin{equation*} y - y_1 = m(x - x_1) \end{equation*}

where \(m \) is the slope and \((x_1, y_1)\) is any point on the line.

Solution

To to graph this line we are going to work backwards a bit to collect the lines structure. Really what we need to graph the line is the same as what we would have needed to write down the equation; the slope and a point.

So using the above note on the equation that we looked at earlier, we can see that the slope, \(m = \frac{2}{3} \) and a point, \((x_1,y_1) = (3,2) \text{.}\)

So to graph this equation, we are going to start at the point we know is on the line, in our case \((3,2) \text{,}\) and then using the slope, we can find a second point on the line.

Once we have these two points, we can draw a line through them. This new line is what is meant by, "graph the line".

In the last example we saw how to plot using a line from the point slope equations. In review, we need to know the slope and a known point on the line. From here we plot the point, and used the slope to plot a second point. By connecting these points, we have a line segment between the two points, and by extending the line beyond the points, we have a the line graphed.

For our next example, we are going to plot a line, given that it is in slope intercept form. We will follow a similar plan to the last example, but we will use the vertical axis intercept as our stating point.

Graph the following line, \(y = \frac{3}{2} x + 2 \text{.}\)

This is a good time to note that, when we are on the \(y\) -axis, the \(x\) value will be zero. So we can verify that 2 is the \(y\) intercept by checking to see what the value will be at \(x=0 \text{.}\) That is,

\begin{equation*} y = \frac{3}{2} 0 + 2 = 2 \end{equation*}

So we can see that the point \((0,2) \) is, in fact, on the line.

Solution

In the next, and final, section, we are going to take a look at an application question involving lines. We will be given a problem described in words, and we will use that to find the equation of a line (actually it will be two lines) and the graph the equations to help answer a application problem.

Subsection 2.4 Application Example

Suppose that you are going to hire a moving company to transport a sensitive piece of machinery. You decide to get a few bids and compare them. Here are the two bids as they were described:

1. Company A will has an upfront cost of $200, and will charge $100 for every 3 hours after that.
2. Company B has no upfront cost, but charges $700 for every 2 hours.

Let's start by looking at where we will be plotting the function and what the quantities will be:

Looking at the blank graph can give us some idea of what we are working with. The plan will be to plot both of the moving companies bids on the coordinate plane. That way we can see what the cost will be for different times for each of them.

Let's start with the statement Company A will has an upfront cost of $200, and will charge $100 for every 3 hours after that. The phase, upfront cost of $200, means that when time is zero, we owe $200. So we know that the point, \((0, 200) \) must be on the graph.

Now we are also given the information, $100 for every 3 hours. This is the slope. It tells us for every 3 hours we move to the right, we are going to move $100 up. So this give us that, \(m_1 = \frac{100}{3}\text{.}\)

So using the point that we know is on the line and the slope that we have here, we can encode the equation for the cost with respect to hours for company A.

Using this any analysis we can determine the equation for the second company.

no upfront cost \(\Rightarrow (0 hours, $0)\) and $700 for every 2 hours \(\Rightarrow m_2 = \frac{700}{2} \text{.}\) leading to,

Now using what we explored from the last section, we can plot both of these equations in the coordinate plane.

Looking over the graph, we can see that Company B is less expensive if we are sure that it can be done in less than around half an hour, after that, Company A is the better choice.

We can determine the exact time that they are the same price by setting the equations equal to one another and solving.