DERIVING THE EQUATION OF A LINE
INTRODUCTION
The objective for this lesson on Deriving the Equation of a Line is, the student will derive the equation of a line and use this form to identify the slope and y-intercept of an equation.
The skills students should have in order to help them in this lesson include an understanding of slope.
We will have three essential questions that will be guiding our lesson. Number one, where is the slope represented in the equation of a line? Number two, where is the y-intercept represented in the equation of a line? And number three, what is the slope-intercept form of the equation of a line? Explain the meaning of the equation.
We will begin by completing the warm-up of identifying the unit rate or slope represented by equations, tables or graphs to prepare for deriving the equation of a line in this lesson.
SOLVE PROBLEM – INTRODUCTION
The SOLVE problem for this lesson is, Mrs. Rosen’s fourth grade class is monitoring the growth of the class tree over the course of the school year. The tree was planted in August and was twenty four inches tall. The students notice that the tree grows at a rate of two inches per month from the monthly measurements taken. What is the equation of the line that represents the height of the tree represented by the variable y, after a specific number of months, represented by the variable x?
We will begin by Studying the Problem. First we need to identify where the question is located within the problem and underline the question. The question for this problem is, what is the equation of the line that represents the height of the tree, represented by the variable y after a specific number of months, represented by the variable x?
Now that we have identified the question, we want to put this question in our own words in the form of a statement. This problem is asking me to find the equation that represents the height of the tree over time.
During this lesson we will learn how to derive the equation of a line and use this form to identify the slope and y-intercept of an equation. We will use this knowledge to complete this SOLVE problem at the end of the lesson.
SLOPE AS RISE OVER RUN
Let’s start by taking a look at the graph of the line seen here. What are we looking for as we examine the line on the graph? We are looking for the slope of the graph or the vertical change over the horizontal change. What is another word for the vertical change? Justify your answer. The rise is the vertical change from one point on the graph to another point on the graph.
Let’s identify two points on the line that are on the graph: The point one, one and the point two, three. What is the vertical change from the point one, one to the point two, three? Let’s find out. We can see what the vertical change is by drawing a line from the point one, one up to the point where we are parallel with the point two, three. This represents the rise. It is two units. The vertical change from the point one, one to the point two, three is two units.
What is another word for the horizontal change? The run is the horizontal change from one point on the graph to another point on the graph.
What is the horizontal change from the point one, one to the point two, three? Explain your answer. Let’s see now. We can move from where we have the rise over to the point two, three to represent the length that is the run. It is one unit. The horizontal change from the point one, one to the point two, three is one unit.
Let’s represent the rise over the run. The rise was two and the run was one. We can represent this with the fraction two over one. Can we simplify two over one? Yes we can. So what is the slope when we simplify the fraction? It is two.
Now let’s take a look at another graph of a line. What are we looking for in this graph? We are looking for the rise over the run.
Let’s identify two points on the line that are on the graph: The point negative three, five and the point zero, three.
What is the vertical change from the point negative three, five to the point zero, three? Let’s see. We need to go down two units. The vertical change is negative two units, since we had to move down to get toward the next point.
And what is the horizontal change from the point negative three, five to the point zero, three? Let’s see. We move over three units. The horizontal change is three units.
So what is the rise over the run? The rise was negative two and the run is three. So the rise over the run is negative two over three.
The rise over the run is one way to identify and write the slope of the line.
In algebra we use the letter m to represent the slope. Let’s write the slope from Questions one and two using the notation “m equals.”
For question one the rise over the run was equal to two over one, which we found we could simplify to two. When we represent this using the letter m we can say that m is equal to two.
In question two the rise over the run was equal to negative two over three. This could not be simplified any further. We can represent the slope as m equals negative two over three.
DERIVE THE EQUATION OF A LINE
Let’s take a look at the table seen here. The table contains three columns. The first column represents the x-variable. The second column represents the function three times x. And the third column represents the y-variable.
Explain how we can complete the table. We need to substitute the x-value into the function to arrive at an output value, y. The function for this table is y equals three x. If we substitute negative two in for x, what is our result? Explain your thinking. When we substitute negative two in for x, we are multiplying three times negative two. Three times negative two equals negative six. Y equals negative six, when x is equal to negative two.
Now complete the rest of the table. Go ahead and do this now. Now let’s review the table. When x is equals to negative one, we multiply three times negative one, which equals negative three. This tells us that y is equal to negative three.
When x equals zero, we multiply three times zero, which equals zero. This tells us that y is equal to zero.
When x is equal to one, we multiply three times one, which equals three. This tells us that y is equal to three, when x is equal to one.
And when x is equal to two, we multiply three times two, which equals six. This tells us when x is equal to two, y is equal to six.
How can we plot the line for this function in Step two? We can graph the time by plotting the points we created.
How do we identify the points from the table? We use the x-coordinate and the y-coordinate to create five ordered pairs. Let’s do this now. The ordered pairs are negative two, negative six; negative one, negative three; zero, zero; one, three; and two, six. We can connect these points to form the line.
What is Step three asking us to find? It asks us to find the rise over the run, or the slope of the line.
What is one way we can find the rise over the run? We can count the vertical and horizontal distance between two points on the graph and write the value as a ratio of rise over run.
What is another way we can find the rise over the run? We can use two points by subtracting their y-coordinates and subtracting their x-coordinates.
If we use the vertical and horizontal distances, we can find the rise over the run. Let’s take a look at two points, the points zero, zero and one, three. We can use these to find the rise over the run. The rise from the point zero, zero to the point one, three is three units. And the run is one unit. The rise over the run is equal to three over one which simplifies to three.
We could also use two points on the line, for example the points two, six and one, three. To find the rise over the run we subtract the y-coordinates, six minus three for the rise. And we subtract the x-coordinates two minus one for the run. Six minus three equals three and two minus one equals one. So the rise over the run is three over one which simplifies to three. Either way we ended up with the same slope for the line, whether we used the rise over the run by counting the spaces from one point to the next or we found the rise over the run by finding the difference in the y-coordinates and the difference in the x-coordinates.
Now what is Step four asking us to find? It is asking us to find where the graph crosses the y-axis. Where does the graph cross the y-axis? It crosses at the point zero, zero.
What is the second part of Step four asking us to find? It asks us to find the y-intercept. What is the meaning of the y-intercept? What does the word “intercept” sound like? It sounds like the word intersect. So what do you predict a y-intercept is? It is a place where the graph intersects the y-axis. So what is the y-intercept? We use the variable b to represent this. Explain your thinking. The y-intercept is zero, because this is the point where the line crosses the y-axis.
According to the last question, what other notation can we use to describe a y-intercept? We can use the variable b to represent the y-intercept. If we substitute negative two in for x, what is our result? Explain your thinking. When we substitute negative two in for x, we have to find three times negative two plus one. Three times negative two is negative six. When we add one to negative six we get negative five. Substituting negative two in for x, y equals negative five. Let’s complete the table when x is equal to negative two.
Now complete the rest of the table. Go ahead and do this now. Let’s review together.
When x is equal to negative one, we replace the x variable in the function with negative one and solve to find y. Three times negative one is negative three, plus one equals negative two. When x equals negative one y equals negative two.
When x equals zero y equals one. When x equals one y equals four. And when x equal two, y equal seven. We have now completed the table.
How can we plot the line in Step two? We can graph the line by plotting the points we created.
How do we identify the points from the table? We use the x-coordinate and the y-coordinate to create five ordered pairs. Let’s plot these points now. We can plot the point negative two, negative five; the point negative one, negative two; the point zero, one; the point one, four and the last point we found was the point two, seven. This point does not fit on the coordinate grid that we have seen here. We would need to extend the coordinate grid up on the y-axis in order to plot this point. That’s ok though, we can use the four points we have plotted to connect them to show the line of this equation.
What is Step three asking us to find? It asks us to find the rise over the run, or the slope. What is one way that we can find the rise over the run? We can count the vertical and horizontal distance between two points on the graph and write the value as a ratio of rise over run.
What is another way we can find the rise over the run? We can use two points by subtracting their y-coordinates and subtracting their x-coordinates. If we use the vertical and horizontal distances, we can find the rise over the run. Let’s do this now. Let’s use the point zero, one and the point one, four. The rise between these two points is three units. And the run between these two points is one unit. The rise over the run is three over one, which simplifies to three. We could also use two points on the line, for example, the point one, four and the point zero, one. We can find the rise over the run by finding the difference in the y-values and the difference in the x-values. For the y-values we find the difference between four and one and for the x-values we find the difference between one and zero. Four minus one is three and one minus zero is one. The rise over the run is three over one, which simplifies to three. Either way we found the slope of the line is three.
What is Step four asking us to find? It asks us to find where the graph crosses the y-axis. Where does the graph cross the y-axis? It crosses at the point zero, one.
What is the second part of Step four asking us to find? It asks us to find the y-intercept. So what is the y-intercept, this could be represented by the variable b? Explain your thinking. The y-intercept is one, because this is the point where the line crosses the y axis.
Now go ahead and complete the charts for Parts C and D based on what we did here with parts A and B. Then, copy your solutions from Parts A through D into the table provided on the next page.
Let’s take a look at the table now. Once completed your table should include the following information. You need to include the equation for each line, its slope and the y-intercept.
Now let’s use the information from this table in order to summarize what we have done so far. What step did we use to find the slope in each one of the problems? We used Step three to find the slope.
And what step did we use the find the y-intercept in each of the four problems? We used Step four to find the y-intercept.
Now looking at the chart that we created, what do you notice about the slopes for Parts A and B? The slope for Part A and Part B are both equal to three. They are the same, since they are both three.
What do you notice about the slopes for Parts C and D? Take a look at the table. You can see that the slope for Parts C and D is negative two for both parts. So they are also the same. They are both negative two.
What do you notice about the equations for Parts A and B? The equation for Part A is y equals three x. And the equation for Part B is y equals three x plus one. We noticed that they both have three as a coefficient for x, the independent value.