# Intermediate Algebra Tutorial 16

Intermediate Algebra
Tutorial 16: Equations of Lines

WTAMU > Virtual Math Lab > Intermediate Algebra

Learning Objectives

After completing this tutorial, you should be able to:
1. Use the slope/intercept form to write a linear equation given the slope and y-intercept.
2. Use the slope/intercept form to graph a linear equation.
3. Use the point/slope equation to set up an equation given any point on the line and the slope.
4. Use the point/slope equation to set up an equation given two points on the line.
5. Use the point/slope equation to set up an equation given a point on the line and a parallel line.
6. Use the point/slope equation to set up an equation given a point on the line and a perpendicular line.

Introduction

In this tutorial we will dive even deeper into linear equations. We will be going over how to come up with our own equations given certain information.  After you finish tutorials 12 - 17,  you will be an old pro at linear equations and graphing.  Let's see what we can do with slopes, y-intercepts and linear equations.

Tutorial

We are going to revisit the slope/intercept form of the line for a moment.  This was initially introduced in Tutorial 15: Slope of a Line.  We are going to use it again to help us come up with equations of lines as well as give us another way to graph lines.

Slope/Intercept Equation of a Line

OR

In this form, m represents the slope and b represents the y-intercept of the line.

Example 1:  Use the slope/intercept form of the linear equation to write the equation of a line with slope - 4; y -intercept (0, ½).

Since we have the two missing pieces of the puzzle,  the slope and the y-intercept, then we can go right into plugging in - 4 for m (slope) and ½ for b (y-intercept) into the slope/intercept form and we will have our equation.

*Slope/intercept form of a line

y = -4x + 1/2 is the equation of the line that has a slope of - 4 and y-intercept of (0, ½).

Nothing to it.

Graphing a Linear Equation
Using the Slope/Intercept Form

Step 1:  Write the linear equation in the slope/intercept form if necessary.

Some equations will already be in the slope/intercept form.  In that case, you do not have to rewrite it.  However, if it isn't in that form, then you need to rewrite it.  Basically, to get it into the slope/intercept form, you solve the linear equation for y.

Step 2: Identify the slope and y-intercept of the linear equation.

This is done exactly as shown in Tutorial 15: The Slope of the Line, where m is the slope and b is the y-intercept when the linear equation is written in the form y = mx + b.

Step 3: Plot the y-intercept point on the graph.

Remember, that when you identify b as your y-intercept that the ordered pair for it is (0, b).

Step 4: Use the slope to find a second point on the graph.

Positive Slope
If your slope is positive, then you are going to go in the same direction for both the rise and the run.  In other words, you either go in a positive direction for rise (up) and a positive direction for run (right) OR a negative direction for rise (down) and a negative direction for run (left).

For example, if the slope is 2/3, then you can rise up 2 and run right 3 from any point that is on the line.  Or, you can go down 2 and run left 3 from any point that is on the line.

Negative Slope
If your slope is negative, then you are going to go in opposite directions for the rise and the run.  In other words, you either go in a negative direction for rise (down) and a positive direction for run (right) OR a positive direction for rise (up) and a negative direction for run (left).

For example, if the slope is -2/3, then you can go down 2 and run right 3 from any point that is on the line.  Or, you can rise 2 and run left 3 from any point that is on the line.

Integer Slopes
If your slope is an integer, remember that the denominator is understood to be 1.  So, the run part of an integer slope is going to be 1.  For example, if the slope is 3, you want to think of it as 3/1.  You would rise up 3 and run right 1 from any point on the line to get another point on the line.

Step 5: Draw the graph.

Note that this gives us another way to graph linear equations. Tutorial 12: Graphing Equations covers graphing by plotting points, Tutorial 14: Graphing Linear Equation covers graphing using intercepts.  This tutorial covers graphing using the slope and the y-intercept. It doesn’t matter which method that you use, they all get the job done.  However, if a test, homework or an instructor asks you to use a certain method make sure you follow the directions and show the work for the asked method.

Let’s look at some examples for this new method.

Example 2: Graph the linear equation using the point/slope form.
3x - 2y = 4

Step 1:  Write the linear equation in the slope/intercept form, if necessary.

*Inverse of add 3x is sub. 3x

*Inverse of mult. by -2 is div. by -2

*Slope/intercept form of the line

Step 2: Identify the slope and y-intercept of the linear equation.

Looking at this equation and lining it up with the slope/intercept form, what do you get for the slope and y-intercept?

I got m = 3/2 and y-intercept = -2.

Step 3: Plot the y-intercept point on the graph.

Step 4: Use the slope to find a second point on the graph.

The slope is 3/2.
Starting on the y-intercept (0, -2), we will rise up 3 and run right 2.

Step 5: Draw the graph.

Example 3: Graph the linear equation using the point/slope form.
y = -x

This linear equation is already in the slope/intercept form.

*Slope/intercept form of the line

Step 2: Identify the slope and y-intercept of the linear equation.

Lining up everything, what do you get for the slope and the y-intercept??

Note how we are missing a constant being added to the x term.  If we are missing that constant, what is it understood to be???

The slope is -1 and the y-intercept is 0.

Step 3: Plot the y-intercept point on the graph.

Step 4: Use the slope to find a second point on the graph.

The slope is -1 or -1/1.
Starting on the y-intercept (0, 0), we will go down 1 and run right 1.

Step 5: Draw the graph.

Point/Slope Form of an Equation

A line going through
the point and
having slope of m
would have the equation

In example 1 above, the information was given to us in a nice neat little package, all ready for us to directly plug it into the slope/intercept form.  As you know, life isn’t always that nice. Sometimes we need to work with the information first before we can write up our linear equation.  Sometimes we need the help of the point/slope equation when we are not given the y-intercept specifically.

We can use this form when we need to come up with a linear equation and we don’t know the y-intercept.

No matter what form that you end up using, keep in mind that you ALWAYS need two pieces of information when you go to write an equation:

1. ANY point on the line
2. Slope

Example 4:  Find the equation of the line with the given slope and containing the given point.  Write the equation in slope/intercept form.
Passes through (1, 7) and has a slope of -2.

What are the two things we need to write an equation of a line????
If you said any point on the line and the slope, you are correct.

Looks like we have all the information we need.  We are ready to put our equation together.  Since we don’t have the y-intercept, we will have to use the point/slope form since that is set up for ANY point (not just the y-int.).

*Point/slope form of the line

Next, we want to write it in the slope/intercept form, which basically means we need to solve for y.

*Dist. the -2 through (  )
*Inverse of sub. 7 is add 7

*Slope/intercept form of the line

The equation of the line that passes through (1, 7)and has a slope of -2 is y = -2x + 9.

Example 5:  Find the equation of a line passing through the given points.  Use function notation to write the equation.
(-5, 3) and (-1, 4).

What are the two things we need to write an equation of a line????
If you said any point on the line and the slope, you are correct.

We have more than enough points. However, what about the slope?  Does this mean we can’t work out the problem?  You are not going to get off that easily.  We do have a way of finding the slope. Tutorial 15: The Slope of a Line shows us how we can get the slope given two points.

Let’s find that slope:

*Slope formula

*Plug in values

*Simplify

OK, now we have our slope, which is 1/4.  Now it is just like example 4 above, we want to put the slope and one point into the point/slope equation.

*Point/slope form of the line

The directions said to write it using function notation. This means we want to solve it for y (get it in the slope/int form) and then put it in function notation.

*Dist. 1/4 through (  )

*Inverse of sub. 3 is add 3

*LCD is 4
*Mult.  3/1 by 4/4 to get 12/4

*Slope/intercept form of the line

*Function notation

The line that passes through (-5, 3) and (-1, 4) in function notation is f(x) = 1/4 x + 17/4.

Example 6:  Find the equation of the line.  Write the equation using function notation.  Passes through (2, 3) and parallel to .

What are the two things we need to write an equation of a line????
If you said any point on the line and the slope you are correct.

We have our point. However, what about the slope?
We need to do a little digging to get it.
Recall that Tutorial 15: The Slope of a Line tells us that parallel lines have the same slope.  So, if we know the slope of the line parallel to our line, we have it made.

Find the slope of the parallel line:

*Slope/intercept form of the line

Now, keep in mind that this is not the equation of our line but of the line parallel to our line.  We needed to write it this way so we could get the slope.  And it looks like the slope is -3. And since our line is parallel to a line that has a slope of -3, our line also has a slope of -3.

OK, now we have our slope, which is -3.  Now it is just like examples 4 and 5 above, we want to put the slope and one point into the point/slope equation.

*Point/slope form of the line

The directions said to write it using function notation. This means we want to solve it for y (get it in the slope/int form) and then put it in function notation.

*Dist. -3 through (   )
*Inverse of sub. 3 is add 3

*Slope/intercept form of the line
*Function notation

The line that passes through (2, 3) and is parallel to y = 5 - 3x is f(x) = -3x + 9.

Example 7:  Find the equation of the line.  Write the equation using function notation.  Passes through (-1, 3) and perpendicular to .

What are the two things we need to write an equation of a line????
If you said any point on the line and the slope, you are correct.

We have our point. However, what about the slope?
We need to do a little work in that department.
Recall that  Tutorial 15: The Slope of a Line tells us that the slopes of perpendicular lines are negative reciprocals of each other.   So, if we know the slope of the line perpendicular to our line, we have it made.

Find the slope of the perpendicular line:

*Inverse of add x is sub. x

*Inverse of mult. by -2 is div. by -2

*Slope/intercept form of the line

In this form, we can tell that the slope of this line is ½.  Since our line is perpendicular to this, we need to take the negative reciprocal of ½ to get our slope.

What did you come up with?
I came up with -2 for the slope of our line.

Now we can go on to the equation of our line:

*Point/slope form of the line

The directions said to write it using function notation. This means we want to solve it for y (get it in the slope/int form) and then put it in function notation.

*Dist. -2 through the (   )
*Inverse of sub. 3 is add 3

*Slope/intercept form of the line
*Function notation

The line that passes through (-1, 3) and is perpendicular to x - 2y = 5 is f(x) = -2x + 1.

Example 8:  Write an equation of the line that is horizontal and passes through (4, -1).

Recall from Tutorial 14: Graphing Equations that a horizontal line is of the form y = c. It is one of our ‘special’ types of lines that does not have both x and y showing in the equation.  So, we will not use the point/slope form, but go right into setting up the equation y = c.

Since it passes through (4, -1), and a horizontal line is in the form y = c, where the y value is ALWAYS equal to the same value throughout, this means our equation would have to be y = -1.

Note that -1 is the y value of the ordered pair given.

Example 9:  Write an equation of the line that has an undefined slope and passes through (2, 3).

Recall from Tutorial 14: Graphing Equations and Tutorial 15: The Slope of the Line that the ‘special’ type of line that has an undefined slope is a vertical line and a vertical line is of the form x = cSo, we will not use the point/slope form, but go right into setting up the equation x = c.

Since it passes through (2, 3), and a vertical line is in the form x = c, where the x value is ALWAYS equal to the same value throughout, this means our equation would have to be x = 2.

Note that 2 is the x value of the ordered pair given.

Practice Problems

These are practice problems to help bring you to the next level.  It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it.  Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument.  In fact there is no such thing as too much practice.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that  problem.  At the link you will find the answer as well as any steps that went into finding that answer.

Practice Problem 1a: Use the slope-intercept form of the linear equation to write the equation of the line.

1a.   Slope 2/3; y-intercept (0, -2)

Practice Problem 2a: Graph the linear equation using the point/slope form.

2a.   3x + 5y = 10

Practice Problem 3a: Find the equation of the line with the given slope and containing the given point.  Write the equation in slope-intercept form.

3a.   Passes through (-3, 2) and has a slope of -1/2.

Practice Problem 4: Find the equation of a line passing through the given points. Use function notation to write the equation.

4a.   (0, 0) and (5, 10)

Practice Problems 5a - 5b: Write an equation of the line.

5a.     Slope 0; passes through (1, 2)
5b.  Vertical; passes through (-1, -2)

Practice Problems 6a - 6b: Find an equation of the line.  Write the equation using function notation.

6a.Passes through (2, 3) and parallel to 5x + 2y = 4

6b.  Passes through (-1, 0) and perpendicular to 3x - y = 2

Need Extra Help on these Topics?

The following are webpages that can assist you in the topics that were covered on this page:

http://www.math.com/school/subject2/lessons/S2U4L3DP.html
This website helps you with graphing linear equations.

http://www.purplemath.com/modules/slopgrph.htm
This website covers graphing linear equations using slopes.

http://www.purplemath.com/modules/slopyint.htm
This website goes over the meaning of slope and y-intercept.

http://www.purplemath.com/modules/strtlneq.htm
This website helps you with writing linear equations.

Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.

Last revised on July 5, 2011 by Kim Seward.