Intermediate Algebra
Tutorial 16:
Equations of Lines
 Learning Objectives
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After completing this tutorial, you should be able to:
- Use the slope/intercept form to write a linear
equation given the slope
and y-intercept.
- Use the slope/intercept form to graph a linear
equation.
- Use the point/slope equation to set up an equation
given any point on
the
line and the slope.
- Use the point/slope equation to set up an equation
given two points on
the line.
- Use the point/slope equation to set up an equation
given a point on the
line and a parallel line.
- Use the point/slope equation to set up an equation
given a point on the
line and a perpendicular line.
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Introduction
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| 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
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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. |
| Step 4: Use the slope to find a
second point on the
graph. |
Recall in
Tutorial 15: The
Slope of the Line that the phrase "rise over run" is used in
conjunction
with slope.
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.
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Example
2: Graph the linear equation using the point/slope form.
3x - 2y = 4 |
 |
*Inverse of add 3x
is sub. 3x
*Inverse of mult. by -2 is div.
by -2
*Slope/intercept form of the
line
|
| 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.
|
The slope is 3/2.
Starting on the y-intercept (0, -2), we will rise up 3 and run
right
2.
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Example
3: Graph the linear equation using the point/slope form.
y = -x |
 |
*Slope/intercept form of the
line |
| 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.
|
The slope is -1 or -1/1.
Starting on the y-intercept (0, 0), we will go down 1 and run right
1.
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Point/Slope Form of an Equation
A line going through
the point  and
having slope of m
would have the equation
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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:
- ANY point on the line
- 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  . |
 |
*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 = c.
So, 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.
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Practice Problems
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| 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.
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Practice
Problem 2a:
Graph the linear equation using the
point/slope form.
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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.
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Practice
Problem 4a:
Find the equation of a line passing
through the given
points.
Use function notation to write the
equation.
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Practice
Problems 5a - 5b:
Write an equation of the line.
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Practice
Problems 6a - 6b:
Find an equation of the line.
Write the equation using function
notation.
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Need Extra Help on These Topics?
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All contents copyright (C) 2001 - 2008, WTAMU and Kim Seward. All rights reserved.
Last revised on Jan. 7, 2002 by Kim Seward. |
