Chapter 5
Equations of Lines
5.1
Cartesian Coordinate System
View the Video Tutorial for this section here.
Suppose you work for a window manufacturer wants to make windows of the following shape.
You are in charge of programming the computer controlled saw that will cut the glass to shape. The question
is, how do you describe the shape to the computer in order to have it cut the glass to the correct shape and
dimensions? Note that these types of machines with robotic arms exist and are used in manufacturing. So we
are not talking about some futuristic AI (articial intelligence) machine. The famous French mathematician
and philosopher René Descartes supposedly had a similar problem. The story goes, that while looking up
at a ceiling, he observed a y crawling around. In trying to describe the y's position, he came up with
what we now refer to as the Cartesian coordinate system. The idea is that we draw two lines, one going
horizontally (left and right), called the x-axis, and the other going vertically (up and down), called the y-axis,
and describe the location of a point by its horizontal and vertical distance from the point where the two
lines intersect, called the origin. So we refer to the point by these two distances, which together are called
the coordinates of the point. The notation we use to describe a point is called an ordered pair, (x,y). Here
are some examples of points on the Cartesian coordinate plane.
We note the following about the above graph.
• The tick marks (the griding) must be evenly spaced in each of the x and y directions. They do not have
to be the same in both directions. For example the tick marks along the x-axis might each represent
1
CHAPTER 5.
2
EQUATIONS OF LINES
5 feet, while the tick marks along the y-axis might each represent
1
2
foot.
• The point A has coordinates (2, 3). The rst number, 2, is the horizontal distance from the origin and
the second number, 3, is the vertical distance from the origin. So, to get to point A from the origin,
we go 2 units to the right and up 3.
• Positive means go up or to the right and negative means go down or to the left. So point C with
coordinates (−3, −4) means from the origin go left 3 and down 4 to get to C.
• To gure out the coordinates of point B, if we start from B and go down vertically to the x-axis,
the number on the x-axis represents the x-coordinate. For B, we see that it is -5. From B, if we go
horizontally until we hit the y-axis, we get the y-coordinate, which in this case is 2. So, we know the
coordinates of B are (−5, 2).
• We refer to the coordinates as an ordered pair, because the order in which the numbers appear makes
a dierence. Notice that the point B has coordinates (−5, 2), but if we switch the ordering, we get
(2, −5) which are the coordinates of point D.
So, going back to our original problem of the window, if we set the origin be at the center bottom point of
the window, then label each point where the window makes a new angle and setup the tick marks as shown
below, we can nd the coordinates of those points.
A = (0, 0), B = (−5, 2), C = (−4, 6), D = (−2, 9), E = (2, 9), F = (4, 6), G = (5, 2). To cut the glass, these
coordinates would be given to the computer and the computer told to cut straight across from point A to
B to C to . . . It should be noted here that all of our discussion above was in 2 dimensions and everything
that we will do in this class will always be in two dimensions. Remember that in 2 dimensions every point
is identied by two coordinate values. If we wish to talk about points in 3 dimensions, then we would need
another axis and hence another coordinate, 4 dimensions would require 4 coordinate values, etc. . .
CHAPTER 5.
5.2
3
EQUATIONS OF LINES
Graphing Lines by Points
View the Video Tutorial for this section here.
Recall that a solution to an equation is the value that if you substitute in for the variable, it makes the
equation true, e.g. the equation x + 3 = 5 has x = 2 as the solution because if you replace x with 2, you get
(2) + 3 = 5 which is true. Now suppose we have an equation with two variables, such as
x+y =5
In this case, the solution to the equation must involve two numbers. One value for x and one for y . So there
are multiple combinations of numbers that will work, in fact there are an innite number of answers. For
example, if x = 5 and y = 0 or x = 1 and y = 4 or x = 0.5 and y = 4.5 or x = −1 and y = 6, etc. . . If we
write the solutions as ordered pairs, (5, 0), (1, 4), (0.5, 4.5), (−1, 6), and plotted all the possible solutions on
the Cartesian coordinate plane, all the points together would form a line.
Again, keep in mind that when we draw the line, we are saying that all the points along the line are solutions
to the equation. Students often think that this means just the points with integer coordinates, such as (0, 5)
and (1, 4) in the above example. However, you should understand that there are more points, on the line,
between (0, 5) and (1, 4) that there are points with just integer coordinate values all along the line.
In general, we have the following denition of the equation of lines.
Denition 5.2.1. Equation of Lines in Two Variables (General Form)
In two dimensions, an equation of the form
Ax + By + C = 0
where A, B, and C are real numbers and A and B are not both zeros, the solutions of
such an equation, (x, y), will graph as a line. Conversely, the equation of any line can
be written in this form.
Example 5.2.1. Verify that the given equation represents a line by putting in the general form, Ax + By +
C = 0, and identifying A, B , and C . Then graph it by nding two points on the line.
1) 2x − 3y = 12
2) y = 32 x − 1
3) 2x = 8
Solutions:
1)
2x − 3y
=
−12
2x + (−3)y + (−12)
=
12
−12
0
So A = 2, B = −3, C = −12 and from Denition 5.2.1, we know that 2x − 3y = 12 must graph as
a line. From geometry, we know that two points dene a line. Therefore, if we can nd two points
or solutions of the equation 2x − 3y = 12, then we can graph the line by connecting the two points.
Note that we can use whatever points we want, as long as they are solutions of the equation. So the
CHAPTER 5.
4
EQUATIONS OF LINES
question is, how do we nd two points? You could just try random guessing for the x and y values,
but that would not be very ecient. Instead, we note that since the points must be solutions of the
given equation, if we were to let x be some value, say x = 0, then plugging this into the equation would
allow us to solve for a y value.
2(0) − 3y
=
12
−3y
=
12
y
= −4
This tells us that the point (0,-4) is a solution of the given equation. Now to nd the second point, we
could use x = 1, but note that if you do, the y values will be a fraction y = − 10
3 . Although there
is nothing wrong with this, the fraction will be a little harder to graph. So, instead of using x = 1,
what if we let y = 0? Remember that our only condition is that the x and y values satisfy the given
equation. There is nothing that says that we have to set the x value rst. So, if y = 0, then
2x − 3(0)
=
12
2x =
12
x =
6
We now have two points, (0,-4) and (6,0). So if we graph them, the line that goes through them will
be the line representing the equation 2x − 3y = 12.
When graphing, be sure to label the axes and the tick marks at the minimum.
2) The equation y = 23 x − 1, will also graph as a line. We know this because
y
− 23 x+1
2
− x+y+1
3
2
x− 1
+1
32
=
−3x
=
0
So A = − 23 , B = 1, C = 1, which according to Denition 5.2.1 graphs as a line. Since the equation is already solved for y , this makes it easier for us to pick the x values and then solve for the y values. Again,
the choice of x values to use is up to us. So we want to try to use numbers that will make the calculations
easier. Also, this form of the equation makes it easier to use what we call a T chart to list the solutions.
x
0
3
So the line is
y or 23 x − 1
2
3 (0)
2
3 (3)
− 1 = −1
−1=2−1=1
(x,y)
(0,-1)
(3,1)
CHAPTER 5.
5
EQUATIONS OF LINES
3) Since this equation has no y in it, we can view the y term as being 0y , as 0y is really just zero and
adding zero does not change anything.
2x + 0y
=
−8
2x + 0y + (−8)
=
8
−8
0
So A = 2, B = 0, C = −8. We note here that this in not the most helpful way to look at the equation.
To better understand how this particular equation graphs, it is more helpful to solve this equation for
x,
1
1
(2x) =
(8)
2
2
x = 4
Now this is where it is important for us to remember that we are working in 2-dimensions. That means
every point must have both an x and a y value. Since this equation only species that the x-coordinate
value must always be 4 and says nothing about the y values, we can use whatever y value we want,
as long as we use 4 for the x value. For example the points (4,-2) and (4,2) are both solutions to the
equation. If we plot all the points that have an x-coordinate of 4, we end up with the following vertical
line.
CHAPTER 5.
5.3
6
EQUATIONS OF LINES
Slope of a Line
View the Video Tutorial for this section here.
One of the characteristic features of lines is the concept called the slope of a line. Most of us already
have an intuitive idea of what a slope is. For example, if you go skying at a mountain resort, there is the
Bunny Slope (gentle incline) for beginners and the Widow Maker (steep incline) for the more advance skier
and/or crazy person. In order for the mathematical denition of slope to match our intuitive idea of slope,
we dene it as follows.
Denition 5.3.1. Slope of a line.
Let A and B be two points on the line. Then the slope of a line, denoted with the
letter m, is dened as the following ratio.
m=
vertical distance from point A to B (or from B to A)
horizontal distance from point A to B (or from B to A)
• some other ways this is dened is
m=
rise change in y
=
run change in x
• positive value means go up or to the right and a negative value means go down
or to the left.
• Slope Formula: given two points (x1 , y1 ) and (x2 , y2 ), the formula to calculate
slope is
m=
y2 − y1
x2 − x1
There are two intuitive concepts about slopes that we want to be sure that our denition above captures.
1) Slope of the steeper line should be greater than the slope of a less steep line.
2) The slope of a line should be constant and so it should not matter what two points you use to nd the
slope and it should also not matter in which direction you go between the two chosen points.
We use the following graph of two lines to explore these two concepts.
Two points on the widow maker are (2, 12) and (4, 24). To calculate the slope we can just count o how
much we need to up and to the right to go from (2, 12) to (4, 24).
m
=
m
=
m
=
go up 12
go right 2
12
2
6
CHAPTER 5.
7
EQUATIONS OF LINES
x1
y1
x2
y2
z}|{ z}|{
z}|{ z}|{
or we can also use the formula for the slope with ( 2 , 12 ) and ( 4 , 24 )
y2 − y1
x2 − x1
24 − 12
=
4−2
12
=
2
= 6
m
=
In either case, we see that the slope for the widow maker is 6.
For the bunny slope, we know that it goes through the points (0, 0) and (8, 4). So, from the origin to
(8, 4), we go up 4 and to the right 8.
m
=
=
4
8
1
2
Notice that the bunny slope has a slope of 12 which is smaller than the slope of the Widow Maker, m = 6.
This coincides with our notion of steepness of the two lines.
Now, to see that the order in which we use the points does not matter when calculating the slope, note
that we went from the point (2, 12) to (4, 24) when calculating the slope of the Widow Maker. If we go in
the reverse direction, from the point (4, 24) to (2, 12), we get
m
go down 12
go left 2
=
−12
−2
6
=
=
x1
y1
x2
y2
z}|{ z}|{
z}|{ z}|{
or using the formula with ( 4 , 24 ) and ( 2 , 12 )
m
y2 − y1
x2 − x1
12 − 24
=
2−4
−12
=
−2
= 6
=
So the slope is still 6. See the blue lines in the graph below.
To see that the choice of points does not matter, we calculate the slope going from the point (2, 12) to the
CHAPTER 5.
8
EQUATIONS OF LINES
new point (6, 36), see purple lines in the gure above.
m
=
=
=
go up 24
go right 4
24
4
6
So, we see that it does not matter in what direction or what two points you use to calculate the slope of the
line, as long as both points are on the line.
Example 5.3.1. Graph the line, given a point on the line and its slope. You must graph the line with in
the provided grid.
1) m = − 43 , (−3, 2)
2) m = 65 , (5, 4)
Solutions:
1) One way to view slope is as the directions for going from one point on the line to another where the
numerator tells you how far to go up or down and the denominator tells you how far to go right or
left. The slope here is
m=−
−3
3
3
=
=
4
4
−4
If we view it as −3
4 , then this means we go down by 3 and to the right 4, where as if we the slope as
3
,
we
go
up
by
3 and to the left by 4. So, using m = −3
−4
4 , and starting from the point (−3, 2), we
have the following.
2) The problem here is that when we start from the point (5, 4), we do not have enough space to go up
by 6 and across the right by 5. Therefore, we will view the slope as the ratio of -6 to -5 instead of 6 to
5.
m=
6
−6
=
5
−5
CHAPTER 5.
9
EQUATIONS OF LINES
Which means that if we go down by 6 then we must go left by 5.
Example 5.3.2. Find the slope of the line given the following information.
1) The line goes through the points (5, −3) and
(−5, −7).
3) The line goes through the points (−4, −9) and
(−6, −9).
2) The equation of the line is 2x − 3y = 12.
4) The equation of the line is x = −3
Solutions:
1) Given two points, we can use the slope formula in Denition 5.3.1. Note that it does not matter which
we consider the rst and the second point. So we will let (5, −3) be the rst point and (−5, −7) be the
second point.
y1
x1
x2
(5,−3)
m
=
=
=
=
y2
(− 5 , − 7 )
−7 − (−3)
−5 − 5
−7 + 3
−10
−4
−10
2
5
So the slope is m = 25 .
2) Given the equation, one way to nd the slope is to nd two points on the line. Note that we will later
learn another way to nd the slope, given the equation. This is the same line (equation) in Example
5.2.1, from which we know that the points (0, −4) and(6, 0) are on the line. Using these two points,
we have
m
=
=
=
−4 − 0
0−6
−4
−6
2
3
where we let (6,0) be the rst point and (0,-4) be the second point. So the slope is m = 23 .
3) Here we are again given two points. We let (-4,-9) be the rst point and (-6,-9) be the second point.
m
−9 − (−9)
−6 − (−4)
−9 + 9
=
−6 + 4
0
=
−2
= 0
=
CHAPTER 5.
10
EQUATIONS OF LINES
So the slope is m = 0. Note that if you use the two points to graph the line, it is a horizontal line.
This makes sense since the slope measures the incline of a line and a horizontal line is at, i.e. it has
no incline.
4) Given the equation of the line, x = −3, we need to nd two points on the line. Since this equation has
no restriction on y , we can use whatever value we want for the y-coordinate. So (-3,0) and (-3,2) are
two solutions to the equation and hence points on the line. Using the slope formula
m =
=
=
2−0
−3 − (−3)
2
−3 + 3
2
0
Recall that a fraction with zero in the denominator is undened. This means that the slope of the line
is undened. Note that x = −3 graphs as a vertical line. In general, a vertical line has an undened
slope. Some may think that since slope measures steepness and a vertical line is innitely steep, the
slope of the vertical line must be innite. The problem with this reasoning is that slope not only
indicates steepness, but also direction. If you take a horizontal line, which has slope zero and rotate it
counter-clockwise, the slope of the line becomes a greater and greater positive number (see gure below.)
Now, if you take the same horizontal line and rotate it clock-wise, the lines become more and more
vertical, but the slopes get more negative (see gure below.)
So, rotating counter-clockwise, it looks like the slope of a vertical line should be positive innity, but
if you rotate clockwise, it looks like the slope of a vertical line should be negative innity. Since the
slope cannot be both positive and negative at the same time, we say that the slope of a vertical line is
undened.
Comment on negative slope and steepness
From Example 5.3.1, we can see that a positive slope means that as you look at the line from left to right,
the line is increasing (going up.) Negative slope means, as you look from left to right, the line is decreasing
(going down.) Also, if you have a sharp eye, you may have noticed that for negative sloped lines, the steeper
you decrease, the more negative the slope (the value of the slope is decreasing.) This seems to go against
our intuitive idea that steeper means greater slope. Remember that -7 is smaller than -1. The problem is
that when we think of steepness, we are only thinking of the incline and not the direction in which the line
inclines. However, slope takes into account both the incline and direction, positive means an increasing line
and a negative slope means a decreasing line. To ignore the direction and just talk about the steepness, we
look at the absolute value of the slope, |m|. For example, a line with slope m = −7 versus a line with slope
m = −1. Since | − 7| = 7 which is bigger than | − 1| = 1, would tell us that the line with a slope of −7 is
steeper than the line with a slope of −1.
CHAPTER 5.
EQUATIONS OF LINES
11
CHAPTER 5.
5.4
12
EQUATIONS OF LINES
Equations of Lines
In sec. 5.2, we discussed the relationship between an equation and the graph that represents the equation.
Denition 5.2.1 told us that any line can be written in the form Ax + By = C . In this section, we will
discuss two other ways of writing the equation of a line that are more useful when you are trying to nd the
equation of a line.
5.4.1
X and Y Intercepts
View the Video Tutorial for this section here.
Before talking about the other ways to write the equation of a line, we mention here two special types
of points called x and y intercepts. In general, the graph of an equation can have more than one x and y
intercepts. But for lines, there will be at most one x-intercept and one y-intercept.
The x-intercept is where the graph intersects the x-axis. Recall that for any point on the x-axis, the
y-coordinate is 0. Therefore, to nd the x-intercept on a line, we would set y = 0 and solve for x. Likewise,
the y-intercept is where the graph intersects the y-axis. So in order to nd the y-intercept, we set x = 0 and
solve for y.
Denition 5.4.1. X and Y Intercepts
Given any equation and its graph (does not have to be a line),
in terms of the graph
• The x-intercept(s) is a point(s) where the graph crosses the x-axis.
• The y-intercept(s) is a point(s) where the graph crosses the y-axis.
in terms of the equation
• Find the x-intercept(s) by setting y = 0 and solving for x.
• Find the y-intercept(s) by setting x = 0 and solving for y .
Note that for lines, there will only be one x-intercept and one y-intercept.
Example 5.4.1. Find the intercepts of the following lines.
1) x − 3y = 6
2) y = 2x
3) y = −2
Solutions:
1) To nd the x-intercept, we set y = 0.
x − 3(0)
=
6
x =
6
So the x-intercept is (6, 0). Note that since the intercepts are points, you must give the ordered pair
as your answer.
To nd the y-intercept, we set x = 0.
(0) − 3y
=
6
−3y
=
6
y
=
y
=
6
−3
−2
So the y-intercept is (0, −2). Note that you were not asked to graph this. The graph is shown to help
visualize what we have done.
CHAPTER 5.
13
EQUATIONS OF LINES
2) x-intercept:
(0) = 2x
0
= x
2
0 = x
So we get (0, 0) as the x-intercept. Note that this must also be the y-intercept as the x-coordinate is
also 0. The origin, (0, 0), is the only point on the Cartesian Coordinate system that can be both the
x and y intercept.
3) y = −2 means that the y-value must always be -2 but that the x-value can be anything. So, since the
y-value can never be 0, there can be no x-intercept. However, to nd the y-intercept, we set x = 0 but
the y value is still -2. So the point (0, −2) is the y-intercept. Note that the line y = −2 will graph as
a horizontal line.
CHAPTER 5.
5.4.2
14
EQUATIONS OF LINES
Slope-Intercept form of the Line
View the Video Tutorial for this section here.
Finding the equations of lines is a key topic in math because of its many uses in real life applications.
To nd the equation of a line will always require two things.
1) The slope of the line.
2) A point on the line.
Given these two things, we can nd the equation of the line in the form called the slope-intercept form of
the line.
Denition 5.4.2. Slope-Intercept form of the Line
y = mx + b
is called the slope-intercept form of the line, and any line (other than the vertical line) can be written in this
form. This is called the slope-intercept form because the slope of the line and the y-intercept can be read
directly from the equation.
• m is the slope of the line
• (0, b) is the y-intercept of the line.
Example 5.4.2. Show that in the slope intercept form of the line, y = mx + b, m is the slope of the line.
Solution:
To calculate the slope using the slope formula, Denition 5.3.1, we need two points. One of the points
we will use will be the y-intercept. Recall that to nd the y-intercept, you set x = 0 and solve for y .
y
=
m(0) + b
y
=
b
So, we can see that (0, b) is the y-intercept. To see that the number m in front of x must represent
slope, we note that since y = mx + b this means that we can replace y with mx + b. So any point (x, y)
y
z }| {
on this line can be written as (x, mx + b). Since we also know that (0, b) is a point on the line, we have
our two points, (x, mx + b) and (0, b), and can use the slope formula, Denition 5.3.1, to calculate the
slope.
(mx + b) − b
x−0
mx
=
x
= m
slope =
We have just shown that the line y = mx + b has m as its slope.
Example 5.4.3. Find the equation of the line given the following information.
1) The line goes through the point (2, 1) with a
slope of 32 .
3)
2) Goes through the points (−1, 2) and (9, −3).
Solutions:
CHAPTER 5.
15
EQUATIONS OF LINES
1) Using the slope-intercept form of the line, y = mx + b, we have found the line if we can nd the values
for m and b. From Denition 5.4.2, we know that m is the slope of the line. So here m = 32 . That
means our equation so far is
y=
2
x+b
3
Now to nd b, we use that fact that (2, 1) is a point on the line. This means that (2, 1) is a solution
of the above equation, i.e. if we replace x with 2 and y with 1, the above equation must be true.
(1)
=
1
=
− 34
2
(2) + b
3
4
+b
34
−3
4
3
3 4
−
3 3
1
−
3
1−
So the equation of the line is
y=
=
b
=
b
=
b
1
2
x−
3
3
2) As stated above, to nd the equation of a line, we always need the slope of the line and a point on the
line. Here, we are given two points instead of the slope and a point. However, given two points we can
calculate the slope using the slope formula, given in Denition 5.3.1. So given the two points (−1, 2)
and (9, −3), we have
−3 − 2
9 − (−1)
−5
=
10
1
= −
2
m
and our line so far is
=
1
y =− x+b
2
To nd b, we can use either of the two given points. We will use the point (−1, 2) just because the
numbers look to be easier to deal with. Again, as in the previous problem, since (−1, 2) is a solution
of the equation, we can replace the x and y with -1 and 1.
(2)
2
− 21
1
2
4 1
−
2 2
3
2
2−
So the line is
1
= − (−1) + b
2
1
=
+b
21
−2
= b
= b
= b
1
3
y =− x+
2
2
3) Since we are just given the graph of the line, we must make some reasonable assumptions. Namely
that we can read o the coordinates of the following two points, A and B, from the graph as being
integer coordinates.
CHAPTER 5.
16
EQUATIONS OF LINES
So A = (−2, 0) and B = (1, 1). Using the slope formula, we get as the slope
m
=
=
1−0
1 − (−2)
1
3
Note that we can also just read the slope o the graph. However, we need to be careful when reading
the slope from the graph since in this case the vertical tick marks (grids) represent 21 units while the
horizontal tick marks represent 1 unit. So, although to go from point A to B, you go up 2 grid lines,
that only represents 1 unit, while going 3 grids to the right represent 3 units. So the slope is again
m = 13 .
The line that we want is then
y=
1
x+b
3
and using point A = (−2, 0), we have
(0)
=
0
=
+ 23
2
3
+3
=
The equation of the line is then
y=
5.4.3
Point-Slope form of the Line
View the Video Tutorial for this section here.
1
(−2) + b
3
2
− +b
3
2
b
1
2
x+
3
3
CHAPTER 5.
17
EQUATIONS OF LINES
As state before, to nd the equation of a line, you need the slope and a point on the line. In the previous
subsection, we found the equation of the line using the slope-intercept form, Denition 5.4.2. Here, we will
discuss another way to write the equation of a line, called the Point-Slope form or formula.
We begin by showing how the Point-Slope form comes from the slope formula, Denition 5.3.1. Recall
−y1
, lets you nd the slope of the line using any two points on the line. So
that the slope formula, m = xy22 −x
1
given the slope, m, and a point (x1 , y1 ) on the line, if we let (x, y) be any other point on the line, then, using
the slope formula, we have
y − y1
m=
x − x1
(x − x1 )m =
y − y1
· (x − x1 )
x − x1
Note that (x, y) is not just any arbitrary point, but an arbitrary
point on the line. So, since both (x, y) and (x1 , y1 ) are points on
the same line, the dierence in the y's over the dierence in the
x's must be a constant, which we call the slope of the line.
Get rid of the denominator on the right side by multiplying
both sides by (x − x1 )
m(x − x1 ) = y − y1
y − y1 = m(x − x1 )
So for any point (x, y) on the line, the above equation must be true. Another way to view it is that any
solution (x, y) to the equation is a point on the line. Hence, the above is the equation of the line with slope
m and going through the point (x1 , y1 ).
Denition 5.4.3. Point-Slope form of the Line
Given the slope m and a point (x1 , y1 ) on the line, the equation of the line is
y − y1 = m(x − x1 )
Note that like the slope-intercept form of the line, the point-slope form can represent
any line except the vertical line.
Example 5.4.4. Find the equation of the line given the following information.
2) Goes through the points (−1, 2) and (9, −3).
1) The line goes through the point (2, 1) with a
slope of 32 .
Solutions:
1) We have already solved this problem using the slope-intercept form of the line. Here we will use the
point-slope form of the line. We are given m = 32 , so
y − (1)
=
y−1
=
2
(x − (2))
3
2
(x − 2)
3
Note that we can put this in slope-intercept form by solving for y .
y−1
=
y− 1
=
+1
2
(x − 2)
3
2
4
x−
3
3
+1
y
=
2
1
x−
3
3
CHAPTER 5.
18
EQUATIONS OF LINES
This is the same as in Example 5.4.3.
2) Since we are given two points, we can nd the slope using the slope formula.
m
−3 − 2
9 − (−1)
−5
=
10
1
= −
2
=
Using the point (−1, 2) (note it does not matte which point is used, as they are both on the same line),
y−2
=
y−2
=
1
− (x − (−1))
2
1
− (x + 1)
2
$
'
Summary of the Equation of Lines
We have discussed three dierent ways of writing the equation of a line. Each has its usefulness in dierent
situations.
• General Form:
Ax + By + C = 0
Any line can be represented in general form, including the vertical line.
Useful form when using the elimination method to solve systems of equations (discussed in the
next chapter.)
Form is not unique, meaning dierent equations can represent the same line. For example
x + y + 1 = 0 and 2x + 2y + 2 = 0 both represent the same line.
• Slope-Intercept Form:
y = mx + b
Every line except the vertical line can be represented in this form.
Can identify the slope and y-intercept immediately, which lets us graph the line easily.
This form is unique, meaning if two equations in this form dier in either m or b, then they
represent two dierent lines.
• Point_Slope Form:
y − y1 = m(x − x1 )
Every line except the vertical line can be represented in this form.
Quickest way to write the equation of a line.
This form is not unique, meaning two dierent equations in this form can represent the same line.
(All you have to do is use a dierent point on the line for (x1 , y1 ).)
&
%
CHAPTER 5.
5.5
EQUATIONS OF LINES
19
Parallel and Perpendicular Lines
View the Video Tutorial for this section here.
Geometrically, we say that two lines are parallel if they never intersect each other. Algebraically, we say
that two lines are parallel if they have the same slope. This makes sense since the slope indicates incline and
direction and you would expect two parallel lines to be going in the same direction with the same incline,
see graph below.
Perpendicular lines are lines that intersect each other at 90 degrees. Algebraically, we say that two lines
are perpendicular if the product of the slopes equals -1 or another way of saying this is that the slopes are
negative reciprocals of each other. For example, if the slope of one line is 23 , then the slope of the perpendicular line is − 32 .
So to summarize, we have
Proposition 5.5.1.
Slope of parallel and perpendicular lines.
• Parallel lines have the same slope.
• Perpendicular lines
The product of their slopes equals -1. (This is useful when you know the
slopes and want to check to see if the lines are perpendicular.)
The slopes are negative reciprocals of each other, i.e. if the slope of one of
the lines is ab , then the slope of the other line is − ab . (This is useful when
you want to nd the slope of the perpendicular line.)
Horizontal and vertical lines are perpendicular to each other.
Before going on to some examples, we will rst show that the above two properties of perpendicular lines
are equivalent. To show this, we start with the property that the product of the slopes to two perpendicular
lines equals -1. If we let the slope of the two perpendicular lines be m1 and m2 , then
m1 · m2
1
m1 (m1 · m2 )
m2 =
= −1
= m11 (−1)
− m11
Solve for m2 .
This says that the slope of one line is the negative reciprocal of
the other.
CHAPTER 5.
20
EQUATIONS OF LINES
To make the above look exactly like Proposition 5.5.1, use ab for m1 and solve for m2 .
Example 5.5.1. Find the equation of the line described below and write it in slope-intercept form, if
possible.
1) The line goes through the point (-2,3) and is
parallel to the line 3x − 4y = 5
and intersecting that line at the y-intercept.
3) The line going through the point (-3,-4) that is
perpendicular to the line y = 5.
2) The line perpendicular to the line 5x + 3y = 9
Solutions:
1) Since the line we want is parallel to the line 3x − 4y = 5, that means that they have the same slope.
To nd the slope, we rewrite the given equation in slope-intercept form, i.e. we solve for y.
3x − 4y
−3x
=
5
−3x
−4y = −3x + 5
1
1
− (−4y) = − (−3x + 5)
4
4
3
5
y =
x−
4
4
So the slope of the line we want is m = 34 . Now that we have the slope and a point (-2,3) we can nd
the line by using either the point-slope form (Denition 5.4.3) or the slope-intercept form (Denition
5.4.2.) We will use the slope-intercept form since the instructions are to put the line in this form at
the end.
y = 43 x + b
3 = 34 (−2) + b
33 = − 23 + b
+2
Since we found the slope to be 43 .
To nd b, we plug in the point (-2,3) and solve for b.
+ 23
9
2
=b
So the line is
y=
3
9
x+
4
2
2) The rst thing we want to do is rewrite the given equation in slope-intercept form so that we can read
o the slope.
5x + 3y
=
3y
=
y
=
y
=
−5x
9
−5x
−5x + 9
1
(−5x + 9)
3
5
− x+3
3
So the slope of this line is − 35 and its y-intercept is (0, 3). Since the line we want is perpendicular
to this line, that means that the slope of the line that we want must be the negative reciprocal,
m = − − 35 = 53 . So the line we want is
y=
3
x+b
5
To nd b, we note that this line is to have the same y-intercept as the given line, which is (0, 3). In
other words, b is still 3. So the line that we want is
y=
3
x+3
5
3) The line y = 5 is a horizontal line which means that the any vertical line will be perpendicular to it.
Since we get a vertical line by xing the x value and we want the vertical line to go through the point
(-3,-4), the line that we want is
x = −3
CHAPTER 5.
21
EQUATIONS OF LINES
Example 5.5.2. Determine if the following lines are perpendicular.
Solution:
We assume that the points A, B, C, and D are on the two lines and all fall on the grid lines and hence
have the following integer coordinates.
A = (−4, −3), B = (1, 4), C = (−3, 4), D = (1, −1)
Using points A and B, the slope of Line 1 is
m1
=
=
4 − (−3)
1 − (−4)
7
5
Using point C and D, the slope of Line 2 is
m2
=
=
−1 − 4
1 − (−3)
−5
4
Since the product of the slopes is not -1
m1 · m2
the lines are not perpendicular.
=
7
5
7
· −
= − 6= −1
5
4
4
CHAPTER 5.
5.6
22
EQUATIONS OF LINES
Applications of Lines
In studying lines, we have seen that the equation of lines involves a relationship between two variables. To
understand this relationship, we use the slope formula as follows. Let (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) . . . be points
on the same line where the x-coordinates are all equally spaced (where we call the equal spacing h), i.e.
x2 − x1 = x3 − x2 = · · · = h
Then using the slope formula, we have the following:
Using the slope formula with points (x1 , y1 ), (x2 , y2 ).
m=
y2 − y1
x2 − x1
m(x2 − x1 ) = y2 − y1
m · h = y2 − y1
m=
y3 − y2
x3 − x2
m(x3 − x2 ) = y3 − y2
m · h = y3 − y2
y3 − y2 = y2 − y1
Clear the fraction by multiplying both sides by x2 − x1 .
Since we picked our x-values so that x2 − x1 = h and so we can
replace x2 − x1 with h.
Now we calculate the slope using points (x2 , y2 ), (x3 , y3 ). Note
that it will still lead to the same slope value m.
Clear the fractions as before by multiplying both sides by
x3 − x2 .
Since they both equal m · h.
What we have just shown is that if the dierence in the x-values are constant then the dierence in the
y-values are also constant. This means that if the relationship between two quantities is linear, then as one
increases by a constant amount the other also increases (or decreases) by a constant amount.
So being told that the relationship between two quantities is linear implies two things.
• As one quantity increases by a constant amount, the other also increases (or decreases) by a constant
amount.
• The formula connecting the two quantities is the equation of a line.
Example 5.6.1. If a car is traveling at a constant speed of 50 mph, then the relationship between the
distance traveled and the time spent traveling is linear. Find the equation that relates the distance to time
and calculate how long it took to travel 413 miles.
Solution:
We let the x-coordinate be time,t, and the y-coordinate be the distance, d. Since the relationship between
the distance and time is linear, this means that if we were to plot (t, d), the points would graph as a line. To
write the equation of a line, we always need two things, the slope of the line and a point on the line. Now
at time 0, since the car has not yet moved, the distance is 0, and so we have our rst point, (0,0). After
one hour, t = 1, the car will have traveled 50 miles and so we get our second point (1,50). Using these two
points, we can calculate the slope.
m
=
=
50 − 0
1−0
50
and since (0, 0) is the y-intercept, writing the equation in slope-intercept form, we have
y
= mx + b
d =
50t + 0
d =
50t
CHAPTER 5.
23
EQUATIONS OF LINES
So the equation that relates distance to time when traveling at a constant speed of 50 mph, is d = 50t. To
calculate how long it took to travel 413 miles, we simply replace d with 413 and solve for t.
413 = 50t
413
= t
50
8.26 = t
So it took 8.26 hours to travel 413 miles at 50 mph.
Example 5.6.2. A 1000 gallon septic tank is being drained by a pump at a constant rate. It takes 45
minutes to remove 423 gallons. How long will it take to drain the tank completely? Round your answer to
two decimal places.
Solution:
If we let A, the amount of liquid in the septic tank, be the y-coordinate and t, the time the pump is on,
be the x-coordinate, then since the liquid is being drained at a constant rate, the relationship between A
and t is linear. In other words, the points (t, A) would graph as a line. We are given two points on this line.
At time 0, the tank is full, so (0,1000) is one point on the line. After 45 minutes, 423 gallons are removed,
which means that 1000 − 423 = 577 gallons remain in the tank. So the second point is (45,577). Given the
two points, we can nd the equation of the line that relates A and t. First we nd the slope using the two
points.
577 − 1000
45 − 0
423
= −
45
m
=
The point (0,1000) is the y-intercept, so if we write the equation of the line in slope-intercept form, we get
y
=
mx + b
423
t + 1000
A = −
45
To calculate the time it takes to empty the tank, we set A = 0 and solve for t.
0
=
−1000
=
−1000
−
45
(−1000) =
423
106.38 ≈
423
t +1000
45 −1000
423
−
t
45 45
423
−
−
t
423
45
t
−
So it takes approximately 106.38 minutes to drain the tank completely.
Example 5.6.3. The relationship between temperature measured in degrees Fahrenheit and degrees Celsius
is linear. Also, water freezes at 32◦ Fahrenheit or 0◦ Celsius and boils at 212◦ Fahrenheit or 100◦ Celsius at
sea level. Find the formula for converting degrees Celsius to degrees Fahrenheit.
Solution:
Since we want to convert Celsius into Fahrenheit, we will let C , degrees in Celsius, be the x-coordinate
and F , degrees in Fahrenheit, be the y-coordinate. Then we know that the points (C, F ), will graph as a line
since we are told that the relationship between them is linear. The information about when water freezes
and boils gives us two points on this line, (0, 32) and (100, 212). So the slope of the line is
m
=
=
=
212 − 32
100 − 0
180
100
9
5
Since (0, 32) is the y-intercept, the equation in slope-intercept form is
y
F
=
mx + b
9
=
C + 32
5
CHAPTER 5.
5.7
24
EQUATIONS OF LINES
Linear Inequalities in Two Variables
View the Video Tutorial for this section here.
Linear inequalities in two variables is of a similar form to the equations of lines with the dierence being
that the equal sign is replaced by an inequality sign. So in general, a linear inequality in two variables can
be written in the form
<
≤
Ax + By + C
0
>
≥
As with the linear inequalities in one variable, the answers to these inequalities cannot be written down
explicitly and so we graph the solutions. There are two techniques to graph the solutions to these inequalities.
One involves an understanding of what the equation and inequalities are saying on a graph and the other
technique is a more straight forward plugging in values and calculating method. Both are valid techniques
and both have their usefulness. The student is encouraged to try to understand both techniques, but the
method you need to learn will depend on your instructor and/or you.
Example 5.7.1. Graph the solutions of the following inequalities.
1) 2x + 3y − 1 > 0
2) 2x − 3y + 1 > 0
3) −4x ≤ 2y + 6
4) x + 3 > 0
Solutions:
1) Method 1: First solve the inequality for y .
2x + 3y−1
−2x
>
+1
0
−2x+1
1
(−2x + 1)
3
2
1
> − x+
3
3
1
(3y) >
3
y
Now we graph the line y = − 23 x + 13 and note that for any point, (x, y), on the line, y and − 32 x + 13
will always be equal to each other. For example, point A = (−1, 1) is on the line and so if we use these
x and y values, we get
y
1
1
− 32 x + 31
− 23 (−1) +
2
1
3 + 3
1
3
1
so y , which in this case is 1 is equal to − 32 x + 13 , for x = −1. Now if we draw a vertical line, say at
x = −1, any point on this line will have the same x value.
This means that for points A, B and C in the above graph, − 23 x + 13 , will be the same value for all
three points. In the case of point A, the y-coordinate of A will be equal to this value. However, for
CHAPTER 5.
25
EQUATIONS OF LINES
point B, the y coordinate will be greater than that of A and hence greater than − 23 x + 13 , while for
point C, the y coordinate will be less than that of A and hence less than − 32 x + 13 .
For B=(-1,2)
y
2
2
2>1
− 23 x + 13
− 23 (−1) + 13
1
or y > − 23 x +
Since we want
For C=(-1,-2)
y
−2
−2
−2 < 1
1
3
− 23 x + 13
− 32 (−1) + 13
1
or y < − 23 x +
1
3
1
2
y >− x+
3
3
We want to include all the points that are above the line, such as point B. So we shade the region
above the line. Note that we use a dashed line to indicate that we do not include any of the points on
the line y = − 23 x + 13 . So the graph of the solution looks like the following.
Method 2: The idea of method 2 is that once we have graphed the line, the line separates the plane
into two pieces and the solution to the inequality must be one of the pieces. To gure out which one,
we pick a test point that we know for certain is on one side of the line and plug it into the inequality. If
the inequality is true, then we shade the side of the line that contains our test point. If the inequality
is false, then we shade the side that does not contain the test point. So if we use as our test point,
(-1,2), then we have
2x + 3y − 1
>
0
2(−1) + 3(2) − 1
>
0
−2 + 6 − 1
>
0
3
>
0
which is true and so we shade the side of the graph that contains this test point. This will give us the
same graph we got using method 1.
Some notes on the 2 methods:
• For method 2 it is not necessary to solve the inequality for y rst. All that is required is that you
graph the related line in some way (nd two points, slope and a point, etc. . .)
• Although method 1 requires more calculations and is more conceptually involved, it is this con-
ceptual complexity that makes this method more useful in helping you to understand another
interpretation of what the equation of a line means. Also, the concept of what is above and below
a graph is an important one, and knowing how to identify them using a vertical line will be of use
later on in this class as well as other math classes.
2) We will focus on the use of method 1 and leave method 2 for students to try out. So the rst thing we
CHAPTER 5.
26
EQUATIONS OF LINES
need to do is to solve the inequality for y.
2x − 3y+1
−2x
>
−1
0
−2x−1
1
1
− (−3y) > − (−2x − 1)
3
3
2
1
y <
x+
3
3
Note that inequality sign changed direction since we multiplied both sides by a negative number, − 13 .
Now we graph the line y = 23 x + 31 and draw a vertical line at x = −2. Since point A is on the line, that
means its x and y values will make the two sides of the inequality equal to each other. Since we want y to
be less than 32 x+ 13 , that means we want to be below the line and in the region where point B is located.
3) First, we must solve for y.
−4x
−6
≤
2y + 6
−6
1
(−4x − 6) ≤ 12 ( 2y)
2
−2x − 3
≤
y
y
≥
−2x − 3
Now we graph the line y = −2x − 3 and we graph it as a solid line since the inequality says that y can
equal as well as be greater than −2x − 3. To determine the greater than part, we draw an arbitrary
vertical line, in this case we use x = 1.
Along the vertical line, x = 1, point A is above the line y = −2x−3 which means that y > −2x−3 when
x and y are replaced with the x and y coordinates of point A. So we shade the side that contains point A.
CHAPTER 5.
27
EQUATIONS OF LINES
Note that point B is below the line y = −2x − 3 and so if you use the coordinates of point B, you will
get y < −2x − 3.
4) Since this inequality only contains one variable, we will rst solve for it.
x+ 3
>
x
>
−3
0
−3
−3
When dealing with equations that have only one variable, we must always keep in mind what dimension we are working in. In one dimension the solution of x > −3 would graph on the number line
as we did in a previous chapter. However, in this chapter we are solving in two dimensions and so
there is an implied y coordinate value. We know that in two dimensions x = −3 graphs as a vertical line.
We use a dashed line since x cannot equal -3. Here, we cannot use the vertical line method we used in
the other problems. However, if we step back for a moment and think about what this problem wants,
it is fairly straightforward to see what the answer must be. We want to shade the region where the x
coordinate values are greater than -3. That means we want to shade all the points on the right side of
the vertical line x = −3.