Equations of Lines 1
Section 2.1
2.1 Equations of Lines
The Slope-Intercept Form
Recall the formula for the slope of a line. Let’s assume that the dependent variable
is y and the independent variable is x and we have a line passing through the points
P (x1 , y1 ) and Q(x2 , y2 ), as shown in Figure 1.
y
Q(x2 ,y2 )
∆y=y2 −y1
x
P (x1 ,y1 )
∆x=x2 −x1
Figure 1. Determining the slope of a line
through two points.
Note that the change in x is ∆x = x2 − x1 and the change in y is ∆y = y2 − y1 . Thus,
the slope of the line is determined by the formula
Slope =
∆y
y2 − y1
.
=
∆x
x2 − x1
(2.1)
We can use the slope formula to find the equation of the line with slope m and y −
intercept given by (0, b). In doing so we arrive at the following.
The Slope-Intercept Form of a Line. If the line L intercepts the y-axis at the
point (0, b) and has slope m, then the equation of the line is
y = mx + b.
(2.2)
This form of the equation of a line is called the slope-intercept form. The
function defined by the equation
f (x) = mx + b
is called a linear function.
It is important to note two key facts about the slope-intercept form y = mx + b.
•
•
The coefficient of x (the m in y = mx + b) is the slope of the line.
The constant term (the b in y = mx + b) is the y-coordinate of the y-intercept (0, b).
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2
Chapter 2
Let’s review an example of how this form of the line is used.
I Example 1
What is the equation of the line having slope −2/3 and y-intercept at (0, 3)? Sketch
the line on graph paper.
The equation of the line is
y = mx + b.
(2.3)
We’re given that the slope m = −2/3 and we’re given that the line intercepts the y-axis
at the point (0, 3). Substitute m = −2/3 and b = 3 in equation (2.3), obtaining
2
y = − x + 3.
3
To sketch the graph of the line, first locate the y-intercept at P (0, 3), as shown in
Figure 2. Starting from the y-intercept at P (0, 3), move 3 units to the right and 2
units downward to the point Q(3, 1). The required line passes through the points P
and Q.
Note that the line “intercepts” the y-axis at 3 and slants downhill, in accordance
with the fact that the slope is negative in this example.
5
y
∆x=3
P (0,3)
∆y=−2
Q(3,1)
5
x
Figure 2. The line has y-intercept at
(0, 3) and slope −2/3.
I Example 2
Given the graph of the line in Figure 3(a), determine the equation of the line.
First, locate the y-intercept of the line, which we’ve labeled P (0, −1) in Figure 3(b).
In the formula y = mx + b, recall that b represents the y-coordinate of the y-intercept.
Thus, b = −1.
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Equations of Lines 3
Section 2.1
Secondly, we need to determine the slope of the line. In Figure 3(b), start at the
point P , move 2 units to the right, then 3 units upward to the point Q(2, 2). This
makes the slope
3
∆y
= .
m=
∆x
2
Substitute m = 3/2 and b = −1 into the slope-intercept form y = mx + b to obtain
3
y = x − 1,
2
which is the desired equation of the line.
5
y
5
y
Q(2,2)
5
∆y=3
x
5
x
P (0,−1)
∆x=2
(a)
Figure 3.
(b)
Determining the equation of a line from its graph.
Horizontal and Vertical Lines
Recall the standard form of the line Ax + By = C.
The Standard Form of a Line. The graph of the equation Ax + By = C is a
line. The form
Ax + By = C
(2.4)
is called the standard form of a line.
The case where A and B are simultaneously equal to zero is not very interesting.
However, the following two cases are of interest.
1. If we let A = 0 and B 6= 0 in the standard form Ax + By = C, then By = C, or
equivalently y = C/B. Note that this has the form y = b, where b is some constant.
2. Similarly, if we let B = 0 and A 6= 0 in the standard form Ax + By = C, then
Ax = C, or equivalently, x = C/A. Note that this has the form x = a, where a is
some constant.
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Chapter 2
The lines having the form x = a and y = b are two of the easiest lines to plot. Let’s
review an example of each.
I Example 3
Sketch the graph of the equation x = 3.
The direction “sketch the graph of the equation x = 3” can be quite vexing unless
one remembers that a graph of an equation is the set of all points that satisfy the
equation. Thus, the direction is better posed if we say “sketch the set of all points
(x, y) that satisfy x = 3,” or equivalently, “sketch the set of all points (x, y) that have
an x-value of 3.” Then it is an easy matter to sketch the vertical line shown in Figure 4.
5
y
5
x
x=3
Figure 4. The graph
of the equation x = 3.
Note that each point on the line has an x-value equal to 3. Also, note that the slope
of this vertical line is undefined.
I Example 4
Sketch the graph of the equation y = 3.
This direction is better posed if we say “sketch the set of all points (x, y) that satisfy
y = 3,” or equivalently, “sketch the set of all points (x, y) that have a y-value of 3.”
Then it is an easy matter to sketch the horizontal line shown in Figure 5.
Note that each point on the line has a y-value equal to 3. Also, note that a horizontal line has slope zero.
The Point-Slope Form of a Line
In the last section, we developed the slope-intercept form of a line (y = mx + b). The
slope-intercept form of a line is applicable when you’re given the slope and y-intercept
of the line. However, there will be times when the y-intercept is unknown.
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Section 2.1
5
Equations of Lines 5
y
y=3
5
x
Figure 5. The graph
of the equation y = 3.
y
Q(x,y)
x
P (x0 ,y0 )
Figure 6.
slope m.
A line through (x1 , y1 ) with
Suppose for example, that you are asked to find the equation of a line that passes
through a particular point P (x1 , y1 ) with slope = m. This situation is pictured in
Figure 6.
Using the slope formula it can be shown that the line that results from this is given
by.
The Point-Slope Form of a Line. If line L passes through the point (x1 , y1 )
and has slope m, then the equation of the line is
y − y1 = m(x − x1 ).
(2.5)
This form of the equation of a line is called the point-slope form.
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Chapter 2
Let’s review an example of how to use this form of the line.
I Example 5
Draw the line that passes through the point P (−3, −2) and has slope m = 1/2. Use
the point-slope form to determine the equation of the line.
First, plot the point P (−3, −2), as shown in Figure 7(a). Starting from the point
P (−3, −2), move 2 units to the right and 1 unit up to the point Q(−1, −1). The line
through the points P and Q in Figure 7(a) now has slope m = 1/2.
y
y
x
x
Q(−1,−1)
R(0,−0.5)
∆y=1
P (−3,−2)
∆x=2
(a) The line through P (−3, −2)
with slope m = 1/2.
(b) Checking the y-intercept.
Figure 7.
To determine the equation of the line in Figure 7(a), we will use the point-slope
form of the line
y − y1 = m(x − x1 ).
(2.6)
The slope of the line is m = 1/2 and the given point is P (−3, −2), so (x1 , y1 ) =
(−3, −2). In equation (2.6), set m = 1/2, x1 = −3, and y1 = −2, obtaining
1
y − (−2) = (x − (−3)),
2
or equivalently,
y+2=
1
(x + 3).
2
(2.7)
This is the equation of the line in Figure 7(a).
Let’s look at another example.
I Example 6
Find the equation of the line passing through the points P (−3, 2) and Q(2, −1). Place
your final answer in slope-intercept form.
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Section 2.1
Equations of Lines 7
Use the slope formula to determine the slope of the line through the points P (−3, 2)
and Q(2, −1).
m=
−1 − 2
3
∆y
=
=−
∆x
2 − (−3)
5
We’ll use the point-slope form of the line
y − y1 = m(x − x1 ).
(2.8)
Let’s use point P (−3, 2) as the given point (x1 , y1 ). That is, (x1 , y1 ) = (−3, 2) (note
that it doesn’t matter which of the given points we chose). Substitute m = −3/5,
x1 = −3, and y1 = 2 in equation (2.8), obtaining
3
y − 2 = − (x − (−3)).
(2.9)
5
This is the equation of the line passing through the points P and Q in point-slope form.
We now need to rewrite it in slope-intercept form.
If we start with equation (2.9) and distribute the slope,
3
y − 2 = − (x − (−3))
5
3
9
y−2=− x− .
5
5
Now add 2 to both sides
3
y =− x−
5
3
y =− x+
5
9
+2
5
1
5
Parallel Lines
Recall that slope controls the “steepness” of a line. Consequently, if two lines are
parallel, they must have the same “steepness” or slope. Let’s look at an example of
parallel lines.
I Example 7
Find the equation of the line that passes through the point P (−2, 2) that is parallel to
the line passing through the points Q(−3, −1) and R(2, 1).
First, we draw the line through the points Q(−3, −1) and R(2, 1), then plot the
point P (−2, 2), as shown in Figure 8(a).
We can use the slope formula to calculate the slope of the line passing through the
points Q(−3, −1) and R(2, 1).
m=
∆y
1 − (−1)
2
=
=
∆x
2 − (−3)
5
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Chapter 2
y
y
T (3,4)
P (−2,2)
∆y=2
P (−2,2)
∆x=5
R(2,1)
R(2,1)
x
Q(−3,−1)
x
Q(−3,−1)
(b) The line through P (−2, 2)
that is parallel to the
line through Q and R.
(a) The line through
Q(−3, −1) and R(2, 1).
Figure 8.
We now draw a line through the point P (−2, 2) that is parallel to the line through
the points Q and R. Parallel lines must have the same slope, so we start at the point
P (−2, 2), “run” 5 units to the right, then “rise” 2 units up to the point T (3, 4), as
shown in Figure 8(b).
We seek the equation of the line through the points P and T . We’ll use the pointslope form of the line
y − y1 = m(x − x1 ).
(2.10)
We’ll use (x1 , y1 ) = (−2, 2) as the given point. The line through P has slope 2/5.
2
y − 2 = (x − (−2)).
5
(2.11)
Let’s place the equation (2.11) in slope-intercept form.
2
x+
5
2
y = x+
5
y−2=
4
5
14
5
or if we wished to have it in standard form,
2x − 5y = −14.
Perpendicular Lines
Suppose that two lines L1 and L2 have slopes m1 and m2 , respectively. Recall that if L1
and L2 are perpendicular, then the product of their slopes is m1 m2 = −1. Alternatively,
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Equations of Lines 9
Section 2.1
the slope of the first line is the negative reciprocal of the second line, and vice-versa;
i.e., m1 = −1/m2 and m2 = −1/m1 . Let’s look at an example of perpendicular lines.
I Example 8
Find the equation of the line passing through the point P (−4, −4) that is perpendicular
to the line 4x + 3y = 12.
Let’s first determine the slope of the line 4x + 3y = 12 by placing this equation in
slope-intercept form (i.e., solve the equation 4x + 3y = 12 for y).
4x + 3y = 12
3y = −4x + 12
4
y =− x+4
3
If two lines are perpendicular, then their slopes are negative reciprocals of one
another. Therefore, the slope of the line that is perpendicular to the line 4x + 3y = 12
(which has slope −4/3) is m = 3/4. Our second line must pass through the point
P (−4, −4).
Let’s use this information to draw the given line 4x +3y = 12 and the perpendicular
line we seek.
y
y
S(0,4)
R(3,0)
x
x
Q(0,−1)
∆y=3
m=−4/3
P (−4,−4)
∆x=4
(b) A line through P (−4, −4) that
is perpendicular to 4x + 3y = 12.
(a) Plot the x and
y-intercepts of 4x + 3y = 12.
Figure 9.
To determine the equation of the perpendicular line we seek (though the points P and
Q), we will use the point-slope form of the line, namely
y − y1 = m(x − x1 ).
(2.12)
The slope of the line we seek is m = 3/4. If we let (x1 , y1 ) = (−4, −4). Set m = 3/4,
x1 = −4, and y1 = −4 in equation (2.12), obtaining
y − (−4) =
3
(x − (−4)),
4
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Chapter 2
or equivalently,
y+4=
3
(x + 4).
4
(2.13)
Alternatively, we could use the slope-intercept form of the line. We solve equation (2.13)
for y,
3
y + 4 = (x + 4)
4
3
(2.14)
y+4= x+3
4
3
y = x − 1.
4
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