Download Equation of a Line

College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
2
Coordinates
and Graphs
2.4 Lines
Lines
In this section, we find equations
for straight lines lying in a coordinate
plane.
• The equations will depend on how the line
is inclined.
• So, we begin by discussing the concept
of slope.
The Slope of a Line
Slope of a Line
We first need a way to measure
the “steepness” of a line.
• This is how quickly it rises (or falls)
as we move from left to right.
Slope of a Line
We define ‘run’ to be the distance we move
to the right and ‘rise’ to be the corresponding
distance that the line rises (or falls).
The slope of a line is the ratio of rise to run:
rise
slope =
run
Slope of a Line
The figure shows situations where slope
is important.
• Carpenters use the term pitch for the slope of a roof
or a staircase.
• The term grade is used for the slope of a road.
Run & Rise
If a line lies in a coordinate plane, then:
• Run is the change in the x-coordinate between
any two points on the line.
• Rise is the corresponding change in the y-coordinate.
Slope of a Line—Definition
That gives us the following definition of slope.
The slope m of a nonvertical line that passes
through the points A(x1, y1) and B(x2, y2)
is:
rise y 2  y1
m

run x2  x1
• The slope of a vertical line is not defined.
Slope of a Line
The slope is independent
of which two points are chosen
on the line.
Slope of a Line
We can see that this is true from the similar
triangles in the figure:
y 2  y1 y 2  y1

x2  x1 x2  x1
Slope of a Line
The figure shows several lines labeled
with their slopes.
Positive & Negative Slopes
Notice that:
• Lines with
positive slope
slant upward
to the right.
• Lines with
negative slope
slant downward
to the right.
Steep & Horizontal Lines
The steepest lines are those for which
the absolute value of the slope is the largest.
A horizontal line
has slope zero.
E.g. 1—Finding Slope of a Line Through Two Points
Find the slope of the line that passes
through the points
P(2, 1) and Q(8, 5)
• Since any two different points determine a line,
only one line passes through these two points.
E.g. 1—Finding Slope of a Line Through Two Points
From the definition, the slope is:
y 2  y1 5  1 4 2
m

 
x2  x1 8  2 6 3
• This says that, for every 3 units we move
to the right, the line rises 2 units.
E.g. 1—Finding Slope of a Line Through Two Points
The line is drawn here.
Point-Slope Form of the
Equations of a Line
Equations of Lines
Now let’s find the equation of the line
that passes through a given point
P(x1, y1) and has slope m.
Equations of Lines
A point P(x, y) with x ≠ x1 lies on this line
if and only if:
• The slope of the line
through P1 and P
is equal to m,
that is,
y  y1
m
x  x1
Equations of Lines
That equation can be rewritten in
the form
y – y1 = m(x – x1)
• Note that the equation is also satisfied
when x = x1 and y = y1.
• Therefore, it is an equation of the given line.
Point-Slope Form of the Equation of a Line
An equation of the line that passes
through the point (x1, y1) and has
slope m is:
y – y1 = m(x – x1)
E.g. 2—Equation of a Line with Given Point and Slope
(a) Find an equation of the line through
(1, –3 ) with slope –½.
(b) Sketch the line.
E.g. 2—Equation of a Line
Example (a)
Using the point-slope form with
m = –½ , x1 = 1, y1 = –3,
we obtain an equation of the line as:
y  3   21  x  1
2y  6   x  1
x  2y  5  0
E.g. 2—Equation of a Line
Example (b)
The fact that the slope is –½ tells us
that:
• When we move to the right 2 units,
the line drops 1 unit.
E.g. 2—Equation of a Line
Example (b)
This enables us to sketch the line
here.
E.g. 3—Equation of a Line Through Two Points
Find an equation of the line through
the points
(–1, 2) and (3, –4)
• The slope of the line is:
4  2
6
3
m
 
3   1
4
2
E.g. 3—Equation of a Line Through Two Points
Using the point-slope form with x1 = –1
and y1 = 2, we obtain:
y  2   32  x  1
2 y  4  3 x  3
3 x  2y  1  0
Slope-Intercept Form of the
Equations of a Line
Equations of Lines
Suppose a nonvertical line has slope m
and y-intercept b.
• This means
the line intersects
the y-axis at
the point (0, b).
Slope-Intercept Form of the Equation of a Line
So, the point-slope form of the equation
of the line, with x = 0 and y = b,
becomes:
y – b = m(x – 0)
• This simplifies to y = mx + b.
• This is called the slope-intercept form
of the equation of a line.
Slope-Intercept Form of the Equation of a Line
An equation of the line that has slope m
and y-intercept b is:
y = mx + b
E.g. 4—Lines in Slope-Intercept Form
(a) Find the equation of the line with
slope 3 and y-intercept –2.
(b) Find the slope and y-intercept
of the line 3y – 2x = 1.
E.g. 4—Slope-Intercept Form
Example (a)
Since m = 3 and b = –2, from the slopeintercept form of the equation of a line,
we get:
y = 3x – 2
E.g. 4—Slope-Intercept Form
Example (b)
We first write the equation in the form
y = mx + b:
3y  2x  1
3y  2x  1
y  32 x  31
• From the slope-intercept form of the equation
of a line, we see that:
the slope is m = ⅔ and the y-intercept is b = ⅓.
Vertical and Horizontal
Lines
Equation of Horizontal Line
If a line is horizontal, its slope is m = 0.
So, its equation is:
y=b
where b is
the y-intercept.
Equation of Vertical Line
A vertical line does not have a slope.
Still, we can write its equation as:
x=a
where a is
the x-intercept.
• This is because
the x-coordinate
of every point
on the line is a.
Vertical and Horizontal Lines
An equation of the vertical line through
(a, b) is:
x=a
An equation of the horizontal line through
(a, b) is:
y=b
E.g. 5—Vertical and Horizontal Lines
(a) An equation for the vertical line through
(3, 5) is x = 3.
(b) The graph of the
equation x = 3 is
a vertical line with
x-intercept 3.
E.g. 5—Vertical and Horizontal Lines
(a) The equation for the horizontal line
through (8, –2) is y = –2.
(b) The graph of the
equation y = –2 is
a horizontal line
with y-intercept –2.
General Equation of a Line
Linear Equation
A linear equation is an equation of
the form
Ax + By + C = 0
where:
• A, B, and C are constants.
• A and B are not both 0.
Equation of a Line
The equation of a line is a linear equation:
• A nonvertical line has the equation
y = mx + b or –mx + y – b = 0
This is a linear equation with
A = –m, B = 1, C = –b
• A vertical line has the equation
x = a or x – a = 0
This is a linear equation with
A = 1, B = 0, C = –a
Graph of a Linear Equation
Conversely, the graph of a linear equation
is a line:
A
C
• If B ≠ 0, the equation becomes: y   x 
B
B
This is the slope-intercept form of the
equation of a line (with m = –A/B and b = –C/B).
• If B = 0, the equation becomes: Ax + C = 0
or x = –C/A
This represents a vertical line.
General Equation of a Line
We have proved the following.
• The graph of every linear equation
Ax + By + C = 0 (A, B not both zero)
is a line.
• Conversely, every line is the graph of a linear
equation.
E.g. 6—Graphing a Linear Equation
Sketch the graph of the equation
2x – 3y – 12 = 0
E.g. 6—Graphing a Linear Equation Solution 1
Since the equation is linear, its graph is
a line.
To draw the graph, it is enough to find
any two points on the line.
• The intercepts are the easiest points to find.
E.g. 6—Graphing a Linear Equation Solution 1
x-intercept:
• Substitute y = 0.
• You get: 2x – 12 = 0.
• Thus, x = 6.
y-intercept:
• Substitute x = 0.
• You get: –3y – 12 = 0.
• Thus, y = –4.
E.g. 6—Graphing a Linear Equation Solution 1
With those points, we can sketch
the graph as shown.
E.g. 6—Graphing a Linear Equation Solution 2
We write the equation in slope-intercept
form:
2x  3 y  12  0
2 x  3 y  12
 3y  2 x  12
y  32 x  4
• This equation is in the form y = mx + b.
• So, the slope is m = ⅔ and the y-intercept
is b = –4.
E.g. 6—Graphing a Linear Equation Solution 2
To sketch the graph, we:
1. Plot the y-intercept.
2. Move 3 units to the right and 2 units up.
Parallel and
Perpendicular Lines
Parallel Lines
Since slope measures the steepness
of a line, it seems reasonable that parallel
lines should have the same slope.
• In fact, we can prove this.
Parallel Lines
Two nonvertical lines are parallel
if and only if:
• They have the same slope.
Parallel Lines—Proof
Let the lines l1 and l2
have slopes m1 and
m2.
• If the lines are parallel,
the right triangles ABC
and DEF are similar.
• Thus,
m1 
d  B,C 
d  A,C 

d  E, F 
d  D, F 
 m2
Parallel Lines—Proof
Conversely, if the
slopes are equal, then
the triangles will be
similar.
• So, BAC  EDF
and the lines are
parallel.
E.g. 7—Equation of a Line Parallel to a Given Line
Find an equation of the line through
the point (5, 2) that is parallel to the line
4x + 6y + 5 = 0.
E.g. 7—Equation of a Line Parallel to a Given Line
First, we write the equation of the given
line in slope-intercept form:
4 x  6y  5  0
6y  4 x  5
y   32 x  56
• So, the line has slope m = –⅔.
• Since the required line is parallel to
the given line, it also has slope m = –⅔.
E.g. 7—Equation of a Line Parallel to a Given Line
From the point-slope form of the equation
of a line, we get:
y  2   32  x  5 
3 y  6  2 x  10
2 x  3 y  16  0
• The equation of the required line is:
2x + 3y – 16 = 0
Perpendicular Lines
The condition for perpendicular
lines is not as obvious as that for
parallel lines.
Perpendicular Lines
Two lines with slopes m1 and m2 are
perpendicular if and only if m1m2 = –1,
that is, their slopes are negative reciprocals:
1
m2  
m1
• Also, a horizontal line (slope 0) is
perpendicular to a vertical line (no slope).
Perpendicular Lines—Proof
Here, we show two lines intersecting
at the origin.
• If the lines intersect at some
other point, we consider lines
parallel to these that intersect
at the origin.
• These lines have the same
slopes as the original lines.
Perpendicular Lines—Proof
If the lines l1 and l2 have slopes m1 and m2,
then their equations are:
y = m1x and y = m2x
Notice that:
• A(1, m1) lies on l1.
• B(1, m2) lies on l2.
Perpendicular Lines—Proof
By the Pythagorean Theorem and
its converse, OA ┴ OB if and only if:
d O, A   d O, B   d  A, B 
2
2
2
Perpendicular Lines—Proof
By the Distance Formula, this becomes:
1
2
m
2
1
  1
2
m
2
2
  1 1   m
2
2
 m1 
2  m  m  m  2m1m2  m
2
1
2
2
2
2
2  2m1m2
m1m2  1
2
1
2
E.g. 8—Perpendicular Lines
Show that the points
P(3, 3), Q(8, 17), R(11, 5)
are the vertices of a right triangle.
E.g. 8—Perpendicular Lines
The slopes of the lines containing PR and
QR are, respectively,
53 1
m1 

11  3 4
5  17
m2 
 4
11  8
• As m1m2 = –1, these lines are perpendicular.
• So, PQR is a right triangle.
E.g. 8—Perpendicular Lines
The triangle is
sketched here.
E.g. 9—Equation of Line Perpendicular to Given Line
Find an equation of the line that is
perpendicular to the line 4x + 6y + 5 =0
and passes through the origin.
• In Example 7, we found that the slope
of the line 4x + 6y + 5 = 0 is –2/3.
• Thus, the slope of a perpendicular line
is the negative reciprocal—that is, 3/2.
E.g. 9—Equation of Line Perpendicular to Given Line
Since the required line passes through
(0, 0), the point-slope form gives:
y – 0 = 3/2(x – 0)
y = 3/2x
E.g. 10—Graphing a Family of Lines
Use a graphing calculator to graph
the family of lines
y = 0.5x + b
for b = –2, –1, 0, 1, 2.
• What property do the lines share?
E.g. 10—Graphing a Family of Lines
The lines are graphed here in the
viewing rectangle [–6, 6] by [–6, 6].
• They all have
the same slope.
• So, they are
parallel.
Modeling with Linear Equations:
Slope as Rate of Change
Rate of Change
When a line is used to model the relationship
between two quantities, the slope of the line
is the rate of change of one quantity with
respect to the other.
Rate of Change
For example, this graph gives the amount
of gas in a tank that is being filled.
• The slope between
the indicated points is:
6 gallons
m
3 minutes
 2 gal/min
• The slope is the rate
at which the tank is
being filled—2 gallons
per minute.
Rate of Change
Here, the tank is being drained at the rate
of 0.03 gallon per minute.
The slope is –0.03.
Slope as Rate of Change
The next two examples give
other situations where the slope
of a line is a rate of change.
E.g. 11—Slope as Rate of Change
A dam is built on a river to create a reservoir.
The water level w in the reservoir is given
by
w = 4.5t + 28
where:
• t is the number of years since the dam
was constructed.
• w is measured in feet.
E.g. 11—Slope as Rate of Change
(a) Sketch a graph of this equation.
(b) What do the slope and w-intercept
of this graph represent?
E.g. 11—Slope as Rate of Change
Example (a)
This equation is linear.
So, its graph is a line.
• Since two points determine a line,
we plot two points that lie on the graph
and draw a line through them.
E.g. 11—Slope as Rate of Change
Example (a)
When t = 0, w = 4.5(0) + 28 = 28
• So, (0, 28) is on the line.
When t = 2, w = 4.5(2) + 28 = 37
• So, (2, 37) is on the line.
E.g. 11—Slope as Rate of Change
Example (a)
The line determined by those points is
shown here.
E.g. 11—Slope as Rate of Change
Example (b)
The slope is m = 4.5.
It represents the rate of change of water
level with respect to time.
• This means that the water level increases
4.5 ft per year.
E.g. 11—Slope as Rate of Change
Example (b)
The w-intercept is 28, and occurs
when t = 0.
• This represents the water level
when the dam was constructed.
E.g. 12—Temperature-Elevation Linear Relationship
(a) As dry air moves upward, it expands
and cools.
• If the ground temperature is 20°C and
the temperature at a height of 1 km is 10°C,
express the temperature T (in °C) in terms
of the height h (in kilometers).
• Assume that the relationship between T and h
is linear.
E.g. 12—Temperature-Elevation Linear Relationship
(b) Draw the graph of the linear equation.
What does its slope represent?
(c) What is the temperature at a height
of 2.5 km?
E.g. 12—Temperature-Elevation
Example (a)
Because we are assuming a linear
relationship between T and h, the equation
must be of the form
T = mh + b
where m and b are constants.
E.g. 12—Temperature-Elevation
Example (a)
When h = 0, we are given that T = 20.
• Thus,
20 = m(0) + b
b = 20
• So, we have:
T = mh + 20
E.g. 12—Temperature-Elevation
Example (a)
When h = 1, we have T = 10.
• So,
10 = m(1) + 20
m = 10 – 20 = –10
• The required expression is:
T = –10h + 20
E.g. 12—Temperature-Elevation
Example (b)
The graph is sketched here.
• The slope is:
m = –10°C/km
• This represents the rate
of change of temperature
with respect to distance
above the ground.
• So, the temperature
decreases 10°C per
kilometer of height.
E.g. 12—Temperature-Elevation
Example (c)
At a height of h = 2.5 km,
the temperature is:
T = –10(2.5) + 20
= –25 + 20
= –5°C