The Cartesian Coordinate System

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The Cartesian Coordinate System

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Title: The Cartesian Coordinate System

1
1
Straight Lines and Linear Functions

The Cartesian Coordinate System
Straight Lines
Linear Functions and Mathematical Models
Intersection of Straight Lines
The Method of Least Squares

2
1.1

The Cartesian Coordinate System

3
The Cartesian Coordinate System

We can represent real numbers geometrically by
points on a real number, or coordinate, line
This line includes all real numbers.
Exactly one point on the line is associated with
each real number, and vice-versa (one dimensional
space).

Origin
Positive Direction
Negative Direction
p
4
The Cartesian Coordinate System

The Cartesian coordinate system extends this
concept to a plane (two dimensional space) by
adding a vertical axis.

4 3 2 1 1 2 3 4
5
The Cartesian Coordinate System

The horizontal line is called the x-axis, and the
vertical line is called the y-axis.

y
4 3 2 1 1 2 3 4
x
6
The Cartesian Coordinate System

The point where these two lines intersect is
called the origin.

y
4 3 2 1 1 2 3 4
Origin
x
7
The Cartesian Coordinate System

In the x-axis, positive numbers are to the right
and negative numbers are to the left of the
origin.

y
4 3 2 1 1 2 3 4
Positive Direction
Negative Direction
x
8
The Cartesian Coordinate System

In the y-axis, positive numbers are above and
negative numbers are below the origin.

y
4 3 2 1 1 2 3 4
Positive Direction
x
Negative Direction
9
The Cartesian Coordinate System

A point in the plane can now be represented
uniquely in this coordinate system by an ordered
pair of numbers (x, y).

y
( 2, 4)
4 3 2 1 1 2 3 4
(4, 3)
x
(3, 1)
(1, 2)
10
The Cartesian Coordinate System

The axes divide the plane into four quadrants as
shown below.

y
4 3 2 1 1 2 3 4
Quadrant I (, )
Quadrant II (, )
x
Quadrant IV (, )
Quadrant III (, )
11
The Distance Formula

The distance between any two points in the plane
can be expressed in terms of the coordinates of
the points.

Distance formula
The distance d between two points P1(x1, y1) and
P2(x2, y2) in the plane is given by

12
Examples

Find the distance between the points ( 4, 3) and
(2, 6).
Solution
Let P1( 4, 3) and P2(2, 6) be points in the
plane.
We have
x1 4 y1 3 x2 2 y2 6
Using the distance formula, we have

Example 1, page 4
13
Examples

Let P(x, y) denote a point lying on the circle
with radius r and center C(h, k). Find a
relationship between x and y.
Solution
By the definition of a circle, the distance
between P(x, y) and C(h, k) is r.
With the distance formula
we get
Squaring both sides gives

y
P(x, y)
C(h, k)
r
k
x
h
Example 3, page 4
14
Equation of a Circle

An equation of a circle with center C(h, k) and
radius r is given by

15
Examples

Find an equation of the circle with radius 2 and
center (1, 3).
Solution
We use the circle formula with r 2, h 1, and
k 3

y
(1, 3)
3
2
x
1
Example 4, page 5
16
Examples

Find an equation of the circle with radius 3 and
center located at the origin.
Solution
We use the circle formula with r 3, h 0, and
k 0

y
3
x
Example 4, page 5
17
1.2

Straight Lines

18
Slope of a Vertical Line

Let L denote the unique straight line that passes
through the two distinct points (x1, y1) and (x2,
y2).
If x1 x2, then L is a vertical line, and the
slope is undefined.

y
L
(x1, y1)
(x2, y2)
x
19
Slope of a Nonvertical Line

If (x1, y1) and (x2, y2) are two distinct points
on a nonvertical line L, then the slope m of L is
given by

y
L
(x2, y2)
y2 y1 ?y
(x1, y1)
x2 x1 ?x
x
20
Slope of a Nonvertical Line

If m gt 0, the line slants upward from left to
right.

y
L
m 1
?y 1
?x 1
x
21
Slope of a Nonvertical Line

If m gt 0, the line slants upward from left to
right.

y
L
m 2
?y 2
?x 1
x
22
Slope of a Nonvertical Line

If m lt 0, the line slants downward from left to
right.

y
m 1
?x 1
?y 1
x
L
23
Slope of a Nonvertical Line

If m lt 0, the line slants downward from left to
right.

y
m 2
?x 1
?y 2
x
L
24
Examples

Sketch the straight line that passes through the
point (2, 5) and has slope 4/3.

Solution
Plot the point (2, 5).
A slope of 4/3 means that if x increases by 3,
y decreases by 4.
Plot the resulting point (5, 1).
Draw a line through the two points.

y
6 5 4 3 2 1
?x 3
(2, 5)
?y 4
(5, 1)
x
1 2 3 4 5 6
L
25
Examples

Find the slope m of the line that goes through
the points (1, 1) and (5, 3).
Solution
Choose (x1, y1) to be (1, 1) and (x2, y2) to be
(5, 3).
With x1 1, y1 1, x2 5, y2 3, we find

Example 2, page 11
26
Examples

Find the slope m of the line that goes through
the points (2, 5) and (3, 5).
Solution
Choose (x1, y1) to be (2, 5) and (x2, y2) to be
(3, 5).
With x1 2, y1 5, x2 3, y2 5, we find

Example 3, page 11
27
Examples

Find the slope m of the line that goes through
the points (2, 5) and (3, 5).
Solution
The slope of a horizontal line is zero

y
6 4 3 2 1
(3, 5)
(2, 5)
L
m 0
x
2 1 1 2 3 4
Example 3, page 11
28
Parallel Lines

Two distinct lines are parallel if and only if
their slopes are equal or their slopes are
undefined.

29
Example

Let L1 be a line that passes through the points
(2, 9) and (1, 3), and let L2 be the line that
passes through the points ( 4, 10) and (3, 4).
Determine whether L1 and L2 are parallel.
Solution
The slope m1 of L1 is given by
The slope m2 of L2 is given by
Since m1 m2, the lines L1 and L2 are in fact
parallel.

Example 4, page 12
30
Equations of Lines

Let L be a straight line parallel to the y-axis.
Then L crosses the x-axis at some point (a, 0) ,
with the x-coordinate given by x a, where a is
a real number.
Any other point on L has the form (a, ), where
is an appropriate number.
The vertical line L can therefore be described as
x a

y
L
(a, )
(a, 0)
x
31
Equations of Lines

Let L be a nonvertical line with a slope m.
Let (x1, y1) be a fixed point lying on L, and let
(x, y) be a variable point on L distinct from
(x1, y1).
Using the slope formula by letting (x, y) (x2,
y2), we get
Multiplying both sides by x x1 we get

32
Point-Slope Form

An equation of the line that has slope m and
passes through point (x1, y1) is given by

33
Examples

Find an equation of the line that passes through
the point (1, 3) and has slope 2.
Solution
Use the point-slope form
Substituting for point (1, 3) and slope m 2, we
obtain
Simplifying we get

Example 5, page 13
34
Examples

Find an equation of the line that passes through
the points (3, 2) and (4, 1).
Solution
The slope is given by
Substituting in the point-slope form for point
(4, 1) and slope m 3/7, we obtain

Example 6, page 14
35
Perpendicular Lines

If L1 and L2 are two distinct nonvertical lines
that have slopes m1 and m2, respectively, then L1
is perpendicular to L2 (written L1 - L2) if and
only if

36
Example

Find the equation of the line L1 that passes
through the point (3, 1) and is perpendicular to
the line L2 described by
Solution
L2 is described in point-slope form, so its slope
is m2 2.
Since the lines are perpendicular, the slope of
L1 must be
m1 1/2
Using the point-slope form of the equation for L1
we obtain

Example 7, page 14
37
Crossing the Axis

A straight line L that is neither horizontal nor
vertical cuts the x-axis and the y-axis at, say,
points (a, 0) and (0, b), respectively.
The numbers a and b are called the x-intercept
and y-intercept, respectively, of L.

y
y-intercept
(0, b)
x-intercept
x
(a, 0)
L
38
Slope-Intercept Form

An equation of the line that has slope m and
intersects the y-axis at the point (0, b) is
given by
y mx b

39
Examples

Find the equation of the line that has slope 3
and y-intercept of 4.
Solution
We substitute m 3 and b 4 into y mx b
and get
y 3x 4

Example 8, page 15
40
Examples

Determine the slope and y-intercept of the line
whose equation is 3x 4y 8.
Solution
Rewrite the given equation in the slope-intercept
form.
Comparing to y mx b, we find that m ¾ and
b 2.
So, the slope is ¾ and the y-intercept is 2.

Example 9, page 15
41
Applied Example

Suppose an art object purchased for 50,000 is
expected to appreciate in value at a constant
rate of 5000 per year for the next 5 years.
Write an equation predicting the value of the art
object for any given year.
What will be its value 3 years after the
purchase?
Solution
Let x time (in years) since the object was
purchased
y value of object (in dollars)
Then, y 50,000 when x 0, so the y-intercept
is b 50,000.
Every year the value rises by 5000, so the slope
is m 5000.
Thus, the equation must be y 5000x 50,000.
After 3 years the value of the object will be
65,000
y 5000(3) 50,000 65,000

Applied Example 11, page 16
42
General Form of a Linear Equation

The equation
Ax By C 0
where A, B, and C are constants and A and B are
not both zero, is called the general form of a
linear equation in the variables x and y.

43
General Form of a Linear Equation

An equation of a straight line is a linear
equation conversely, every linear equation
represents a straight line.

44
Example

Sketch the straight line represented by the
equation
3x 4y 12 0
Solution
Since every straight line is uniquely determined
by two distinct points, we need find only two
such points through which the line passes in
order to sketch it.
For convenience, lets compute the x- and
y-intercepts
Setting y 0, we find x 4 so the x-intercept
is 4.
Setting x 0, we find y 3 so the y-intercept
is 3.
Thus, the line goes through the points (4, 0) and
(0, 3).

Example 12, page 17
45
Example

Sketch the straight line represented by the
equation
3x 4y 12 0
Solution
Graph the line going through the points (4, 0)
and (0, 3).

y
L
1 1 2 3 4
(4, 0)
x
1 2 3 4 5 6
(0, 3)
Example 12, page 17
46
Equations of Straight Lines

Vertical line x a
Horizontal line y b
Point-slope form y y1 m(x x1)
Slope-intercept form y mx b
General Form Ax By C 0

47
1.3

Linear Functions and Mathematical Models

Real
-
world
Mathematical
Formulate
problem
model
Solve
Test
Solution of real
-
Solution of
world Problem
mathematical model
Interpret
48
Mathematical Modeling

Mathematics can be used to solve real-world
problems.
Regardless of the field from which the real-world
problem is drawn, the problem is analyzed using a
process called mathematical modeling.
The four steps in this process are

Real-world problem
Mathematical model
Formulate
Solve
Test
Solution of real- world Problem
Solution of mathematical model
Interpret
49
Functions

A function f is a rule that assigns to each value
of x one and only one value of y.
The value y is normally denoted by f(x),
emphasizing the dependency of y on x.

50
Example

Let x and y denote the radius and area of a
circle, respectively.
From elementary geometry we have
y px2
This equation defines y as a function of x, since
for each admissible value of x there corresponds
precisely one number y px2 giving the area of
the circle.
The area function may be written as
f(x) px2
To compute the area of a circle with a radius of
5 inches, we simply replace x in the equation by
the number 5
f(5) p(52) 25p

51
Domain and Range

Suppose we are given the function y f(x).
The variable x is referred to as the independent
variable, and the variable y is called the
dependent variable.
The set of all the possible values of x is called
the domain of the function f.
The set of all the values of f(x) resulting from
all the possible values of x in its domain is
called the range of f.
The output f(x) associated with an input x is
unique
Each x must correspond to one and only one value
of f(x).

52
Linear Function

The function f defined by
where m and b are constants, is called a linear
function.

53
Applied Example U.S. Health-Care Expenditures

Because the over-65 population will be growing
more rapidly in the next few decades, health-care
spending is expected to increase significantly in
the coming decades.
The following table gives the projected U.S.
health-care expenditures (in trillions of
dollars) from 2005 through 2010
A mathematical model giving the approximate U.S.
health-care expenditures over the period in
question is given by
where t is measured in years, with t 0
corresponding to 2005.

Year 2005 2006 2007 2008 2009 2010
Expenditure 2.00 2.17 2.34 2.50 2.69 2.90
Applied Example 1, page 29
54
Applied Example U.S. Health-Care Expenditures

We have
Sketch the graph of the function S and the given
data on the same set of axes.
Assuming that the trend continues, how much will
U.S. health-care expenditures be in 2011?
What is the projected rate of increase of U.S.
health-care expenditures over the period in
question?

Year 2005 2006 2007 2008 2009 2010
Expenditure 2.00 2.17 2.34 2.50 2.69 2.90
Applied Example 1, page 29
55
Applied Example U.S. Health-Care Expenditures

We have
Solution
The graph of the given data and of the function S
is

Year 2005 2006 2007 2008 2009 2010
Expenditure 2.00 2.17 2.34 2.50 2.69 2.90
S(t)
3.0 2.8 2.6 2.4 2.2 2.0
t
1 2 3 4 5
Applied Example 1, page 29
56
Applied Example U.S. Health-Care Expenditures

We have
Solution
The projected U.S. health-care expenditures in
2011 is
or approximately 3.06 trillion.

Year 2005 2006 2007 2008 2009 2010
Expenditure 2.00 2.17 2.34 2.50 2.69 2.90
Applied Example 1, page 29
57
Applied Example U.S. Health-Care Expenditures

We have
Solution
The function S is linear, so the rate of increase
of the U.S. health-care expenditures is given by
the slope of the straight line represented by S,
which is approximately 0.18 trillion per year.

Year 2005 2006 2007 2008 2009 2010
Expenditure 2.00 2.17 2.34 2.50 2.69 2.90
Applied Example 1, page 29
58
Cost, Revenue, and Profit Functions

Let x denote the number of units of a product
manufactured or sold.
Then, the total cost function is
C(x) Total cost of manufacturing x units of
the product
The revenue function is
R(x) Total revenue realized from the sale of x
units of the product
The profit function is
P(x) Total profit realized from manufacturing
and selling x units of the product

59
Applied Example Profit Function

Puritron, a manufacturer of water filters, has a
monthly fixed cost of 20,000, a production cost
of 20 per unit, and a selling price of 30 per
unit.
Find the cost function, the revenue function, and
the profit function for Puritron.
Solution
Let x denote the number of units produced and
sold.
Then,

Applied Example 2, page 31
60
1.4

Intersections of Straight Lines

61
Finding the Point of Intersection

Suppose we are given two straight lines L1 and L2
with equations
y m1x b1 and y m2x b2
(where m1, b1, m2, and b2 are constants) that
intersect at the point P(x0, y0).
The point P(x0, y0) lies on the line L1 and so
satisfies the equation y m1x b1.
The point P(x0, y0) also lies on the line L2 and
so satisfies y m2x
b2 as well.
Therefore, to find the point of intersection
P(x0, y0) of the lines L1 and L2, we solve for x
and y the system composed of the two equations
y m1x b1 and y m2x b2

62
Example

Find the point of intersection of the straight
lines that have equations
y x 1 and y 2x 4
Solution
Substituting the value y as given in the first
equation into the second equation, we obtain
Substituting this value of x into either one of
the given equations yields y 2.
Therefore, the required point of intersection is
(1, 2).

Example 1, page 40
63
Example

Find the point of intersection of the straight
lines that have equations
y x 1 and y 2x 4
Solution
The graph shows the point of intersection (1, 2)
of the two lines

y
5 4 3 2 1
L1
(1, 2)
x
1 1 2 3 4 5
L2
Example 1, page 40
64
Applied Example Break-Even Level

Prescott manufactures its products at a cost of
4 per unit and sells them for 10 per unit.
If the firms fixed cost is 12,000 per month,
determine the firms break-even point.
Solution
The revenue function R and the cost function C
are given respectively by
Setting R(x) C(x), we obtain

Applied Example 2, page 41
65
Applied Example Break-Even Level

Prescott manufactures its products at a cost of
4 per unit and sells them for 10 per unit.
If the firms fixed cost is 12,000 per month,
determine the firms break-even point.
Solution
Substituting x 2000 into R(x) 10x gives
So, Prescotts break-even point is 2000 units of
the product, resulting in a break-even revenue of
20,000 per month.

Applied Example 2, page 41
66
Applied Example Market Equilibrium

The management of ThermoMaster, which
manufactures an indoor-outdoor thermometer at
its Mexico subsidiary, has determined that the
demand equation for its product is
where p is the price of a thermometer in dollars
and x is the quantity demanded in units of a
thousand.
The supply equation of these thermometers is
where x (in thousands) is the quantity that
ThermoMaster will make available in the market at
p dollars each.
Find the equilibrium quantity and price.

Applied Example 6, page 44
67
Applied Example Market Equilibrium

Solution
We need to solve the system of equations
for x and p.
Lets solve the first equation for p in terms of
x

Applied Example 6, page 44
68
Applied Example Market Equilibrium

Solution
We need to solve the system of equations
for x and p.
Now we substitute the value of p into the second
equation

Applied Example 6, page 44
69
Applied Example Market Equilibrium

Solution
We need to solve the system of equations
for x and p.
Finally, we substitute the value x 5/2 into the
first equation that we already solved

Applied Example 6, page 44
70
Applied Example Market Equilibrium

Solution
We conclude that the equilibrium quantity is 2500
units and the equilibrium price is 5.83 per
thermometer.

Applied Example 6, page 44
71
1.5

The Method of Least Squares

72
The Method of Least Squares

In this section, we describe a general method
known as the method for least squares for
determining a straight line that, in a sense,
best fits a set of data points when the points
are scattered about a straight line.

73
The Method of Least Squares

Suppose we are given five data points
P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4),
and P5(x5, y5)
describing the relationship between two
variables x and y.
By plotting these data points, we obtain a
scatter diagram

y
P5
10 5
P3
P2
P4
P1
x
5 10
74
The Method of Least Squares

Suppose we try to fit a straight line L to the
data points P1, P2, P3, P4, and P5.
The line will miss these points by the amounts
d1, d2, d3, d4, and d5
respectively.

y
L
10 5
x
5 10
75
The Method of Least Squares

The principle of least squares states that the
straight line L that fits the data points best is
the one chosen by requiring that the sum of the
squares of d1, d2, d3, d4, and d5, that is
be made as small as possible.

y
L
10 5
x
5 10
76
The Method of Least Squares

Suppose we are given n data points
P1(x1, y1), P2(x2, y2), P3(x3, y3), . . . ,
Pn(xn, yn)
Then, the least-squares (regression) line for the
data is given by the linear equation
y f(x) mx b
where the constants m and b satisfy the
equations
and
simultaneously.
These last two equations are called normal
equations.

77
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Here, we have n 5 and
x1 1 x2 2 x3 3 x4 4 x5 5
y1 1 y2 3 y3 4 y4 3 y5 6

Before using the equations it is convenient to
summarize these data in the form of a table

x y x2 xy
1 1 1 1
2 3 4 6
3 4 9 12
4 3 16 12
5 6 25 30
15 17 55 61
Example 1, page 53
78
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Here, we have n 5 and
x1 1 x2 2 x3 3 x4 4 x5 5
y1 1 y2 3 y3 4 y4 3 y5 6
Using the table to substitute in the second
equation we get

Example 1, page 53
79
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Here, we have n 5 and
x1 1 x2 2 x3 3 x4 4 x5 5
y1 1 y2 3 y3 4 y4 3 y5 6
Using the table to substitute in the first
equation we get

Example 1, page 53
80
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Now we need to solve the simultaneous equations
Solving the first equation for b gives

Example 1, page 53
81
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Now we need to solve the simultaneous equations
Substituting b into the second equation gives

Example 1, page 53
82
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Now we need to solve the simultaneous equations
Finally, substituting the value m 1 into the
first equation that we already solved gives

Example 1, page 53
83
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Now we need to solve the simultaneous equations
Thus, we find that m 1 and b 0.4.
Therefore, the required least-squares line is

Example 1, page 53
84
Example

Find the equation of the least-squares line for
the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5,
6)
Solution
Below is the graph of the required least-squares
line
y x 0.4

y
L
6 5 4 3 2 1
x
1 2 3 4 5
Example 1, page 53
85
Applied Example U.S. Health-Care Expenditures

Because the over-65 population will be growing
more rapidly in the next few decades, health-care
spending is expected to increase significantly in
the coming decades.
The following table gives the U.S. health
expenditures (in trillions of dollars)
from 2005 through 2010
Find a function giving the U.S. health-care
spending between 2005 and 2010, using the
least-squares technique.

Year, t 0 1 2 3 4 5
Expenditure, y 2.00 2.17 2.34 2.50 2.69 2.90
Applied Example 3, page 55
86
Applied Example U.S. Health-Care Expenditures

Solution
The calculations required for obtaining the
normal equations are summarized in the following
table
Use the table to obtain the second normal
equation

t y t2 ty
0 2.00 0 0
1 2.17 1 2.17
2 2.34 4 4.68
3 2.50 9 7.50
4 2.69 16 10.76
5 2.90 25 14.50
15 14.60 55 39.61
Applied Example 3, page 55
87
Applied Example U.S. Health-Care Expenditures

Solution
The calculations required for obtaining the
normal equations are summarized in the following
table
Use the table to obtain the first normal equation

t y t2 ty
0 2.00 0 0
1 2.17 1 2.17
2 2.34 4 4.68
3 2.50 9 7.50
4 2.69 16 10.76
5 2.90 25 14.50
15 14.60 55 39.61
Applied Example 3, page 55
88
Applied Example U.S. Health-Care Expenditures