10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
2.1 - 1
2.1
Rectangular Coordinates and
Graphs
Ordered Pairs
The Rectangular Coordinate System
The Distance Formula
The Midpoint formula
Graphing Equations
2.1 - 2
Ordered Pairs
An ordered pair consists of two components,
written inside parentheses, in which the
order of the components is important.
2.1 - 3
Example 1
WRITING ORDERED PAIRS
Use the table to
write ordered
pairs to express
the relationship
between each
type of
entertainment
and the amount
spent on it.
Type of
Entertainment
VHS rentals/
sales
DVD
rentals/sales
Amount
Spent
$23
$71
CDs
$20
Theme parks
$43
Sports tickets
$50
Movie tickets
$30
2.1 - 4
Example 1
WRITING ORDERED PAIRS
a. DVD rental sales
Solution:
Type of
Entertainment
VHS rentals/
sales
Use the data in the
second row:
DVD rental/sales
(DVD rental sales, $71).
Amount
Spent
$23
$71
CDs
$20
Theme parks
$43
Sports tickets
$50
Movie tickets
$30
2.1 - 5
Example 1
WRITING ORDERED PAIRS
b. Movie tickets
Solution:
Use the data in the last
row:
(Movie tickets, $30).
Type of
Entertainment
VHS rentals/
sales
DVD
rentals/sales
Amount
Spent
$23
$71
CDs
$20
Theme parks
$43
Sports tickets
$50
Movie tickets
$30
2.1 - 6
The Rectangular Coordinate
System
y-axis
Quadrant I
Quadrant II
P(a, b)
b
a
Quadrant III
x-axis
0
Quadrant IV
2.1 - 7
The Distance Formula
Using the coordinates of ordered pairs, we
can find the distance between any two
y
points in a plane.
Horizontal side of
the triangle has
length
d (P,Q )  8  ( 4)  12
P(– 4, 3)
Q(8, 3)
x
R(8, – 2)
Definition of distance.
2.1 - 8
The Distance Formula
y
Vertical side of the
triangle has length…
P(– 4, 3)
Q(8, 3)
x
d (P,Q )  3  ( 2)  5
R(8, – 2)
2.1 - 9
The Distance Formula
y
By the Pythagorean
theorem, the length of
the remaining side of
the triangle is
P(– 4, 3)
12  5  144  25  169  13
2
2
Q(8, 3)
x
R(8, – 2)
2.1 - 10
The Distance Formula
y
So the distance
between (–4, 3) and
(8, –4) is 13.
P(– 4, 3)
Q(8, 3)
x
R(8, – 2)
2.1 - 11
The Distance Formula
To obtain a general formula
for the distance between two
points in a coordinate plane,
let P(x1, y1) and R(x2, y2) be
any two distinct points in a
plane. Complete a triangle
by locating point Q with
coordinates (x2, y1). The
Pythagorean theorem gives
the distance between P and
R as
y
P(x1, y1)
Q(x2, y1)
x
R(x2, y2)
d (P, R )  ( x2  x1 )2  ( y 2  y1 )2
2.1 - 12
The Distance Formula
Note Suppose that P(x1, y1) and
R(x2, y2) are two points in a coordinate
plane. Then the distance between P and
R, written d(P, R) is given by
d (P, R )  ( x2  x1 )  ( y 2  y1 )
2
2
2.1 - 13
USING THE DISTANCE
FORMULA
Example 2
Find the distance between P(–8, 4) and
Q(3, –2.)
Solution:
d (P,Q ) 
2
3

(

8
)

(

2

4
)


2
 11  ( 6)
2
 121  36
2
Be careful when
subtracting a
negative number.
 157
2.1 - 14
Using the Distance Formula
If the sides a, b, and c of a
triangle satisfy a2 + b2 = c2, then
the triangle is a right triangle with
legs having lengths a and b and
hypotenuse having length c?
2.1 - 15
Example 3
DETERMINING WHETHER THREE
POINTS ARE THE VERTICES OF A
RIGHT TRIANGLE
Are points M(–2, 5), N(12, 3), and
Q(10, –11) the vertices of a right triangle.
Solution This triangle is a right triangle if
the square of the length of the longest side
equals the sum of the squares of the lengths
of the other two sides.
2.1 - 16
Example 3
d (M , N ) 
DETERMINING WHETHER THREE
POINTS ARE THE VERTICES OF A
y
RIGHT TRIANGLE
12  ( 2)
2
 (3  5 )
2
M(– 2, 5)
N(12, 3)
 196  4
 200
x
Q(10, – 11)
2.1 - 17
Example 3
d (M,Q ) 
DETERMINING WHETHER THREE
POINTS ARE THE VERTICES OF A
y
RIGHT TRIANGLE
2
10

(

2
)

(

11

5
)


2
M(– 2, 5)
N(12, 3)
 144  256
 400
x
Q(10, – 11)
 20
2.1 - 18
Example 3
DETERMINING WHETHER THREE
POINTS ARE THE VERTICES OF A
y
RIGHT TRIANGLE
d (N,Q )  (10  12)2  ( 11  3)2
M(– 2, 5)
N(12, 3)
 4  196
 200
x
Q(10, – 11)
2.1 - 19
Example 3
DETERMINING WHETHER THREE
POINTS ARE THE VERTICES OF A
y
RIGHT TRIANGLE
The longest side has length
20 units.Since...

200
 
2
200

2
 400  202,
the triangle is a
right triangle with
hypotenuse joining
M and Q.
M(– 2, 5)
N(12, 3)
x
R(10, – 11)
2.1 - 20
Colinear
We can tell if three points are collinear, that
is, if they lie on a straight line, using a similar
procedure
Three points are collinear if the sum of the
distances between two pairs of points is
equal to the distance between the remaining
pair of points.
2.1 - 21
Example 4
DETERMINING WHETHER THREE
POINTS ARE COLLINEAR
Are the points (–1, 5), (2,–4), and (4, –10)
collinear?
Solution The distance between (–1, 5) and
(2,–4) is
( 1  2)  5  ( 4)  9  81  90  3 10
2
2
2.1 - 22
Example 4
DETERMINING WHETHER THREE
POINTS ARE COLLINEAR
Are the points (–1, 5), (2,–4), and (4, –10)
collinear?
Solution The distance between (2,–4) and
(4,–10) is
(2  4)   4  ( 10)  4  36  40  2 10
2
2
2.1 - 23
Example 4
DETERMINING WHETHER THREE
POINTS ARE COLLINEAR
Are the points (–1, 5), (2,–4), and (4, –10)
collinear?
Solution The distance between the
remaining pair of points (–1, 5) and (4,–10)
is
( 1  4)  5  ( 10)  25  225  250  5 10
2
2
Because 3 10  2 10  5 10,
the three points are collinear.
2.1 - 24
Midpoint Formula
The midpoint of the line segment with
endpoints (x1, y1) and (x2, y2) has
coordinates
 x1  x2 y1  y 2 
,

.
2 
 2
2.1 - 25
Example 5
USING THE MIDPOINT FORMULA
Use the midpoint formula to determine the
following.
a. Find the coordinates of M of the segment
with endpoints (8, –4) and (–6,1).
Solution
 8  ( 6)  4  1  3 
,

   1,  
2
2   2

Substitute in the
midpoint formula.
2.1 - 26
Example 5
USING THE MIDPOINT FORMULA
Use the midpoint formula to determine the
following.
b. Find the coordinates of the other endpoint B
of a segment with one endpoint A(–6, 12) and
midpoint M(8, –2).
Solution
Let (x, y) represent the coordinates of B. Use
the midpoint formula twice.
2.1 - 27
Example 5
USING THE MIDPOINT FORMULA
Let (x, y) represent the coordinates of B. Use
the midpoint formula twice.
x-value of A
Substitute
carefully.
x-value of M
x  ( 6)
8
2
x  6  16
x-value of A
y-value of M
y  12
 2
2
y  12  4
y  16
x  22
The coordinates of endpoint B are (22, –16).
2.1 - 28
Example 6
APPLYING THE MIDPOINT
FORMULA TO DATA
The graph indicates the increase of
McDonald’s restaurants worldwide from 20,000
in 1996 to 31,000 in 2006.
Use the midpoint formula and the two given
endpoints to estimate the number of
restaurants in 2001, and compare it to the
actual (rounded) figure of 30,000
2.1 - 29
Example 6
APPLYING THE MIDPOINT
FORMULA TO DATA
Number of McDonald’s Restaurants
Worldwide (in thousands)
35
30
25
20
2006
1996
Year
2.1 - 30
Example 6
APPLYING THE MIDPOINT
FORMULA TO DATA
Solution: 2001 is halfway between 1996
and 2006. Find the coordinates of the
midpoint of the segment that has endpoints
(1996, 20) and (2006, 31).
 1996  2006 20  31
,

  (2001, 25.5)
2
2 

Our estimate is 25,500, which is well
below the actual figure of 30,000.
2.1 - 31
Example 7
FINDING ORDERED PAIRS THAT
ARE SOLUTIONS OF EQUATIONS
For the equation, find at least three ordered
pairs that are solutions
a. y  4 x  1
Solution: Choose any real number for x or y
and substitute in the equation to get the
corresponding value of the other variable.
2.1 - 32
Example 7
a. y  4 x  1
FINDING ORDERED PAIRS THAT
ARE SOLUTIONS OF EQUATIONS
Solution:
y  4x  1
y  4x  1
y  4( 2)  1 Let x = –2.
3  4x  1
y  8  1
4  4x
Multiply.
Let y = 3.
Add 1.
y  9
1  x Divide by 4.
This gives the ordered pairs (–2, –9) and (1, 3).
Find one more ordered pair that works!
2.1 - 33
Example 7
FINDING ORDERED PAIRS THAT
ARE SOLUTIONS OF EQUATIONS
b. x  y  1
Solution
x  y 1
Given equation
1 y 1
Let x = 1
1 y 1
Square both sides
2y
One ordered pair is (1, 2).
Find two more ordered pair!
2.1 - 34
Example 7
FINDING ORDERED PAIRS THAT
ARE SOLUTIONS OF EQUATIONS
c. A table provides an organization method
for determining ordered pairs.
Solution:
x
–2
–1
0
1
2
y
0
–3
–4
–3
0
y  x 4
2
Five ordered pairs
are listed as
solutions for this
equation.
2.1 - 35
Graphing an Equation by
Point Plotting
Step 1 Find the intercepts.
Step 2 Find as many additional ordered
pairs as needed.
Step 3 Plot the ordered pairs from Steps 1
and 2.
Step 4 Connect the points from Step 3
with a smooth line or curve.
2.1 - 36
Example 8
GRAPHING EQUATIONS
a. Graph the equation y  4 x  1
Solution:
Step 1 Let y = 0 to find the x-intercept, and
let x = 0 to find the y-intercept.
y  4x  1
y  4x  1
0  4x  1
y  4(0)  1
1  4x
y  0 1
1
y  1
 x 1 
4
 ,0  and  0, 1
4 
2.1 - 37
Example 8
GRAPHING EQUATIONS
a. Graph the equation y  4 x  1
Solution:
Step 2 We find some other ordered pairs.
y  4x  1
y  4x  1
y  4( 2)  1 Let x = –2.
3  4x  1
y  8  1
4  4x
Multiply.
Let y = 3.
Add 1.
y  9
1  x Divide by 4.
This gives the ordered pairs (–2, –9) and (1, 3).
2.1 - 38
Example 8
GRAPHING EQUATIONS
a. Graph the equation y  4 x  1
Solution:
Step 3 Plot the four ordered pairs from
y
Steps 1 and 2.
Step 4 Connect
the points with a
straight line.
y  4x  1
x
2.1 - 39
Example 8
GRAPHING EQUATIONS
b. Graph the equation x  y  1
Solution:
y
Plot the ordered
pairs found in
example 7b, and
develop a fourth
point to confirm
the direction the
curve will take
as x increases.
x
2.1 - 40
Example 8
GRAPHING EQUATIONS
c. Graph the equation y  x 2  4
Solution:
y
x
y
–2
0
–1
–3
–4
0
–3
1
2
0
x
This curve is called
a parabola.
2.1 - 41