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Graphs and Linear Equations
List of Objectives
To graph points on a rectangular coordinate
system
To determine a solution of a linear equation in two variables
To solve application problems
To graph an equation of the forry 1
-
To graph an equation of the forrn Ax
mx
*
b
* B! -
C
To find the x- and yintercepts of a straight line
To find the slope of a straight line
To graph a line using the slope and 7-intercept
To find the equation of a line using the equation)
-
mx
*
To find the equation of a line using the point-slope formula
b
UNIT I
SECTION
1
1.1 Objective
293
Graphs and Linear Equations
The Rectangular Coordinate System
To graph points on a rectangular coordinate
system
A rectangular coordinate system
ll
Quadrant
i. ..i.. -i .'..
is
... i
formed by two number lines, one horizontal and one vellical, that intersect
ilir
at the zero point of each line. The
point of intersection is called the origin. The two lines are called the coor-
vertical
hoiizoniaii
''''''t
.
i
!axtsi
dinate axes, or simply axes.
EAD'
The axes determine a plane and divide
the plane into four regions, called
quadranls. The quadrants are numbered counterclockwise from I to lV.
axis
t
:
tzJ4
Ouadrant
Each point in the plane can be identified by a pair of numbers called an ordered pair,
The first number of the pair measures a horizontal distance and is called the ab-
scissa. The second number of the pair measures a vertical distance and is called the
ordinate. The coordinates of a point are the numbers in the ordered pair associated
with the point.
o
O
r
t-
horizontal distance
o
o
1
ordered pair --> (3
j
E
uo
!
C
6
,
vertical distance
4)
t_
ordinate
"ba",ar"
The graph of an ordered pair is a point in
the plane. The graphs of the points (2,3)
and (3,2) are shown at the right. Notice that
they are different points, The order in which
the numbers in an ordered pair appear is
c
o
O
-J(2,3)
-i-1(3,
2)
important.
Example
1
Graph the ordered pairs
(-2,-3), (3,-2), (1,3), and
Example
2
(
- 1 ,3), (1 ,4), (-4,0),
(-2,-1)
(4,1).
Solution
Graph the ordered pairs
and
Your solution
' i : -;1
€
N
\t
ci
o
o
6
at
294
UN
Example
3
lT 8
Graphs and Linear Equations
Find the coordinates of each of
Example
4
the points.
-+-3-2-t
.Ai i
o
i-
Find the coordinates of each of
the points.
34
....i.....t.
oD:
::
t:
Solution
A
(-4,-2)
Your solution
B (4,4)
c
(0,
n /?
?
\v,
Example 5
a)
b)
-3)
2\
Example 6
Name the abscissas of
points A and C.
Name the ordinates of
points B and D.
a)
b)
Name the abscissas of
points A and C.
Name the ordinates of
points B and D.
oo
I
o
@
o
@
E
o
I
!
d
-4
E
o
c
o
o
Solution
a)
b)
Abscissa
Abscissa
Ordinate
Ordinate
of point A:
of point C:
of point 8:
of point D:
a
o
l\
Your solution
tt
ci
0
E
a
o
a
c
o
0
6
o
1.2 Objective
To determine a solution of a linear equation in two variables
An equation of the form
y-mx +b,where mandb
are constants, is a linear
equation in two variables.
Examples of linear equations
are shown at the right. Note
y-3x+4
y-2x-3
Y--!x+t
y--2x
y-x+2
(m-3, b-4)
(m=2, b=-3)
(m:-!,0=t1
(m=-2,b=0)
(m =1, b:2)
that the exponent of each vari-
able is always one.
A solution of an equation in two variables is an ordered pair of numbers (x,y) which
makes the equation a true staiement.
UN
IT 8
295
Graphs and Linear Equations
ls (1,-2; a solution oI y
Replace x with
Replace y with
-
3x
-
f=3x-S
5?
l, the abscissa.
-2,Ihe ordinate.
E
5
Yes, (1 , -2) is a solution
of the equation y - 3x - 5
Compare the results. lf the results are equal, the
given ordered pair is a solution. lf the results are
not equal, the given ordered pair is not a solution.
Besides the ordered pair (1,-2), there are many other ordered pair solutions of the
5. For example, the method used above can be used to show lhat
equation y = 3x
-
(2,1),
(-1,-8), (3 -t)
and
(0,-5)
are also solutions.
ln general, a linear equation in two variables has an infinite number of solutions. By
choosing any value for x and substituting that value into the linear equation, a corresponding value of y can be {ound.
oo
Find the ordered pair solution of
y
Substitute 1 lor x,
Solve for y.
-,
- 2x - 5 corresponding to x = 1.
y-2x-5
-2.1-5
_q
The ordered pair sorution is
I
o
@
o
(1
.
e
-3)
o
o
E
o
L
oc
c
o
c
o
o
Example
7
Solution
ls
Example
(-3,2) a solution of
y-2x+2?
y-2x+2
212(-3)+z
-6+2
8
ls (2,
-4)
a solution of
r- -!x-sz
Your solution
-4
2+-4
Example 9
No,
(-3,2) is not a solution
y,,
o,, -tt L,
o
L^
-
of
Find the ordered pair solution
ot
y'3 - ?^
Example
10
1 corresponding
-
of
Y-5x_a
I
|
I
€
-1
The ordered pair solution
(3,1)
- -I, * 1 correspond-
Your solution
r
--2rat
5\e./ -
y
ingtox=4.
tox=3.
Solution
Find the ordered pair solution
is
t
tt
o
6
c
o
=
oo
UNIT
296
1.3 Obiective
Graphs and Linear Equations
8
To solve application problems
A rectangular coordinate system is frequently used in business, science, and mathematics to show a relationship between two variables. One variable is represented
along the horizontal axis and the other variable is represented along the vertical axis'
For example, a physics student
measured the sPeed of imPact of a
solid ball dropped from various
heights above the ground. The results were recorded as the following
ordered pairs: (1,8), (4,16), (9,24).
The abscissa of an ordered Pair is
the height in feet and the ordinate of
an ordered pair is the speed in feet
per second. For examPle, the ordered pair (4,16) corresponds to
dropping the ball from a height of
4 ft and recording the speed as
16 ftls. The results are graphed at
Ec
o
o
25
aB
o.
zv
qq)
a"z
q)
o
15
10
(
c
0
2 34567BI
10
Height (in feet)
the right.
oo
I
Example
11
Strategy
Solution
A chemist measured the temperature of a chemical reaction
at different times and recorded
the results as the following ordered pairs: (5,10), (10,15),
and (20,25). The abscissa of
an ordered pair is time measured in minutes, and the ordinate of an ordered pair is temperature measured in degrees.
Graph the ordered pairs on the
rectangular coordinate system.
Graph the ordered Pairs (5,10),
(10,15), and (20,25).
(Doe
.r
ta
f (!)
^^
E9 t'
b s', 15
93
E-
o.=
to
',t)'
.
; o(10, 15)
o (5, 10)r
:
5
o
12
@
A biochemist recorded the
number of bacteria in a culture
at different times as the following ordered pairs: (1,8), (3,10),
and (4,11). The abscissa of an
ordered Pair is the time in
hours, and the ordinate of an
ordered Pair is the number of
bacteria. GraPh the ordered
pairs on the rectangular coordi-
o
o
d
E
IE
C
d
c
@
c
nate system.
Your strategy
Your solution
:l::
(3oi
Example
5 10152025
Time (in minutes)
(!
q)
o
(!
o
o
o
_o
E
f
z
o
!t
!i
Time (in hours)
E
o
c
I
t
o
ah
UN
lT 8
297
Graphs and Linear Equations
1.1 Exercises
1.
Graph the ordered pairs (-2,1),
(3,-5), (-2,4), and (0,3).
Graph the ordered pairs
(0,-5), (-3,0), and (0,2).
(0,0),
Graph the ordbred pairs (-1,4),
(0,2), and (4,0).
(-2,-3),
2.
Graph the ordered pairs
(5,-1),
(-3,-3), (-1,0), and (1,-1).
Graph the ordered pairs (-4,5),
(-3,1), (3,-4), and (5,0).
Graph the ordered pairs (5,2),
(-4,-1),
(0,0), and (0,3).
-1 t:2:0
l l , :4-i'z
i t
)
298
UNIT
7.
8
Graphs and Linear Equations
Find the coordinates of each of the
8.
Points.
9.
Find the coordinates of each of the
10.
Find the coordinates of each of the
points.
Find the coordinates of each of the
points.
points.
()o
x
o
@
o
@
d
E
o
L
oc
6
c
o
a
o
o
11.
a)
b)
Name the abscissas of points
A and C.
Name the ordinates of points
B and D.
12.
a)
b)
Name the abscissas of points
A and C.
Name the ordinates of points
B and D,
UN
lT 8
299
Graphs and Linear Equations
1.2 Exercises
13.
ls (3,4) a solution of
y-
-x + 7?
15. ls (-1,2) a
solution
17.
y
14. ls(2,-3)asolution oIy - x + 5?
16.
(1,-3) a
-2x - 1?
ls
y:lx_1?
ls (4,1) a solution ot
-
f,x +
1?
18. ls (-5,3) a
Y
19.
ls (0,4) a solution oI
y
- |x
+
+?
solution
solution
- -!x + 1?
ls
(-2,0)
y=
-+x
-
a
solution
1?
o
O
I
o
o
E
21.
ls (0,0) a solution ot
y
-
3x
+
2?
ls (0,0) a solution oI Y
: -1*?
E
uo
!
c
E
o
o
O
Find the ordered pair solution
y-3x-2
x-3.
corresponding
of
to
I
Find the ordered pair solution of
y=
f,x
X:
24, Find the ordered pair solution of
y - 4x I 1 corresponding to
-1
corresponding to
y:
]x - 2
corresponding to
X=4.
6.
pair solution of
corresPonding to
27. Find the ordered
y--3x+1
Find the ordered pair solution of
x=0.
28. Find the ordered pair solution of
y=
lx - 5
corresponding to
X=0.
Find the ordered pair solution of
y = f,x +
tr
2
corresponding to
Find the ordered pair solution of
Y
- -t, -,
x-12.
corresPonding to
300
UNIT 8
Graphs and Linear Equations
1.3 Application
Problems
Solve.
1. A physics
student, measuring the
amount of frictional force on a
smooth surface, recorded the following ordered pairs: (1 ,2), (2,3),
(3,5), (3,6). The first component of
an ordered pair is the distance in
feet the object slid across the sur-
2.
perature
in a town. The
results
were recorded as the following ordered pairs: (-3,3), (-1,2), (0,4),
(1,1). The first component of an
ordered pair is the temperature in
degrees and the second compo-
face, and the second component is
the force in pounds used to move
the object. Graph the
For ten consecutive winter days, a
meteorologist measured the tem-
nent is the number of days that tem-
perature was recorded during the
ten-day period. Graph the ordered
ordered
pairs.
pairs.
a
E
C
f
Number of
5
days
o
o_ 4
C
o
o
o
e
Temperature
()o
I
o
@
o
(in degrees)
o
2
LL
12345
E
a
L
!
6
c
o
C
o
O
Distance (in feet)
3.
The magnification of an object at
various distances from a lens was
recorded as the following ordered
pairs: (3,5), (s,10), (7,15), (9,20).
The first component of an ordered
pair is the magnification and the
second component is the distance
in centimeters.
Graph the ordered pairs.
from the object
The speed of a ball as it rolls down a
ramp is recorded as the following
ordered pairs: (0,0), (1,2), (2,4),
(3,6) The first component of an
ordered pair is the time in seconds
and the second component is the
speed in feet per second. Graph
the ordered pairs.
p
o
25
E
5
oo)
Q)
:20
:=
o
o
4.
U) 4
o)
QA\
15
a6
C
g10
.2
o5
o
o
J
2
1
357
Magnif ication
12345
Time (in seconds)
UNIT
SECTION 2
2.1 Obiective
8
301
Graphs and Linear Equations
Graphs of Straight Lines
To graph an equation of the fotrn 1t = mx I
b
ol an equation in two variables is a drawing of the ordered pair solutions
of the equatron. For a linear equation in two variables, the graph is a straight line.
To graph a linear equation, find ordered parr solutions of the equation. Do this by
The graph
choosing any value of x and f inding the corresponding value of y. Repeat this procedure, choosing different values for x, unttl you have found the number of solutions
desired.
Since the graph of a linear equation in two variables is a straight line, and a straight Iine
is determined by two points, it is necessary to find only two solutions. However, it is
recommended that at least three points be used to insure accuracy.
Graphy-2x+1.
o
()
I
o
o
o
o
E
uo
!
c
E
o
o
O
Choose any values of x and then find
the corresponding values of Y, The
numbers 0, 2, and -1 were chosen arbitrarily for x. lt is convenient to
record these solutions in a table.
The horizontal axis is the x-axis.
The vertical axis is the y-axis.
Graph the ordered Pair solutions
(0,1), (2 5), and (-1,-1)
Draw a line through the ordered
pair solutions.
0
2
1
2x+1
2.0 +1
2.2 +1
2(-1) +1
1
tr
1
-l
5
4
3
2
2345
1
-2
-3
r4
:5
Remember that a graph is a drawing of the ordered pair solutions of the equation'
Therefore, every point on the graph is a solution of the equation and every solution ol
the equation is a point on the graph.
(-2,-3)
and (1,3) are points
on the graph and that these points are
solutions of the equation Y 2x + 1
Note that
Y=2x+
-
1,3) r
l-1
-Z
!
-4
1:
I
UNIT 8
302
Graph
Step
Step
Graphs and Linear Equations
y=lx-1.
1
2
Find at least three solutions,
When m is a fractron, choose values of x
that will simPlifY the evaluatton.
Display the ordered pairs in a table.
-'l
0
3
0
-3 -2
Graph the ordered pairs on a rectangular
coordinate system and draw a straight line
through the points.
Examplel Graphy-3x-2.
Example
Your
2
Graph
solution
y
-
3x
+
y
-2
0
'l
oo
1
tr
-1
I
4
2
1.
a
o
6
E
uo
!
c
e
Example4Graphy--2x.
Example3 Graphy-2x.
Your
Example5 Graph
Solution x
0
2
I
y=f,x-1.
y
Example6 Graph
v
Your
solution
y
y=lx-3
y
-1
0
-2 -2
4
solution
a
c
o
(-)
1
ct)
a
Gi
c
o
o
c
.9
=
o
at
UNIT 8
2.2 Objective
303
Graphs and Linear Equations
To graph an equation of the forrn Ax
I
By
-
C
An equation in the form Ax I By : Q, 2x a 3y = 6 (A:2,8where A, B, and C are constants, is x - 2y - -4 (A:1,8(A - 2,8 also a linear equation. Examples of
2x I y : g
4x - 5y = 2 (A-4,8these equations are shown at the
right.
3, C: 6)
-2, C : -4)
1, C:0)
-5,C :2)
To graph an equation of the form Ax + By = C, first solve the equation lor y. Then
follow the same procedure used for graphing an equation of the form f - mx * b.
Graph 3x
+ 4y
:
12.
3x-y4y:12
Step 1 Solve the equation for y
4y--3x+12
a
v=-:X+3
o
O
Step
2
Find at least three solutions.
Display the ordered pairs in a table.
Step
3
Graph the ordered pairs on a rectangu-
lar coordinate system and draw
a
straight line through the points.
=
o
@
o
@
d
E
o
I
!
c
7o
C
oo
The graph of an equation in which one of the variables is missing is either a horizontal
or a vertical line.
The equation y =2 could be written 0'x + | = 2.
No matter what value of x is chosen, y is always 2.
Some solutions to the equation are (3,2), 1-1,2),
(0,2), and (-4,2). The graph is shown at the right.
The graph ol
y = b is a horizontal line passing
through point (0,b).
The equatiotl X = -2 could be written x + 0 . y = -2.
No matter what value of y is chosen, x"is always -2,
Some solutions of the equation are (-2,3), (-2,-2),
(-2,0), and (-2,2). The graph is shown at the right.
The graph ol
x = a is a vertical line
through point (a,0).
passing
304
UNIT
Example
7
Graph 2x
-
8
5y
Graphs and Linear Equations
:10.
Example
Solution 2x - 5y :10
8
Graph 5x
-
2y
=
10.
Your solulion
-5y=-2x+10
y=tx-2
yv
-2
0
0
5
-5 -4
Example 9
Solutlon
xq2y=6.
xq2y=6
2y--x+6
Example
Graph
Y
10
Graph
x
-
3y
-
9.
Your solution
- -lx + e
X
0
2
-2
4
3
2
4
o
O
I
1
i
I
:
: -
:-a
@
o
@
d
E
u
Example
11
Solutlon
Example
13
Solution
Graph
y = -2.
The graph of an equation of the
lorm y = b is a horizontal line
passing through point (0,b).
Graph X
=
3.
The graph of an equation of the
form x = a is a vertical line
passing through point (a,O).
Example
12 Graph y :3,
Your solution
Example
oc
o
E
o
F
d
14 Graph X = -4.
Your solutlon
gt
!i
ci
tr
o
o
E
B
!o
o
UNIT 8
305
Graphs and Linear Equations
2.1 Exercises
Graph.
1. Y '2x-3
3.
y=I,
5. y =Z^
4, y--3x
-
1
6. y=]x+2
::4 t-2:
7.
v--f,x+z
v
. ,ll:4
i i i r4l
y--]x+r
306
UNIT I
Graphs and Linear Equations
Graph.
::4 t-2
11.
t
y-2x-4
12.
=a :-?. i
! _"..t
15.
y=-!x+
,'!
i-.?.
:
'-..l '- -i_
_j.._
16. f=5x-4
UNIT 8
307
Graphs and Linear Equations
2.2 Exercises
-4
'-2
0
i. ...i. t. -4
19.
2x + 3y
:6
20. 3x+2y-4
v
v
.
21.
x-2y=4
:... 1.. 1.......1......! a
22. x -3y =6
308
UNIT
I
Graphs and Linear Equations
Graph.
214
27.
29. X=3
'>4 :-2 i0
: -",i,, .-.i'-.'-. n - ;-"-
30.
y=-4
.>4 .-2
-"i-.-:-i-- !
i
-:-C
4x-3y=12
UNIT I
SECTION 3
3.1 Obiective
Graphs and Linear Equations
309
Intercepts and Slopes of Straight Lines
To find the x- and y-intercepts of a straight line
The graph of the equation 2x + 3y = 6 is shown at
the right. The graph crosses the x-axis at the point
(3,0) This point is
called the x-intercept. The
graph also crosses the y-axis at the point (0,2). This
point is called the y-intercept.
Find the x-intercept and the y-intercept of the graph of the equation 2x
To find the x-intercept, let
y: O.
x-coordinate
2x-3y=12
:
12
O.)
2x-3y-12
2(0)-3y:12
-3y - 12
2x-3(0)=12
2x=12
x -G
y
(6,0).
a
O
The x-intercept is
I
o
@
o
Find the y-intercept of
A
-
-a
The y-intercept is
(0,-4).
y
- 3x + 4.
Y:3x+4
o
6
E
3y
To find the y-intercept, let x = 0,
(Any point on the y-axis has
(Any point on the x-axis has
y-coordinate 0.)
Letx=0.
o
I
!
c
-
=3(0)+4
-4
c
The y-intercept is (0,4).
o
O
For any equation of the form
y : mx + b, the y-intercept is (0,b).
A linear equation can be graphed by finding the x- and y-intercepts and then drawing
a line through the two points.
Example
1
Find the x- and y-intercepts for
- 2y - 4. Graph the line.
Example
2
x
Solution
x-intercept:
x-2y-4
x-2(0)=4
X=4
y-intercept:
x-2y=4
0-2y-4
-2y-4
y
(4,0)
-
-L
Find the x- and y-intercepts for
- y - 4. Graph the line.
4x
Your solution
a
(0,-2)
c
o
rt
q
C
o
o
t
oo
UNIT
310
Example
3
Solution
8
Graphs and Linear Equations
Find the x- and y-lntercepts for
- 2x - 4. Graph the line.
Example
4
y
x-intercept:
y-2x-4
0'-2x-4
-2x - -4
y-intercept:
(0,b)
Find the x- and y-intercepts for
Graph the line.
y - 3x - 6.
Your solution
b- -4
(0,
X=2
-4)
(2,0)
4
o
€
ti
ci
c
o
-g
6
at
3.2 Obiective
To find the slope of a straight line
Thegraphs
oIy-!x+l
andy=2xq.1
are
shown at the right. Each graph crosses the y-axis at
the point (0,1), but the graphs have different slants.
The slope of a line is a measure of the slant of a line.
The symbol for slope is m.
The slope oJ a line containing two points is the ratio
of the change in the y values of the two points to the
change in the x values. The line containing the
points (-2,-3) and (6,1) is graphed at the right.
The change in the y values is the difference between
the two ordinates.
Changeiny- 1 -(-3)=4
The change in the x values is the difference between
the two abscissas.
Changeinx:6 -(-2)=B
changeiny 4
changeinx B
Slope=ffi=-.L
1
2
A formula for the slope of a ltne containing
two
points, P, and Pr, whose coordinates are (x,,y-') and
(xr,yr), is given by:
Slooe=m=fz-ft
'
x2-xt
UN
lT I
Graphs and Linear Equations
311
Find the slope of the line containing the points
(-1,1)
Let P., = (- 1 ,1) and P2 = (2,3).
(lt does not matter which point is
named P., or Pr', the slope will be
the same.)
v^-v.
Xz-X,
2
-(-1)
and q,3)
;.i
,,,
j
.
3
positive slope
A line which slants upward to the right always has a positive stope.
Find the slope of the line containing the points
Let P,
- (-3,4)
and
(-3,4)
and
(2,-2).
P2: (2,-Z).
-2-4
^-fz-ltXz-Xt - 2-(_3)
-6
=_=__
5
y
6
S
oo
:I
negative slope
o
@
o
@
d
A line which slants downward to the right always has a negative slope.
E
o
oc
6
C
o
c
o
o
Find the slope of the line containing the points
Let P. =
(-1,3)
and P2
(-1,3)
and (4,3).
= @,3).
3;3,
Jn-fz-Yt
-9-O
Xz-Xr = 4-\-1)
b
zero slope
A horizontal line has zero slope.
Find the slope of a line containing the points
= (2,-2) and Pr: (2,4).
4. (-.2)
, -Yz-lr
Xz-xt- 2-2 -o Nota
(2,-2)
and (2,4).
Let P.,
0
A vertical line has no slope.
l- l-(2,4)^i
i
real
i -i --i
i:.." i
tii
numOei
no slope
UNIT I
312
Example 5
Graphs and Linear Equations
Example
Find the slope of the line containing the points (-2,-1) and
6
= (-2,-1)
Let P.'
P2:
(-1,2)
and
(1,3).
(3,4),
Solution
Find the slope of the line con-
taining the points
Your solution
and
(3,4).
4-(-1)
5
^ '-x._-\-3-GA-5-'
-Yz-lt
-
1
The slope is 1.
Example 7
Find the slope of the line containing the points (-3,1) and
Example
8
(2,-2).
Solution
Let P., =
Find the slope of the line containing the points (1,2) and
(4,
(-3,1)
Ihe slope,t
Your solution
and
P, = (2,-2).
v--v.t1
t2
n
-2-1
- Xz-Xt - 2-(-3)
-5)
-3
5
oo
-;.
I
@
o
o
Example 9
Find the slope of the line containing the points (-1,4) and
(-
Solution
1
Example
10
Find the slope of the line containing the points (2,3) and
(2,7).
,0).
Let P, = (-1,4) and
P, = (-1,0).
0-4
^ -Yz-Yt "'-lf,i--1-i:t-
c
t
o
I
E
C
G
c
s)
oo
Your solution
-4
o
The line has no slope.
Example
11
Find the slope of the line containing the points (-'1,2) and
Example
12
(-5,-3).
(4,2).
Solution
Let
P,
P'
=
-
(-'1,2) and
Find the slope of the line con-
taining the points
(1,-3)
and
Your solution
(4,2).
^=';+=ffi=$=o
The line has zero slope.
o
aI
ct
(a
!t
a
c
o
6
c
:
.9
E
at
UNIT
3.3
Obiective
Graphs and Linear Equations
8
313
To graph a line using the slope and y-intercept
- 3, * t ,.
shown at the right. The points (-3,-1)
The graph of the equation
y
and (3,3) are on the graph. The slope
of the line is:
3-(-1)
,',__
- _ 3-(-3) _46 _23
Note that the slope of the line has the
same value as the coef{icient of x.
| :
For any equation of the form
trlx 1 b, the slope of the line is m, the coefficient of
The y-intercept is (0,b). Thus, the torm y = fftX
b is called the slope-intercept
x.
*
form of a straight line.
Find the slope and the y-intercept of the line
v
'o
o
@
o
d
E
Slope
I!
6
c
Ic
-]x + t.
=filx + [t
*
'= l-* l" L!
I
o
y=
=
n1
The slope
is
3
-4'
l-
1
= -q
y-intercepl = (0, b) = (0, 1)
The y-intercept is (0,1)
o
O
When the equation of a straight line is in the form y = mx * b, the graph can be
drawn using the slope and y-intercept. First locate the y-intercept. Use the slope to
find a second point on the line. Then draw a line through the two points.
Graphy-2x-3.
y-intercept
:
(0,b) = (0,-3)
m_D_?_changeiny
I"-L
-1-;r*q.,
Beginning
right
at the y-intercept,
1 unlt (change in
move
x) and then up 2
units (change in y).
(1
,
- 1) is a second pornt on the graph.
Draw a line through the two points
-3) and (1 , - 1 ).
(0,
(1
, -1)
lup2
J;i
i
314
UN
Example 13
Graph
lT 8
Graphs and Line4r Equations
y- -!x+l
byusing
Example 14
the slope and y-intercePt.
the slope and y-intercePt.
Solution
Your solution
(0,b) = (0,1)
(Move right 3 units,
-3- then down 2 units.)
y-intercept
'It
Graphy=-Ir-lbyusing
=
2
- -3 - -2
v
risht3l
I :
i-ldown
Example 15
Graph
y
- -1r
I
:
OU using the
Example 16
y-intercept
:
,tr--4- 3_-3
y
- -3" OU using the
slope and y-intercePt.
slope and y-intercePt,
Solution
Graph
(0,b) = (0,0)
Your solution
_a
4
t
o
@
-4
6
F
I
1)
c
.....j.;.... i......
i
c
o
c
o
o
0
,:-3 -2: -1
-.i
::i;
i' i i -i-2
<
Example 17
Graph 2x
-
3y
-
6 by using
Example 18
Solutlon
Solve the equation for y.
Graph
x
-
2y
-
4 by using the
slope and y-intercePt.
the slope and y-intercept.
Your solution
2x-3y=6
-3y=-2x+6
v=lx-2
y-intercept
v
= (0,-2)
^ =3
o
l'
ri
c
o
o
t
o
o
o
UN
lT I
Graphs and Linear Equations
315
3.1 Exercises
Find the x- and y-iniercepts
1. X-L=3
2.
4' y=2x+10
5. x-5y=10
6.
7. y-3x+12
8. Y:5x+10
9. 2x-3y-O
10. 3x+4y-0
11.
3x + 4y
:12
y--!x+s
3' l:3x-G
12.
o
O
I
o
o
6
Find the x- and y-intercepts and graph.
E
o
L
!
z
13. 5xa2y=10
@
v
c
o
O
15.
y=|x-3
::q
i2
:-4
i*i
14. x-3y=6
3x + 2y
:12
y:lx-a
UNIT 8
316
Graphs and Linear Equations
3.2 Exercises
Find the slope of the line containing the points.
17. P{4,2), P2G,4)
18.
P1Q,1), P2G,4)
19. P,(-t,3),
20. P{-2,1),
P2Q,2)
21.
P{2,4), P2(4,-1)
22.
23. P{-2,3),
P2(2,1)
24. P{5,-2),
P,(1,0)
P1(1,3),
P2(2,4)
P2(5,-3)
25. Pl(8,-3), P29,1)
o
O
z
-o
@
o
26.
P1(0,3),
PzQ,-1)
27. P{3,-4), P13,5)
28, P,(-r
o
,2), P2G1,3) (
E
Io
!
6
c
c
o
o
32.
P{4,-2), P2(3,-2)
30.
P1(3,0); P2(2,-1)
33. P{-2,3), P2(J,3)
P1(5,1), P2G2,1)
35. P{-2,4), P2G1,-1) 36.
P1(6,
38.
P.,(-1,5), P,(5,1)
Pl(5,1), P2(5,-2)
39.
-4), P2G,-2)
31.
P,(0,
- 1), P2(3,-2)
34. Pr(4,-1), Pr(-3,-1)
37. P,(-2,-3),
40. P,(-
P2(-2,1)
t ,5), P2Q ,1)
UNIT
8
317
Graphs and Linear Equations
3.3 Exercises
Graph by using the slope and y-intercept.
41.
y-3x+1
44. y
45. 2xay=3
47.
x-2y=4
=|x
+
1
46. 3x-y=1
318
s0. , =L^
o
o
I
o
o
o
E
uo
E
C
o
c
oo
56.
UNIT 8
SECTION 4
319
Graphs and Linear Equations
Equations of Straight Lines
4.1 Objective
To
i"d
the equation of a line using the
equation):mx *
b
When the slope of a line and a point on the line are known, the equation of the line can
be written using the slope-intercept form,
y
-
mx
a
b.
Find the equation of a line which has slope 3 and y-intercept (0,2).
The given slope, 3, is
Replace m with 3.
m.
y-mxqb
y-3x+b
l
The given point, (0,2), is the y-intercept.
Replace b with
rr-?vtD
v,\
,/
2.
Find the equation of the line which has slope
I
-
|
L
and contains point
(-2,4)
y-mx1b
The given stope,
o
o
I
],
y:Lr+o
is m.
Reolace
m with 1.
'2
@
o
c
The given poinl, (-2,4) is a solution
of the equation of the line.
Replace x and y in the equation with
the coordinates of the point.
o
O
Solve for b, the y-intercept.
o
6
E
o
L
!
a
6
Ic
4=t(-2)+o
4=-l
5=b
y-mxab
Write the equation of the line by
replacing m and b in the equation
by their values.
Example
1
(3,-3)
Solution
t
Find the equation of the line
which contains the point
and has sloOe
t
l=ix+o
-s:fta)+o
-3=2+b
tr_A
f
+O
Example
2
-
Find the equation of the line
which contains the point
(4,-2)
,
1_
vf | va,
2'\
and has stope
|.
Your solution
(\l
ao
tc
t -
t_
J^'\
tt
c
o
g
o
=o
tt
UNIT
920
4.2
Obiective
8
Graphs and Linear Equations
To find the equation of a line using the point-slope formula
An alternate method for finding the equation of a line, given the slope and a point on
the line, involves use of the point-slope formula. The point-slope formula is derived
from the formula for slope.
Let (x.',/r) be the given point on the line
and(x,l) be any other point on the
line.
r
t1 _ h
Itt
X-Xt
-
Formula for slope
Yr-
Multiply both sides of the
equation by (x - x.,).
Y,r.
X-Xr-
(r -- xr)
-
ntrx
-
x,)
y-yt-m(x-xr)
Simplify.
= m(x 'x,tl.
ln this equation, m is the slope and (xt,y) is the given point.
The point-slope lormula is
y
- lt
Find the equation of a line which passes through point
(x.,,y.) =
(-3,2)
(-3,2)
and has slope
l-h-m(x-xr)
y -z =f t" - <-slt
^ =3
oo
I
o
@
y-2:!1x+3)
I
o
y-2=lx+2
y
The equation of the line is
Example
3
y=
to
line
point
(-2,-l) and has slope f .
Use the point-slope formula
find the equation of a
which passes through
(x.,,y.,):(-2,-1)
y-yt-m(x-xr)
Solution, =*
=!x
+
f,.
G
E
I
!
c
a
c
o
c
o
o
lx
+
A
Example
4
Use the point-slope formula to
find the equation of a line
which passes through point
g,-2)
and has stooe
f
.
Yoursolution
y_(_1)=8tx_(_2)l
r*1=|fr+21
Y+1-|x+s
y=|x+2
The equation of the line
\/
'z
:1x
+
6t
o
$
is
ri
c
o
z.
o
3o
at
UNIT
Graphs and Linear Equations
8
321
4.1 Exercises
Use the slope-iniercept form.
1.
Find the equation of the line which
contains the point (0,2) and has
slope 2.
3.
Find the equation of the line which
contains ihe point (-1 ,2) and has
2.
slope
4.
slope *3.
o
O
5.
Find the equation of the line which
contains the point (3,1) and has
I
o
o
stope
o
Find the equation of the line which
contains the point (0, 1) and has
-
-
2.
Find the equation of the line which
contains the point (2,-3) and has
slope 3.
6.
I
Find the equation of the line which
contains the point (-2,3) and has
slope
].
E
o
L
!
C
c
o
c
o
O
7.
Find the equation of the line which
contains the point (4,-2) and has
stope
9.
11.
-
Find the equation of the line which
contains the point (2,3) and has
stope
f;
Find the equation of the line which
contains the point (5, -3) and has
stope
8.
10.
Find the equation of the line which
contains the point (5, i ) and has
-
stope
f;.
Find the equation of the line which
- |.
12,
|,
contains the point (2,3) and has
Find the equation of the line which
contains the point (-1,2) and has
slope
slope
].
-|
322
UN
lT 8
Graphs and Linear Equations
4.2 Exercises
Use the point-slope formula.
13.
15.
17.
Find the equation of the line which
passes through point (1, -1) and
has slope 2.
14.
Find the equation of the line which
passes through point (-2,1) and
has slope -2.
16.
Find the equation of the line which
18.
Find the equation of the line which
passes through the point (2,3) and
has slope
-.l.
Find the equation of the line which
passes through point
and has slope
(-1,-3)
-3.
Find the equation of the line which
o
passes through point (0,0) and
O
passes through point (0,0) and >t
TI
has slope
has slope
o\
o
f,.
- |.
@
d
E
uo
E
C
6
o
c
o
O
19.
Find the equation of the line which
passes through point (2,3) and
has slope
21.
23.
-f
a
f
.
22,
24.
f,.
Find the equation of the line which
passes through point (-5,0) and
has slooe
.
Find the equation of the line which
passes through point (-2,1) and
has slope
Find the equation of the line whtch
passes through point (3,-1) and
has slope
|.
Find the equation of the line which
passes through point (-4,1) and
has slooe
20.
- ].
Find the equation of the line which
passes through point (3,-2) and
has slope
|.
UNIT 8
Review
SECTION
1
1.1
323
Graphs and Linear Equations
/Test
Graph the ordered pairs
and (0,2).
(-3,1)
Find the ordered pair solulion of
1.2
y
u^
- -|xJ + 2corresponding
v_o
1.3
The costs for an amount of energy were recorded as the follow-
ing ordered pairs: (1 ,5), (2,10),
(4,20), and (5,25), where the
abscissa is the number of kiloI
watt-hours used and the ordinate
is the cost in cents. Graph the
o
ordered pairs.
o
o
a
25
o
20
c
C)
c
u)
o
O
15
10
E
0
12345
Kilowatt-hours
o
E
uo
oC
c
Ic
o
c)
SECTION
2
2.1a Graphy-3x+1.
2.1b
Graphy--]x+s.
2.2b Graphx*3=0.
to
324
UNIT I
Graphs and Linear Equations
Review
SECTION
3
/Test
3.1a Find the x-
and y-intercept for
6x-4y-12.
Y=*x+t'
Find the slope of the line contain-
ing the points
(2,-3)
3.1b Find the x- and y-intercept for
3.2b
and (4,1).
Find the slope of the line containing the points (3, -4) and
(1 ,
3.3a
Graph the line which has slope
- anO y-intercept (0,4).
f
SECTION
4
4.1a Find the equation of the tine
which contains the point (0,-1)
and has slope 3.
4.2a
Find the equation of the line
which contains the point (2,3)
and has stope
|.
3.3b
-4).
Graph the line which has stope 2
and y-intercept -2.
4.1b Find the equation of the line
which contains the point (-3,1)
and has slope
$.
4.2b Find the equation of the line
which contains the point (-1,2)
and has stope
-f
.
UN
lT 8
325
Graphs and Linear Equations
Review/Test
SECTION
1
1.1
Which point
(-2,-3)?
coordinates
has
1.2
a)A
b)B
a)
b)
c)
d)
d)D
1.3
o
o
2
(-2,3)
(-2,-5)
(-2,5)
(-2,-3)
:
A physicist recorded the distances a ball would rolldown a ramp as the following ordered pairs: (1,1), (2,5), (3,7). The abscissa istime in seconds and the
ordinate is distance in feet. What is the distance the ball rolled in 2 s?
b) 3ft
a) 1ft
SECTION
o
A--L.
c)c
:3-2-1ol 1 2 3;
, 1l :
' :
,I::
eQ,,-21
Find the ordered pair solution of
- 2x - 1 corresponding to
y
21a Which is
the
y-2x-1?
a)yb)
graph
c) sft
of
d)
2.1b Which is the graPh
,r-.
Y
-^1
a)yb)Y
v
7t1
t-lJ
I
o
@
d
E
o
I
o
c
6
c
o
o
O
2.2a
a)
Which
is the
2x-3y -6?
v
b)v
graPh
ot
2.2b
a)
Which
ts
y-3=0?
the
b)
graph
v
of
UNIT 8
326
Graphs and Linear Equations
Review/Test
SECTION
3
3.1a
3.2a
Name the x-intercept for the line
2x-3y=12.
a) (6,0)
c) (0, -3)
b)
d)
(0,
-4)
3)
a)l
c)5
b)
7
5
d)
-7
Name the y-intercept for the line
5x+2y=20.
a)
c)
(2,0)
Find the slope of the line containpoints (2, ing
and
(
3,4).
-
the
3.1b
3.2b
(0,2)
(5,0)
b)
d)
(0,10)
(4,0)
Find the slope of the line contain-
ing the points (2,3) and
a)4
c) no slope
(-2,3).
b) -4
d)0
5
Which is the graph of the line
which has slope -1 and y-intercept (0,2)?
vb)
3.3b
Which is the graph of the line
which has stoOe and y-intercept
v
(0,-l)?
I
vb)
a)
0
SECTION
4
4.1a Find the equation of the
which contains the point
and has slope -3.
a) y = -3x a2
b) y-bx-2
c) Y:2x-3
d) l:*2x+3
line
(Q,2)
4.1b Find the equation of the line
which contains the point (2,-1)
and has stope
|.
a) y_rx_1
-
.t
D)
13
y=Zx*z
c),15Y=Zx-fz
..
y_rx_2
d)
1
4.2a Find the equation of the
line
which contains the point (6,1)
and has slope ?
3'
.t
a)
y:ix+1
b) y:tr+a
c) y=lx-3
d) v=?r-t
4.2b Find the equation of the line
which contains the point
and has slope 2.
a) y=2x-1
b) y=2xt2
c) y =2x
d) y=2x-3
(-1,0)