Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 3 A4 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
10
G
lenc
oe G
eom
etry
Enri
chm
ent
Sp
heri
cal
Geo
metr
y
On
a m
ap, l
ongi
tude
an
d la
titu
de a
ppea
r to
be
lin
es. H
owev
er, l
ongi
tude
an
d la
titu
de e
xist
on
a s
pher
e ra
ther
th
an o
n a
fla
t su
rfac
e. I
n o
rder
to
accu
rate
ly a
pply
geo
met
ry t
o lo
ngi
tude
an
d la
titu
de, w
e m
ust
con
side
r sp
her
ical
geo
met
ry.
Th
e fi
rst
fou
r ax
iom
s in
sph
eric
al g
eom
etry
are
th
e sa
me
as t
hos
e in
th
e E
ucl
idea
n
Geo
met
ry y
ou h
ave
stu
died
. How
ever
, in
sph
eric
al g
eom
etry
, th
e m
ean
ings
of
lin
es a
nd
angl
es a
re d
iffe
ren
t.
1. A
str
aigh
t li
ne
can
be
draw
n b
etw
een
an
y tw
o po
ints
.
How
ever,
a s
traig
ht
line in s
pherical geom
etr
y is a
gre
at
cir
cle
. A
gre
at
circle
is a
circle
that
goes a
round t
he s
phere
and c
onta
ins t
he d
iam
ete
r of
the s
phere
.
2. A
fin
ite
lin
e se
gmen
t ca
n b
e ex
ten
ded
infi
nit
ely
in b
oth
dir
ecti
ons.
A lin
e o
f in
finite length
in s
pherical geom
etr
y w
ill g
o a
round its
elf a
n infinite n
um
ber
of
tim
es.
3. A
cir
cle
can
be
draw
n w
ith
an
y ce
nte
r or
rad
ius.
So,
in s
pherical geom
etr
y,
a g
reat
circle
is b
oth
a lin
e a
nd a
circle
.
4. R
igh
t an
gles
can
be
fou
nd
on t
he
sph
ere.
Latitu
de a
nd longitude m
eet
at
right
angle
s o
n a
sphere
.
Th
e fi
fth
axi
om o
f E
ucl
idea
n G
eom
etry
sta
tes
that
giv
en a
ny
stra
igh
t li
ne
and
a po
int
not
on
it,
th
ere
exis
ts o
ne
and
only
on
e st
raig
ht
lin
e th
at p
asse
s th
rou
gh t
hat
poi
nt
and
nev
er
inte
rsec
ts t
he
firs
t li
ne.
Th
e fi
fth
axi
om i
s al
so k
now
n a
s th
e P
aral
lel
Pos
tula
te.
Exer
cise
s 1
. Get
a b
all.
Wra
p tw
o ru
bber
ban
ds a
rou
nd
the
ball
to
repr
esen
t tw
o li
nes
(gr
eat
circ
les)
on
th
e sp
her
e. H
ow m
any
poin
ts o
f in
ters
ecti
on a
re t
her
e?
2. T
ry t
o dr
aw t
wo
lin
es (
grea
t ci
rcle
s) o
r w
rap
two
rubb
er b
ands
aro
un
d a
ball
th
at d
o n
ot
inte
rsec
t. I
s it
pos
sibl
e?
3. M
ake
a co
nje
ctu
re a
bou
t th
e n
um
ber
of p
oin
ts o
f in
ters
ecti
on o
f an
y tw
o li
nes
(gr
eat
circ
les)
in
sph
eric
al g
eom
etry
.
4. D
oes
the
fift
h a
xiom
, or
Par
alle
l P
ostu
late
, hol
d fo
r sp
her
ical
geo
met
ry?
Exp
lain
.
3-1
T
wo
lin
es (
gre
at
cir
cle
s)
will
alw
ays i
nte
rsect
in t
wo
po
ints
in
sp
heri
cal
geo
metr
y.
N
o,
becau
se t
he P
ara
llel
Po
stu
late
s s
tate
s t
hat
the l
ine w
ill
never
inte
rsect
an
d t
hat
is n
ot
po
ssib
le i
n s
ph
eri
cal
geo
metr
y b
ecau
se
two
lin
es (
gre
at
cir
cle
s)
alw
ays i
nte
rsect
in t
wo
po
ints
.
2
no
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Lesson 3-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
11
G
lenc
oe G
eom
etry
3-2
Stud
y G
uide
and
Inte
rven
tion
An
gle
s a
nd
Para
llel
Lin
es
Para
llel L
ines
an
d A
ng
le P
airs
Wh
en t
wo
para
llel
lin
es a
re c
ut
by a
tra
nsv
ersa
l, th
e fo
llow
ing
pair
s of
an
gles
are
con
gru
ent.
• co
rres
pon
din
g an
gles
• al
tern
ate
inte
rior
an
gles
• al
tern
ate
exte
rior
an
gles
Als
o, c
onse
cuti
ve i
nte
rior
an
gles
are
su
pple
men
tary
.
In
th
e fi
gure
, m∠
2 =
75.
Fin
d t
he
mea
sure
s
of t
he
rem
ain
ing
angl
es.
m∠
1 =
105
∠
1 an
d ∠
2 fo
rm a
lin
ear
pair
.m
∠3
= 1
05
∠3
and
∠2
form
a l
inea
r pa
ir.
m∠
4 =
75
∠4
and
∠2
are
vert
ical
an
gles
.m
∠5
= 1
05
∠5
and
∠3
are
alte
rnat
e in
teri
or a
ngl
es.
m∠
6 =
75
∠6
and
∠2
are
corr
espo
ndi
ng
angl
es.
m∠
7 =
105
∠
7 an
d ∠
3 ar
e co
rres
pon
din
g an
gles
.m
∠8
= 7
5 ∠
8 an
d ∠
6 ar
e ve
rtic
al a
ngl
es.
Exer
cise
sIn
th
e fi
gure
, m∠
3 =
102
. Fin
d t
he
mea
sure
of
each
an
gle.
Tel
l w
hic
h p
ostu
late
(s)
or t
heo
rem
(s)
you
use
d.
1. ∠
5 2.
∠6
3. ∠
11
4. ∠
7
5. ∠
15
6. ∠
14
In t
he
figu
re, m
∠9
= 8
0 an
d m
∠5
= 6
8. F
ind
th
e m
easu
re
of e
ach
an
gle.
Tel
l w
hic
h p
ostu
late
(s)
or t
heo
rem
(s)
you
use
d.
7. ∠
12
8. ∠
1
9. ∠
4 10
. ∠3
11. ∠
7 12
. ∠16
pq
m n
12 3
4
65
78
910 11
12
1413
1516
p
m n
12 3
4 65
78
wv
p q
12 3
4
65
78
910 11
12
1413
1516
Exam
ple
102;
Alt
. In
t. A
ng
les T
h.
78;
Co
ns.
Int.
102;
Co
rre.
An
gle
s T
h.
102;
Co
rre.
An
gle
s T
h.
102;
Co
rre.
An
gle
s T
h.
78;
Co
ns.
Int.
An
gle
s T
h;
Co
rre.
An
gle
s T
h.
100;
Su
pp
. A
ng
les
80;C
orr
. A
ng
les
Th
.
100;
Co
ns I
nt.
An
gle
s T
h.
80;
Att
. In
t.A
ng
les T
h.
68;
Vert
ical
An
gle
s T
h.
112;
Vert
ical
An
gle
s T
h;
Co
ns.
Inte
rio
r A
ng
les T
h.
An
gle
s T
h.
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Answers (Lesson 3-1 and Lesson 3-2)
A01_A21_GEOCRMC03_890512.indd A4A01_A21_GEOCRMC03_890512.indd A4 5/27/08 2:42:10 AM5/27/08 2:42:10 AM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 3 A5 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
12
G
lenc
oe G
eom
etry
3-2
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
An
gle
s a
nd
Para
llel
Lin
es
Alg
ebra
an
d A
ng
le M
easu
res
Alg
ebra
can
be
use
d to
fin
d u
nkn
own
val
ues
in
an
gles
for
med
by
a tr
ansv
ersa
l an
d pa
rall
el l
ines
.
If m
∠1
= 3
x +
15,
m∠
2 =
4x
- 5
, an
d m
∠3
= 5
y,
fin
d t
he
valu
e of
x a
nd
y.
p ‖ q
, so
m∠
1 =
m∠
2 be
cau
se t
hey
are
co
rres
pon
din
g an
gles
.
m∠
1 =
m∠
2
3x +
15
= 4
x -
5
3x +
15
- 3
x =
4x
- 5
- 3
x
15
= x
- 5
15
+ 5
= x
- 5
+ 5
20
= x
pq
r s
12
34
r ‖ s
, so
m∠
2 =
m∠
3 be
cau
se t
hey
are
co
rres
pon
din
g an
gles
.
m∠
2 =
m∠
3
75
= 5
y
75
−
5
= 5y
−
5
15
= y
Exer
cise
sF
ind
th
e va
lue
of t
he
vari
able
(s)
in e
ach
fig
ure
. Exp
lain
you
r re
ason
ing.
1.
( 5x
- 5
) °( 6
y-
4) °
( 4x
+ 1
0)°
2.
( 1
5x+
30)
°
( 3y
+ 1
8)°
10x°
90°
3.
( 11x
+ 4
) °
( 13y
- 5
) °( 5y
+ 5
) °
5x°
4.
( 5x
- 2
0)°
3x°
2y° 4y
°
Fin
d t
he
valu
e of
th
e va
riab
le(s
) in
eac
h f
igu
re. E
xpla
in y
our
reas
onin
g.
5.
2y°
106°
x°( 4
z+
6) °
6.
2x°
2y°
90°
x°
z°
Exam
ple
x =
15;
y =
19;
use c
orr
esp
on
din
g
an
d s
up
ple
men
tary
an
gle
s
x =
11;
y =
10;
use
co
nsecu
tive i
nte
rio
r an
gle
s
x =
74;
y =
37;
z =
25;
use c
on
secu
tive i
nte
rio
r, c
orr
esp
on
din
g,
an
d s
up
ple
men
tary
an
gle
s
x =
6;
y =
24;
Use c
on
secu
tive
inte
rio
r an
gle
s
x =
10;
y =
25;
Use c
on
secu
tive
inte
rio
r an
d a
ltern
ate
in
teri
or
an
gle
s
x =
30;
y =
15 ;
z =
150 u
se
su
pp
lem
en
tary
, alt
ern
ate
in
teri
or,
an
d c
on
secu
tive i
nte
rio
r an
gle
s
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Lesson 3-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
13
G
lenc
oe G
eom
etry
Skill
s Pr
acti
ceA
ng
les a
nd
Para
llel
Lin
es
In t
he
figu
re, m
∠2
= 7
0. F
ind
th
e m
easu
re o
f ea
ch a
ngl
e.
1. ∠
3 2.
∠5
3. ∠
8 4.
∠1
5. ∠
4 6.
∠6
In t
he
figu
re, m
∠7
= 1
00. F
ind
th
e m
easu
re o
f ea
ch a
ngl
e.
7. ∠
9 8.
∠6
9. ∠
8 10
. ∠2
11. ∠
5 12
. ∠11
In t
he
figu
re, m
∠3
= 7
5 an
d m
∠10
= 1
05. F
ind
th
e m
easu
re
of e
ach
an
gle.
13. ∠
2 14
. ∠5
15. ∠
7 16
. ∠15
17. ∠
14
18. ∠
9
Fin
d t
he
valu
e of
th
e va
riab
le(s
) in
eac
h f
igu
re. E
xpla
in y
our
reas
onin
g.
19.
(5x)
°
40°
( 3y
- 1
) °
20
.
(7x)
°
( 8x
- 1
0)°
( 6y
+ 2
0)°
21.
(9x
+21
)°
(11x
-1)
°
(5y
-5)
°
22
.
(4y
+4)
°60
°
(3x
-3)
°
q
r s
12 4
3 65
87
sm
ut
12 43
65 8
910 11
127
3-2
x
w
z
y
12
43
65
8
910
1112
1314
1516
7
70
110
110
110
110
70
100
80
100
80
80
100
105
105
75
105
105
75
x =
28,
y =
47;
Use t
he
su
pp
lem
en
tary
an
gle
s t
o f
ind
x.
Th
en
use a
ltern
ate
exte
rio
r an
gle
s t
o f
ind
y.
x =
10,
y =
15;
Use a
ltern
ate
in
teri
or
an
gle
s t
o f
ind
x.
Th
en
use
su
pp
lem
en
tary
an
gle
s t
o f
ind
y.
x =
11,
y =
13;
Use c
orr
esp
on
din
g
an
gle
s t
o f
ind
x.
Th
en
use
su
pp
lem
en
tary
an
gle
s t
o f
ind
y.
x =
21,
y =
29;
Use a
ltern
ate
in
teri
or
an
gle
s t
o f
ind
x.
Th
en
use
su
pp
lem
en
tary
an
gle
s t
o f
ind
y.
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Answers (Lesson 3-2)
A01_A21_GEOCRMC03_890512.indd A5A01_A21_GEOCRMC03_890512.indd A5 5/27/08 2:42:13 AM5/27/08 2:42:13 AM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 3 A6 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
14
G
lenc
oe G
eom
etry
Prac
tice
An
gle
s a
nd
Para
llel
Lin
es
In t
he
figu
re, m
∠2
= 9
2 an
d m
∠12
= 7
4. F
ind
th
e m
easu
re
of e
ach
an
gle.
Tel
l w
hic
h p
ostu
late
(s)
or t
heo
rem
(s)
you
use
d.
1. ∠
10
2. ∠
8
3. ∠
9 4.
∠5
5. ∠
11
6. ∠
13
Fin
d t
he
valu
e of
th
e va
riab
le(s
) in
eac
h f
igu
re. E
xpla
in y
our
reas
onin
g.
7.
3x°( 9
x+
12)
°
( 4y
- 1
0)°
8.
3y°
( 2x
+ 1
3)°
( 5y
- 4
) °
Fin
d x
. (H
int:
Dra
w a
n a
uxi
liar
y li
ne.
)
9.
100°
50°
1
10
.
144°
62° 1
11. P
RO
OF
Wri
te a
par
agra
ph p
roof
of
Th
eore
m 3
.3.
G
iven
: � ||
m, m
|| n
P
rove
: ∠1
� ∠
12
12. F
ENC
ING
A d
iago
nal
bra
ce s
tren
gth
ens
the
wir
e fe
nce
an
d pr
even
ts
it f
rom
sag
gin
g. T
he
brac
e m
akes
a 5
0° a
ngl
e w
ith
th
e w
ire
as s
how
n.
Fin
d th
e va
lue
of t
he
vari
able
.
n
m
rs
12
43 6
5
8
910
1112
1314
1516
7
m n
12
34
65 7
8
�
k
910
1112
50°
y°
3-2
92;
Co
rr.
� T
h.
88;
Co
rr.
� T
h,
Su
pp
∠s
106;
Su
pp
. �
Sam
ple
pro
of:
It i
s g
iven
th
at
� ‖
m,
so
∠1 �
∠8 b
y t
he A
ltern
ate
E
xte
rio
r A
ng
les T
heo
rem
. S
ince i
t is
giv
en
th
at
m ‖
n,
∠8 �
∠12 b
y t
he C
orr
esp
on
din
g A
ng
les P
ostu
late
. T
here
fore
, ∠
1 �
∠12,
sin
ce c
on
gru
en
ce o
f an
gle
s i
s
tran
sit
ive.
130
92;
Vert
. �
106;
Co
ns.
�
106;
Su
pp
. �
130
98
x =
14,
y =
37;
Use
Su
pp
lem
en
tary
an
d a
ltern
ate
exte
rio
r an
gle
s x =
28,
y =
23;
Use c
orr
esp
on
din
g
an
d s
up
ple
men
tary
an
gle
s
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Lesson 3-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
15
G
lenc
oe G
eom
etry
Wor
d Pr
oble
m P
ract
ice
An
gle
s a
nd
Para
llel
Lin
es
1. R
AM
PS A
par
kin
g ga
rage
ram
p ri
ses
to c
onn
ect
two
hor
izon
tal
leve
ls o
f a
park
ing
lot.
Th
e ra
mp
mak
es a
10°
an
gle
wit
h t
he
hor
izon
tal.
Wh
at i
s th
e m
easu
re o
f an
gle
1 in
th
e fi
gure
?
Ram
pLe
vel 2
Leve
l 110
˚1
2. B
RID
GES
A d
oubl
e de
cker
bri
dge
has
tw
o pa
rall
el l
evel
s co
nn
ecte
d by
a
net
wor
k of
dia
gon
al g
irde
rs. O
ne
of t
he
gird
ers
mak
es a
52°
an
gle
wit
h t
he
low
er l
evel
as
show
n i
n t
he
figu
re. W
hat
is
th
e m
easu
re o
f an
gle
1?
52˚
1
3. C
ITY
EN
GIN
EER
ING
Sev
enth
Ave
nu
e ru
ns
perp
endi
cula
r to
bot
h 1
st a
nd
2nd
Str
eets
, wh
ich
are
par
alle
l. H
owev
er,
Map
le A
ven
ue
mak
es a
115
° an
gle
wit
h
2nd
Str
eet.
Wh
at i
s th
e m
easu
re o
f an
gle
1?
Maple Ave.
115 ˚ 1
2nd
St.
1st S
t.
7th Ave.
4. P
OD
IUM
S A
car
pen
ter
is b
uil
din
g a
podi
um
. Th
e si
de p
anel
of
the
podi
um
is
cut
from
a r
ecta
ngu
lar
piec
e of
woo
d.
116 ˚
1
Th
e re
ctan
gle
mu
st b
e sa
wed
alo
ng
the
dash
ed l
ine
in t
he
figu
re. W
hat
is
the
mea
sure
of
angl
e 1?
5. S
ECU
RIT
YA
n i
mpo
rtan
t br
idge
cro
sses
a
rive
r at
a k
ey l
ocat
ion
. Bec
ause
it
is s
o im
port
ant,
rob
otic
sec
uri
ty c
amer
as a
re
plac
ed a
t th
e lo
cati
ons
of t
he
dots
in
th
e fi
gure
. Eac
h r
obot
can
sca
n x
deg
rees
. O
n t
he
low
er b
ank,
it
take
s 4
robo
ts t
o co
ver
the
full
an
gle
from
th
e ed
ge o
f th
e ri
ver
to t
he
brid
ge. O
n t
he
upp
er b
ank,
it
tak
es 5
rob
ots
to c
over
th
e fu
ll a
ngl
e fr
om t
he
edge
of
the
rive
r to
th
e br
idge
.
uppe
r ban
k
low
er b
ank
Bridge
a. H
ow a
re t
he
angl
es t
hat
are
cov
ered
by
th
e ro
bots
at
the
low
er a
nd
upp
er
ban
ks r
elat
ed?
Der
ive
an e
quat
ion
th
at x
sat
isfi
es b
ased
on
th
is
rela
tion
ship
.
b.
How
wid
e is
th
e sc
ann
ing
angl
e fo
r ea
ch r
obot
? W
hat
are
th
e an
gles
th
at
the
brid
ge m
akes
wit
h t
he
upp
er a
nd
low
er b
anks
?
3-2
Th
ey a
re c
on
secu
tive i
nte
rio
r an
gle
s a
nd
are
su
pp
lem
en
tary
. 4x +
5x =
180
x =
20;
up
per
ban
k =
100 a
nd
lo
wer
ban
k =
80
170
52
65
64
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Answers (Lesson 3-2)
A01_A21_GEOCRMC03_890512.indd A6A01_A21_GEOCRMC03_890512.indd A6 5/27/08 2:42:19 AM5/27/08 2:42:19 AM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 3 A7 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
16
G
lenc
oe G
eom
etry
Enri
chm
ent
Van
ish
ing
Po
int
If y
ou l
ook
dow
n a
roa
d th
at d
oes
not
hav
e an
y cu
rves
or
ben
ds i
n i
t, t
he
side
s of
th
e ro
ad
that
are
par
alle
l ap
pear
to
mee
t at
a s
ingl
e po
int.
Th
is i
s ca
lled
th
e va
nis
hin
g po
int
and
has
be
en u
sed
in a
rtw
ork
sin
ce t
he
1400
s.
Th
e pi
ctu
re b
elow
sh
ows
a st
raig
ht
road
goi
ng
into
th
e di
stan
ce. T
he
para
llel
lin
es o
f th
e le
ft a
nd
righ
t si
des
of t
he
road
hav
e be
en t
race
d to
sh
ow t
he
van
ish
ing
poin
t.
NEXT
REST
STO
P64
mile
s
In t
he
foll
owin
g p
ictu
res,
dra
w l
ines
to
fin
d t
he
van
ish
ing
poi
nt
or p
oin
ts.
1.
2.
3-2
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-3
Cha
pte
r 3
17
G
lenc
oe G
eom
etry
3-3
Stud
y G
uide
and
Inte
rven
tion
Slo
pes o
f Lin
es
Slo
pe
of
a Li
ne
Th
e sl
ope
m o
f a
lin
e co
nta
inin
g tw
o po
ints
wit
h c
oord
inat
es (
x 1, y 1)
and
(x2,
y 2) i
s gi
ven
by
the
form
ula
m =
y 2 -
y1
−
x 2 -
x1 , w
her
e x 1
≠ x
2.
F
ind
th
e sl
ope
of e
ach
lin
e.
For
lin
e p,
su
bsti
tute
(1,
2)
for
(x1,
y 1) a
nd
(-2,
-2)
for
(x 2,
y 2).
m =
y 2
- y
1 −
x 2
- x
1
= -
2 -
2
−
-2
- 1
or
4 −
3
For
lin
e q,
su
bsti
tute
(2,
0)
for
(x1,
y 1) a
nd
(-3,
2)
for
(x2,
y 2).
m =
y 2
- y
1 −
x 2
- x
1
=
2
- 0
−
-3
- 2
or
- 2 −
5
Exer
cise
sD
eter
min
e th
e sl
ope
of t
he
lin
e th
at c
onta
ins
the
give
n p
oin
ts.
1. J
(0, 0
), K
(-2,
8)
2. R
(-2,
-3)
, S(3
, -5)
3. L
(1, -
2), N
(-6,
3)
4. P
(-1,
2),
Q(-
9, 6
)
5. T
(1, -
2), U
(6, -
2)
6. V
(-2,
10)
, W(-
4, -
3)
Fin
d t
he
slop
e of
eac
h l
ine.
7.
� ��
AB
8.
� ��
CD
9.
� ��
�
EM
10
. �
��
AE
11. �
���
EH
12
. �
���
BM
x
y
O
( 1, 2
)
( –2,
–2)( –
3, 2
)
( 2, 0
)
qp
x
y
O
C( –
2, 2
)
A( –
2, –
2) H
( –1,
–4)
B( 0
, 4)
M( 4
, 2)
E( 4
, –2)
D( 0
, –2)
Exam
ple
un
defi
ned
3 2
−
5
0-2
-
1
−
2
-4
- 5
−
7
0
-
2
−
5
-
1
−
2
13
−
2
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Answers (Lesson 3-2 and Lesson 3-3)
A01_A21_GEOCRMC03_890512.indd A7A01_A21_GEOCRMC03_890512.indd A7 5/27/08 2:42:23 AM5/27/08 2:42:23 AM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 3 A8 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
18
G
lenc
oe G
eom
etry
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Slo
pes o
f Lin
es
Para
llel a
nd
Per
pen
dic
ula
r Li
nes
If
you
exa
min
e th
e sl
opes
of
pair
s of
par
alle
l li
nes
an
d th
e sl
opes
of
pair
s of
per
pen
dicu
lar
lin
es, w
her
e n
eith
er l
ine
in e
ach
pai
r is
ver
tica
l, yo
u
wil
l di
scov
er t
he
foll
owin
g pr
oper
ties
.
Tw
o li
nes
hav
e th
e sa
me
slop
e if
an
d on
ly i
f th
ey a
re p
aral
lel.
Tw
o li
nes
are
per
pen
dicu
lar
if a
nd
only
if
the
prod
uct
of
thei
r sl
opes
is
-1.
Exer
cise
sD
eter
min
e w
het
her
�
⎯⎯
�
MN
an
d �
⎯
�
RS
are
pa
rall
el, p
erp
end
icu
lar,
or
nei
ther
. Gra
ph
ea
ch l
ine
to v
erif
y yo
ur
answ
er.
1. M
(0, 3
), N
(2, 4
), R
(2, 1
), S
(8, 4
) 2.
M(-
1, 3
), N
(0, 5
), R
(2, 1
), S
(6, -
1)
3. M
(-1,
3),
N(4
, 4),
R(3
, 1),
S(-
2, 2
) 4.
M(0
, -3)
, N(-
2, -
7), R
(2, 1
), S
(0, -
3)
Gra
ph
th
e li
ne
that
sat
isfi
es e
ach
con
dit
ion
.
5. s
lope
= 4
, pas
ses
thro
ugh
(6,
2)
6. p
asse
s th
rou
gh H
(8, 5
), p
erpe
ndi
cula
r to
� ��
AG
wit
h A
(−5,
6)
and
G(−
1, −
2)
7. p
asse
s th
rou
gh C
(−2,
5),
par
alle
l to
� ��
LB
wit
h L
(2, 1
) an
d B
(7, 4
)
3-3
D
eter
min
e w
het
her
�
⎯⎯
�
AB
an
d �
⎯
⎯
�
CD
are
pa
rall
el, p
erp
end
icu
lar,
or
nei
ther
for
A(-
1, -
1), B
(1, 5
), C
(1, 2
), D
(5, 4
). G
rap
h e
ach
lin
e to
ver
ify
you
r an
swer
.
Fin
d th
e sl
ope
of e
ach
lin
e.
slop
e of
� ��
AB
= 5
- (
-1)
−
1 -
(-
1)
= 6 −
2 o
r 3
slop
e of
� ��
CD
= 4
- 2
−
5 -
1 =
2 −
4 =
1 −
2
Th
e tw
o li
nes
do
not
hav
e th
e sa
me
slop
e, s
o th
ey a
re n
ot p
aral
lel.
To
dete
rmin
e if
th
e li
nes
are
per
pen
dicu
lar,
fin
d th
e pr
odu
ct o
f th
eir
slop
es
3
( 1 −
2 ) =
3 −
2 o
r 1.
5 P
rodu
ct o
f sl
ope
for
� ��
AB
an
d � �
�
CD
Sin
ce t
he
prod
uct
of
thei
r sl
opes
is
not
–1,
th
e tw
o li
nes
are
n
ot p
erpe
ndi
cula
r.T
her
efor
e, t
her
e is
no
rela
tion
ship
bet
wee
n
� ��
AB
an
d �
��
CD
.
Wh
en g
raph
ed, t
he
two
lin
es i
nte
rsec
t bu
t n
ot a
t a
righ
t an
gle.
Exam
ple
y
x
y
x
( 8, 5
)
( 6, 2
) ( 5,−
2)
para
llel
perp
en
dic
ula
r
neit
her
para
llel
See s
tud
en
ts’
wo
rk
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Lesson 3-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
19
G
lenc
oe G
eom
etry
Skill
s Pr
acti
ceS
lop
es o
f Lin
es
Det
erm
ine
the
slop
e of
th
e li
ne
that
con
tain
s th
e gi
ven
poi
nts
.
1. S
(-1,
2),
W(0
, 4)
2
2. G
(-2,
5),
H(1
, -7)
-
4
3. C
(0, 1
), D
(3, 3
) 2
−
3
4. J
(-5,
-2)
, K(5
, -4)
-
1
−
5
Fin
d t
he
slop
e of
eac
h l
ine.
5.
y
x
3
−
4
6.
y
x
-
2
Det
erm
ine
wh
eth
er �
⎯
⎯
�
AB
an
d �
⎯
⎯
�
MN
are
pa
rall
el, p
erp
end
icu
lar,
or
nei
ther
.G
rap
h e
ach
lin
e to
ver
ify
you
r an
swer
.
7. A
(0, 3
), B
(5, -
7), M
(-6,
7),
N(-
2, -
1)
8. A
(-1,
4),
B(2
, -5)
, M(-
3, 2
), N
(3, 0
)
9. A
(-2,
-7)
, B(4
, 2),
M(-
2, 0
), N
(2, 6
) 10
. A(-
4, -
8), B
(4, -
6), M
(-3,
5),
N(-
1, -
3)
Gra
ph
th
e li
ne
that
sat
isfi
es e
ach
con
dit
ion
.
11. s
lope
= 3
, pas
ses
thro
ugh
A(0
, 1)
12. s
lope
= -
3 −
2 ,
pass
es t
hro
ugh
R(-
4, 5
)
x
y OA( 0
, 1)
x
y O
R( –
4, 5
)
13. p
asse
s th
rou
gh Y
(3, 0
), p
aral
lel
to �
��
DJ
14
. pas
ses
thro
ugh
T(0
, -2)
, per
pen
dicu
lar
wit
h D
(-3,
1)
and
J(3
, 3)
to
� ��
CX
wit
h C
(0, 3
) an
d X
(2, -
1)
x
y O
J(3,
3)
D( –
3, 1
)Y
( 3, 0
)
x
y O
C( 0
, 3)
T(0,
–2)
X( 2
, –1)
3-3
p
ara
llel
neit
her
p
ara
llel
perp
en
dic
ula
r
See s
tud
en
ts’
gra
ph
s.
001_
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6P
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Answers (Lesson 3-3)
A01_A21_GEOCRMC03_890512.indd A8A01_A21_GEOCRMC03_890512.indd A8 5/27/08 2:42:28 AM5/27/08 2:42:28 AM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF Pass
Chapter 3 A9 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
20
G
lenc
oe G
eom
etry
Prac
tice
Slo
pes o
f Lin
es
Det
erm
ine
the
slop
e of
th
e li
ne
that
con
tain
s th
e gi
ven
poi
nts
.
1. B
(-4,
4),
R(0
, 2)
-
1
−
2
2. I
(-2,
-9)
, P(2
, 4)
13
−
4
Fin
d t
he
slop
e of
eac
h l
ine.
3.
� ��
�
LM
2
−
3
4. �
���
GR
-
2
−
5
5. a
lin
e pa
rall
el t
o � �
��
GR
6.
a l
ine
perp
endi
cula
r to
� ��
PS
-
2
−
5
-
1
−
2
Det
erm
ine
wh
eth
er �
⎯
⎯
�
KM
an
d �
⎯
�
ST
are
pa
rall
el, p
erp
end
icu
lar,
or
nei
ther
. G
rap
h e
ach
lin
e to
ver
ify
you
r an
swer
.
7. K
(-1,
-8)
, M(1
, 6),
S(-
2, -
6), T
(2, 1
0)
8. K
(-5,
-2)
, M(5
, 4),
S(-
3, 6
), T
(3, -
4)
9. K
(-4,
10)
, M(2
, -8)
, S(1
, 2),
T(4
, -7)
10
. K(-
3, -
7), M
(3, -
3), S
(0, 4
), T
(6, -
5)
Gra
ph
th
e li
ne
that
sat
isfi
es e
ach
con
dit
ion
.
11. s
lope
= -
1 −
2 ,
con
tain
s U
(2, -
2)
12. s
lope
= 4 −
3 ,
con
tain
s P
(-3,
-3)
x
y OU
( 2, –
2)
x
y O
P( –
3, –
3)
13. c
onta
ins
B(-
4, 2
), p
aral
lel
to �
��
FG
14
. con
tain
s Z
(-3,
0),
per
pen
dicu
lar
to �
��
EK
w
ith
F(0
, -3)
an
d G
(4, -
2)
wit
h E
(-2,
4)
and
K(2
, -2)
x
y
O
B( –
4, 2
)
F( 0
, –3)
G( 4
, –2)
x
y
O
E( –
2, 4
)
Z( –
3, 0
)
K( 2
, –2)
15. P
RO
FITS
Aft
er T
ake
Tw
o be
gan
ren
tin
g D
VD
s at
th
eir
vide
o st
ore,
bu
sin
ess
soar
ed.
Bet
wee
n 2
005
and
2010
, pro
fits
in
crea
sed
at a
n a
vera
ge r
ate
of $
9000
per
yea
r. T
otal
pr
ofit
s in
201
0 w
ere
$45,
000.
If
prof
its
con
tin
ue
to i
ncr
ease
at
the
sam
e ra
te, w
hat
wil
l th
e to
tal
prof
it b
e in
201
4?
x
y O
R
G
S
P
M
L
3-3
n
eit
her
perp
en
dic
ula
r
p
ara
llel
perp
en
dic
ula
r
See s
tud
en
ts’
wo
rk
$81,0
00
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Lesson 3-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
21
G
lenc
oe G
eom
etry
Wor
d Pr
oble
m P
ract
ice
Slo
pes o
f Lin
es
1. H
IGH
WA
YS
A h
igh
way
on
-ram
p ri
ses
15 f
eet
for
ever
y 10
0 h
oriz
onta
l fe
et
trav
eled
. Wh
at i
s th
e sl
ope
of t
he
ram
p?
2. D
ESC
ENT
An
air
plan
e de
scen
ds a
t a
rate
of
300
feet
for
eve
ry 5
000
hor
izon
tal
feet
th
at t
he
plan
e tr
avel
s. W
hat
is
the
slop
e of
th
e pa
th o
f de
scen
t?
3. R
OA
D T
RIP
Jen
na
is d
rivi
ng
400
mil
es
to v
isit
her
gra
ndm
oth
er. S
he
man
ages
to
tra
vel
the
firs
t 10
0 m
iles
of
her
tri
p in
tw
o h
ours
. If
she
con
tin
ues
at
this
rat
e,
how
lon
g w
ill
it t
ake
her
to
driv
e th
e re
mai
nin
g di
stan
ce?
4. W
ATE
R L
EVEL
Bef
ore
the
rain
beg
an,
the
wat
er i
n a
lak
e w
as 2
68 i
nch
es d
eep.
T
he
rain
beg
an a
nd
afte
r fo
ur
hou
rs o
f ra
in, t
he
lake
was
274
in
ches
dee
p. T
he
rain
con
tin
ued
for
on
e m
ore
hou
r at
th
e sa
me
inte
nsi
ty. W
hat
was
th
e de
pth
of
the
lake
wh
en t
he
rain
sto
pped
?
5. C
ITY
BLO
CK
ST
he
figu
re s
how
s a
map
of
par
t of
a c
ity
con
sist
ing
of t
wo
pair
s of
pa
rall
el r
oads
. If
a co
ordi
nat
e gr
id i
s ap
plie
d to
th
is m
ap, F
ord
Str
eet
wou
ld
hav
e a
slop
e of
-3.
Clov
er S
t.
6th St.
Ford St.
B St
.
N
a. T
he
inte
rsec
tion
of
B S
tree
t an
d F
ord
Str
eet
is 1
50 y
ards
eas
t of
th
e in
ters
ecti
on o
f F
ord
Str
eet
and
Clo
ver
Str
eet.
How
man
y ya
rds
sou
th i
s it
?
b.
Wh
at i
s th
e sl
ope
of 6
th S
tree
t?
Exp
lain
.
c. W
hat
are
th
e sl
opes
of
Clo
ver
and
B S
tree
ts?
Exp
lain
.
d.
Th
e in
ters
ecti
on o
f B
Str
eet
and
6th
Str
eet
is 6
00 y
ards
eas
t of
th
e in
ters
ecti
on o
f B
Str
eet
and
For
d S
tree
t. H
ow m
any
yard
s n
orth
is
it?
3-3
3
−
20
-
3
−
50
6 h
ou
rs
275.5
in
.
-3;
Fo
rd S
treet
an
d 6
th
Str
eet
are
para
llel
so
th
ey
have t
he s
am
e s
lop
e.
Bo
th h
ave a
slo
pe o
f 1
−
3 b
ecau
se
bo
th a
re p
erp
en
dic
ula
r to
Fo
rd
an
d 6
th,
an
d t
he s
lop
e o
f a
perp
en
dic
ula
r is
giv
en
by t
he
neg
ati
ve r
ecip
rocal.
200 y
d
450 y
d
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Answers (Lesson 3-3)
A01_A21_GEOCRMC03_890512.indd A9A01_A21_GEOCRMC03_890512.indd A9 5/27/08 2:42:33 AM5/27/08 2:42:33 AM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 3 A10 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
22
G
lenc
oe G
eom
etry
3-3
Enri
chm
ent
Slo
pes a
nd
Po
lyg
on
sIn
coo
rdin
ate
geom
etry
, th
e sl
opes
of
two
lin
es d
eter
min
e if
th
e li
nes
are
pa
rall
el o
r pe
rpen
dicu
lar.
Th
is k
now
ledg
e ca
n b
e u
sefu
l w
hen
wor
kin
g w
ith
pol
ygon
s.
1. T
he
coor
din
ates
of
the
vert
ices
of
a tr
ian
gle
are
A(-
6, 4
), B
(8, 6
), a
nd
C(4
, -4)
. Gra
ph �
AB
C.
2. J
, K, a
nd
L a
re m
idpo
ints
of
−−
AB
, −−
− B
C , a
nd
−−
AC
, re
spec
tive
ly. F
ind
the
coor
din
ates
of
J, K
, an
d L
. D
raw
�J
KL
.
3. W
hic
h s
egm
ents
app
ear
to b
e pa
rall
el?
4. S
how
th
at t
he
segm
ents
nam
ed i
n E
xerc
ise
3 ar
e pa
rall
el b
y fi
ndi
ng
the
slop
es o
f al
l si
x se
gmen
ts.
Th
e co
ord
inat
es o
f th
e ve
rtic
es o
f ri
ght
�P
QR
are
giv
en. F
ind
th
e sl
ope
of e
ach
si
de
of t
he
tria
ngl
e. T
hen
nam
e th
e h
ypot
enu
se.
5. P
(5, 1
), Q
(1, -
1), R
(-2,
5)
6. P
(-2,
-3)
, Q(5
, 1),
R(2
, 3)
sl
ope
of −−
− P
Q =
1
−
2
sl
ope
of
−−−
PQ
=
4
−
7
sl
ope
of −
−−
QR
=
- 2
slop
e of
−−
−
QR
=
-
2
−
3
sl
ope
of −
−
PR
=
-
4
−
7
sl
ope
of −
−
PR
=
3
−
2
h
ypot
enu
se:
−−
P
R
h
ypot
enu
se:
−−
P
Q
Th
e co
ord
inat
es o
f q
uad
rila
tera
l P
QR
S a
re g
iven
. Gra
ph
qu
adri
late
ral
PQ
RS
an
d
fin
d t
he
slop
es o
f th
e d
iago
nal
s. S
tate
wh
eth
er t
he
dia
gon
als
are
per
pen
dic
ula
r.
7. P
(-2,
6),
Q(4
, 0),
R(1
, -4)
, S(-
5, 2
) 8.
P(0
, 6),
Q(3
, 0),
R(-
4, -
2), S
(-5,
4)
S
P
Q
R
x
y
O
S
P
Q
R
x
y
OA
B
C
K
J
Lx
y
O
J
(1,
5),
K(6
, 1),
L(-
1,
0)
−−
A
B a
nd
−−
LK
, −
−
B
C a
nd
−−
JL ,
−−
A
C a
nd
−−
JK
−−
A
B :
1
−
7 ;
−−
LK
: 1
−
7 ;
−−
B
C :
5
−
2 ;
−−
JL :
5
−
2 ;
−−
A
C :
- 4
−
5 ;
−−
JK
: -
4
−
5
−−
P
R :
-
10
−
3 ;
−−
S
Q :
-
2
−
9 ;
no
−−
P
R :
2;
−−
S
Q :
-
1
−
2 ;
yes
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-3
Cha
pte
r 3
23
G
lenc
oe G
eom
etry
Spre
adsh
eet
Act
ivit
yS
lop
es o
f Lin
es
You
can
use
a s
prea
dsh
eet
to i
nve
stig
ate
the
slop
e of
a l
ine.
U
se a
sp
read
shee
t to
fin
d t
he
slop
e of
a l
ine
that
con
tain
s th
e p
oin
ts (
-2,
3)
and
(4,
1).
Ste
p 1
U
se t
he
firs
t ce
ll o
f th
e sp
read
shee
t fo
r th
e x
valu
e of
th
e fi
rst
poin
t. U
se c
ell
B1
for
the
y va
lue
of t
he
firs
t po
int.
Use
cel
l C
1 fo
r th
e x
valu
e of
th
e se
con
d po
int
and
use
cel
l D
1 fo
r th
e y
valu
e of
th
e se
con
d po
int.
Ste
p 2
In
cel
l E
1, e
nte
r an
equ
als
sign
fol
low
ed b
y th
e ex
pres
sion
for
th
e n
um
erat
or o
f th
e sl
ope,
wh
ich
is
D1
- B
1. T
hen
pre
ss E
NT
ER
to
retu
rn t
he
nu
mer
ator
of
the
slop
e of
th
e li
ne.
Ste
p 3
In
cel
l F
1, e
nte
r an
equ
als
sign
fo
llow
ed b
y th
e ex
pres
sion
for
th
e de
nom
inat
or o
f th
e sl
ope,
wh
ich
is
C1
- A
1. T
hen
pre
ss E
NT
ER
to
retu
rn t
he
den
omin
ator
of
the
slop
e.
Th
e n
um
erat
or o
f th
e sl
ope
is -
2 an
d th
ede
nom
inat
or i
s 6.
So,
the
slop
e is
- 2 −
6 o
r -
1 −
3 .
U
se a
sp
read
shee
t to
det
erm
ine
wh
eth
er �
⎯
⎯
�
PQ
an
d �
⎯
⎯
�
UV
are
pa
rall
el,
per
pen
dic
ula
r, o
r n
eith
er f
or P
(-1,
2),
Q(-
3, 6
), U
(0, 1
), a
nd
V(2
, 2).
Ste
p 1
E
nte
r th
e or
dere
d pa
irs
for
� ��
PQ
in
row
2 a
nd
the
orde
red
pair
s fo
r � �
��
UV
in
row
3 a
s ab
ove.
Ste
p 2
W
ith
cel
l E
1 se
lect
ed, c
lick
on
th
e bo
ttom
rig
ht
corn
er o
f ce
ll E
1 an
d dr
ag i
t to
E3.
T
his
ret
urn
s th
e n
um
erat
ors
of t
he
slop
es. W
ith
cel
l F
1 se
lect
ed, c
lick
on
th
e bo
ttom
rig
ht
corn
er o
f ce
ll F
1 an
d dr
ag i
t to
F3.
Th
is r
etu
rns
the
den
omin
ator
s of
th
e sl
opes
.
For
� ��
PQ
, th
e n
um
erat
or o
f th
e sl
ope
is 4
an
d th
e de
nom
inat
or i
s -
2. S
o, t
he
slop
e
is
4 −
-
2 or
-2.
For
� ���
UV
, th
e n
um
erat
or o
f th
e sl
ope
is 1
an
d th
e de
nom
inat
or i
s 2.
So,
th
e sl
ope
is 1 −
2 .
Sin
ce t
he
prod
uct
of
the
slop
es i
s -
1, t
he
lin
es a
re p
erpe
ndi
cula
r.
Exer
cise
sU
se a
sp
read
shee
t to
fin
d t
he
slop
es o
f th
e li
nes
th
at c
onta
in t
he
give
n p
oin
ts.
1. (
2, 4
), (
1, 7
)
2. (-
2, 8
), (
3, -
5)
3.
(0, 4
), (
7, 0
)
4. (
3, 5
), (
-1,
9)
5.
(3, -
1), (
9, -
6)
6. (-
2, 5
), (
-7,
-2)
Use
a s
pre
adsh
eet
to d
eter
min
e w
het
her
�
⎯⎯
�
PQ
an
d �
⎯
⎯
�
UV
are
pa
rall
el,
per
pen
dic
ula
r, o
r n
eith
er.
7. P
(22,
3),
Q(3
, 1),
U(0
, 3),
an
d V
(5, 5
) 8.
P(3
, 5),
Q(1
, 22)
, U(2
3, 2
4), a
nd
V(2
1, 3
)
3-3
Exam
ple
1
AB
CD
1 432
EF
Sh
eet
1S
hee
t 2
Sh
eet
3
�2
�1 0
3 2 1
4�3 2
1 6 2
�2 4 1
6�2 2
Exam
ple
2
023_
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:20:
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Answers (Lesson 3-3)
A01_A20_GEOCRMC03_890512.indd A10A01_A20_GEOCRMC03_890512.indd A10 6/19/08 2:21:14 PM6/19/08 2:21:14 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF 2nd
Chapter 3 A11 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
24
G
lenc
oe G
eom
etry
Stud
y G
uide
and
Inte
rven
tion
Eq
uati
on
s o
f Lin
es
Wri
te E
qu
atio
ns
of
Lin
es Y
ou c
an w
rite
an
equ
atio
n o
f a
lin
e if
you
are
giv
en a
ny
of
the
foll
owin
g:•
the
slop
e an
d th
e y-
inte
rcep
t,•
the
slop
e an
d th
e co
ordi
nat
es o
f a
poin
t on
th
e li
ne,
or
• th
e co
ordi
nat
es o
f tw
o po
ints
on
th
e li
ne.
If m
is
the
slop
e of
a l
ine,
b i
s it
s y-
inte
rcep
t, a
nd
(x1,
y 1) i
s a
poin
t on
th
e li
ne,
th
en:
• th
e sl
ope-
inte
rcep
t fo
rm o
f th
e eq
uat
ion
is
y =
mx
+ b
,•
the
poi
nt-
slop
e fo
rm o
f th
e eq
uat
ion
is
y -
y1
= m
(x -
x1)
.
W
rite
an
eq
uat
ion
in
sl
ope-
inte
rcep
t fo
rm o
f th
e li
ne
wit
h
slop
e -
2 an
d y
-in
terc
ept
4.
y =
mx
+ b
S
lope-inte
rcept
form
y =
-2x
+ 4
m
= -
2,
b =
4
Th
e sl
ope-
inte
rcep
t fo
rm o
f th
e eq
uat
ion
of
the
lin
e is
y =
-2x
+ 4
.
W
rite
an
eq
uat
ion
in
p
oin
t-sl
ope
form
of
the
lin
e w
ith
slo
pe
- 3 −
4 th
at c
onta
ins
(8, 1
).
y -
y1
= m
(x -
x1)
P
oin
t-slo
pe f
orm
y -
1 =
- 3 −
4 (x
- 8
) m
= -
3
−
4 ,
(x1,
y 1)
= (
8,
1)
Th
e po
int-
slop
e fo
rm o
f th
e eq
uat
ion
of
the
lin
e is
y -
1 =
- 3 −
4 (x
- 8
).
Exer
cise
sW
rite
an
eq
uat
ion
in
slo
pe-
inte
rcep
t fo
rm o
f th
e li
ne
hav
ing
the
give
n s
lop
e an
d
y-in
terc
ept
or g
iven
poi
nts
. Th
en g
rap
h t
he
lin
e.
1. m
: 2, b
: -3
2. m
: - 1 −
2 ,
b: 4
y
= 2
x -
3
y =
-
1
−
2 x
+ 4
3. m
: 1 −
4 ,
b: 5
4.
m: 0
, b: -
2
y
= 1
−
4 x
+ 5
y =
-2
5. m
: - 5 −
3 ,
(0 ,
1 −
3 )
6. m
: -3,
(1,
-11
)
y
= -
5
−
3 x
+ 1
−
3
y =
-3x -
8
Wri
te a
n e
qu
atio
n i
n p
oin
t-sl
ope
form
of
the
lin
e h
avin
g th
e gi
ven
slo
pe
that
co
nta
ins
the
give
n p
oin
t. T
hen
gra
ph
th
e li
ne.
7. m
= 1 −
2 ,
(3, -
1)
8. m
= -
2, (
4, -
2)
y
+ 1
= 1
−
2 (x
- 3
) y +
2 =
-2(x
- 4
)
9. m
= -
1, (
-1,
3)
10. m
= 1 −
4 ,
(-3,
-2)
y
- 3
= -
(x +
1)
y +
2 =
- 1
−
4 (x
+ 3
)
11. m
= -
5 −
2 ,
(0, -
3)
12. m
= 0
, (-
2, 5
)
y
+ 3
= -
5
−
2 x
y -
5 =
0
3-4
Exam
ple
1Ex
amp
le 2
See s
tud
en
ts’
wo
rk
See s
tud
en
ts’
wo
rk
023_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-4
Cha
pte
r 3
25
G
lenc
oe G
eom
etry
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Eq
uati
on
s o
f Lin
es
Wri
te E
qu
atio
ns
to S
olv
e Pr
ob
lem
s M
any
real
-wor
ld s
itu
atio
ns
can
be
mod
eled
u
sin
g li
nea
r eq
uat
ion
s.
D
onn
a of
fers
com
pu
ter
serv
ices
to
smal
l co
mp
anie
s in
her
cit
y. S
he
char
ges
$55
per
mon
th f
or m
ain
tain
ing
a w
eb s
ite
and
$45
per
hou
r fo
r ea
ch
serv
ice
call
.a.
Wri
te a
n e
qu
atio
n t
o re
pre
sen
t th
e to
tal
mon
thly
cos
t, C
, for
m
ain
tain
ing
a w
eb s
ite
and
for
h h
ours
of
serv
ice
call
s.
For
eac
h h
our,
th
e co
st
incr
ease
s $4
5. S
o th
e ra
te
of c
han
ge, o
r sl
ope,
is
45.
Th
e y-
inte
rcep
t is
loc
ated
w
her
e th
ere
are
0 h
ours
, or
$55
.C
= m
h +
b =
45h
+ 5
5
b.
Don
na
may
ch
ange
her
cos
ts t
o re
pre
sen
t th
em
by
the
equ
atio
n C
= 2
5h +
125
, wh
ere
$125
is
the
fixe
d m
onth
ly f
ee f
or a
web
sit
e an
d t
he
cost
per
h
our
is $
25. C
omp
are
her
new
pla
n t
o th
e ol
d o
ne
if
a c
omp
any
has
5 1 −
2 h
ours
of
serv
ice
call
s. U
nd
er
w
hic
h p
lan
wou
ld D
onn
a ea
rn m
ore?
Fir
st p
lan
For
5 1 −
2 h
ours
of
serv
ice
Don
na
wou
ld e
arn
C =
45h
+ 5
5 =
45 (
5 1
−
2
) + 5
5
= 2
47.5
+ 5
5 or
$30
2.50
Sec
ond
Pla
n
For
5 1 −
2 h
ours
of
serv
ice
Don
na
wou
ld e
arn
C =
25h
+ 1
25 =
25(
5.5)
+ 1
25=
137
.5 +
125
or
$262
.50
Don
na
wou
ld e
arn
mor
e w
ith
th
e fi
rst
plan
.
Exer
cise
sF
or E
xerc
ises
1–4
, use
th
e fo
llow
ing
info
rmat
ion
.
Jerr
i’s c
urre
nt s
atel
lite
tel
evis
ion
serv
ice
char
ges
a fl
at r
ate
of $
34.9
5 pe
r m
onth
for
the
bas
ic
chan
nels
and
an
addi
tion
al $
10 p
er m
onth
for
eac
h pr
emiu
m c
hann
el. A
com
peti
ng s
atel
lite
te
levi
sion
ser
vice
cha
rges
a f
lat
rate
of
$39.
99 p
er m
onth
for
the
bas
ic c
hann
els
and
an
addi
tion
al $
8 pe
r m
onth
for
eac
h pr
emiu
m c
hann
el.
1. W
rite
an
equ
atio
n i
n s
lope
-in
terc
ept
form
th
at m
odel
s th
e to
tal
mon
thly
cos
t fo
r ea
ch s
atel
lite
ser
vice
, wh
ere
p is
th
e n
um
ber
of p
rem
ium
ch
ann
els.
C
urr
en
t serv
ice:
C =
10p
+ 3
4.9
5
C
om
peti
ng
serv
ice:
C =
8p
+ 3
9.9
9
3. A
th
ird
sate
llit
e co
mpa
ny
char
ges
a fl
at
rate
of
$69
for
all
chan
nel
s, i
ncl
udi
ng
the
prem
ium
ch
ann
els.
If
Jerr
i w
ants
to
add
a fo
urt
h p
rem
ium
ch
ann
el, w
hic
h s
ervi
ce
wou
ld b
e le
ast
expe
nsi
ve?
t
he t
hir
d c
om
pan
y
2. I
f Je
rri
wan
ts t
o in
clu
de t
hre
e pr
emiu
m
chan
nel
s in
her
pac
kage
, wh
ich
ser
vice
w
ould
be
less
, her
cu
rren
t se
rvic
e or
th
e co
mpe
tin
g se
rvic
e?
c
om
peti
ng
serv
ice
4. W
rite
a d
escr
ipti
on o
f h
ow t
he
fee
for
the
nu
mbe
r of
pre
miu
m c
han
nel
s is
ref
lect
ed
in t
he
equ
atio
n.
Th
e f
ee f
or
the
nu
mb
er
of
pre
miu
m c
han
nels
re
pre
sen
ts t
he r
ate
of
ch
an
ge,
or
slo
pe,
of
the e
qu
ati
on
.
3-4
Exam
ple
023_
042_
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Answers (Lesson 3-4)
A01_A20_GEOCRMC03_890512.indd A11A01_A20_GEOCRMC03_890512.indd A11 6/10/08 4:40:36 PM6/10/08 4:40:36 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 3 A12 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
26
G
lenc
oe G
eom
etry
Skill
s Pr
acti
ceE
qu
ati
on
s o
f Lin
es
Wri
te a
n e
qu
atio
n i
n s
lop
e-in
terc
ept
form
of
the
lin
e h
avin
g th
e gi
ven
slo
pe
and
y-
inte
rcep
t. T
hen
gra
ph
th
e li
ne.
See s
tud
en
ts’
gra
ph
s.
1. m
: -4,
b: 3
2.
m: 3
, b: -
8
y
= -
4x +
3
y =
3x -
8
3. m
: 3 −
7 ,
(0, 1
)
4. m
: - 2 −
5 ,
(0, -
6)
y
= 3
−
7 x
+ 1
y =
- 2
−
5 x
- 6
Wri
te e
qu
atio
ns
in p
oin
t-sl
ope
form
of
the
lin
e h
avin
g th
e gi
ven
slo
pe
that
co
nta
ins
the
give
n p
oin
t. T
hen
gra
ph
th
e li
ne.
See s
tud
en
ts’
gra
ph
s.
5. m
= 2
, (5,
2)
6. m
= -
3, (
2, -
4)
y
- 2
= 2
(x -
5)
y +
4 =
-3(x
- 2
)
7. m
= -
1 −
2 ,
(-2,
5)
8. m
= 1 −
3 ,
(-3,
-8)
y
- 5
= -
1
−
2 (x +
2)
y +
8 =
1
−
3 (x
+ 3
)
Wri
te a
n e
qu
atio
n i
n s
lop
e-in
terc
ept
form
for
eac
h l
ine
show
n o
r d
escr
ibed
.
9. r
y =
x +
3
10. s
y =
-2x +
2
11. t
y =
3x -
3
12. u
y =
1
−
3 x
- 5
13. t
he
lin
e pa
rall
el t
o li
ne
r th
at c
onta
ins
(1, -
1) y
= x
- 2
14. t
he
lin
e pe
rpen
dicu
lar
to l
ine
s th
at c
onta
ins
(0, 0
) y =
1
−
2 x
15. m
= 6
, b =
-2
16. m
= -
5 −
3 ,
b =
0
y
= 6
x -
2
y =
-
5
−
3 x
17. m
= -
1, c
onta
ins
(0, -
6)
18. m
= 4
, con
tain
s (2
, 5)
y
= -
x -
6
y =
4x -
3
19. c
onta
ins
(2, 0
) an
d (0
, 10)
20
. x-i
nte
rcep
t is
-2,
y-i
nte
rcep
t is
-1
y
= -
5x +
10
y =
-
1
−
2 x
- 1
x
y
O
r
s
t
u
3-4
023_
042_
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OC
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-4
Cha
pte
r 3
27
G
lenc
oe G
eom
etry
Prac
tice
Eq
uati
on
s o
f Lin
es
Wri
te a
n e
qu
atio
n i
n s
lop
e-in
terc
ept
form
of
the
lin
e h
avin
g th
e gi
ven
slo
pe
and
y-
inte
rcep
t or
giv
en p
oin
ts. T
hen
gra
ph
th
e li
ne.
1. m
: 2 −
3 ,
b: -
10
2. m
: - 7 −
9 ,
(0,
- 1 −
2 ) 3.
m: 4
.5, (
0, 0
.25)
y
= 2
−
3 x
- 1
0
y =
- 7
−
9 x
- 1
−
2
y =
4.5
x +
0.2
5
Wri
te e
qu
atio
ns
in p
oin
t-sl
ope
form
of
the
lin
e h
avin
g th
e gi
ven
slo
pe
that
co
nta
ins
the
give
n p
oin
t. T
hen
gra
ph
th
e li
ne.
4. m
: 3 −
2 ,
(4, 6
) 5.
m: -
6 −
5 ,
(-5,
-2)
y
- 6
= 3
−
2 (x -
4)
y +
2 =
-
6
−
5 (x +
5)
6. m
: 0.5
, (7,
-3)
7.
m: -
1.3,
(-
4, 4
)
y
+ 3
= 0
.5(x
- 7
),
y -
4 =
-1.3
(x +
4)
Wri
te a
n e
qu
atio
n i
n s
lop
e-in
terc
ept
form
for
eac
h l
ine
show
n o
r d
escr
ibed
.
8. b
y =
-x -
5
9. c
y =
- 2
−
5 x
+ 4
10. p
aral
lel
to l
ine
b, c
onta
ins
(3, -
2) y
= -
x +
1
11. p
erpe
ndi
cula
r to
lin
e c,
con
tain
s (-
2, -
4) y
= 5
−
2 x
+ 1
12. m
= -
4 −
9 ,
b =
2
13. m
= 3
, con
tain
s (2
, -3)
y
= -
4
−
9 x
+ 2
y =
3x -
9
14. x
-in
terc
ept
is -
6, y
-in
terc
ept
is 2
15
. x-i
nte
rcep
t is
2, y
-in
terc
ept
is -
5
y
= 1
−
3 x
+ 2
y =
5
−
2 x
- 5
16. p
asse
s th
rou
gh (
2, -
4) a
nd
(5, 8
) 17
. con
tain
s (-
4, 2
) an
d (8
, -1)
y
= 4
x -
12
y =
- 1
−
4 x
+ 1
18. C
OM
MU
NIT
Y E
DU
CA
TIO
N A
loc
al c
omm
un
ity
cen
ter
offe
rs s
elf-
defe
nse
cla
sses
for
te
ens.
A $
25 e
nro
llm
ent
fee
cove
rs s
upp
lies
an
d m
ater
ials
an
d op
en c
lass
es c
ost
$10
each
. Wri
te a
n e
quat
ion
to
repr
esen
t th
e to
tal
cost
of
x se
lf-d
efen
se c
lass
es a
t th
e co
mm
un
ity
cen
ter.
C =
10x +
25
x
y
O
c
b
3-4
See s
tud
en
ts’
wo
rk
See s
tud
en
ts’
wo
rk
023_
042_
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OC
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Answers (Lesson 3-4)
A01_A20_GEOCRMC03_890512.indd A12A01_A20_GEOCRMC03_890512.indd A12 6/10/08 4:40:40 PM6/10/08 4:40:40 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF 2nd
Chapter 3 A13 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
28
G
lenc
oe G
eom
etry
Wor
d Pr
oble
m P
ract
ice
Eq
uati
on
s o
f Lin
es
1. G
RO
WTH
At
the
sam
e ti
me
each
mon
th
over
a o
ne
year
per
iod,
Joh
n r
ecor
ded
the
hei
ght
of a
tre
e h
e h
ad p
lan
ted.
He
then
cal
cula
ted
the
aver
age
grow
th r
ate
of t
he
tree
. Th
e h
eigh
t h
in
in
ches
of
the
tree
du
rin
g th
is p
erio
d w
as g
iven
by
the
form
ula
h=
1.7
t+ 2
8, w
her
e t
is t
he
nu
mbe
r of
mon
ths.
Wh
at a
re t
he
slop
e an
d y-
inte
rcep
t of
th
is l
ine
and
wh
at d
o th
ey s
ign
ify?
Th
e s
lop
e i
s 1
.7 a
nd
is t
he
avera
ge n
um
ber
of
inch
es
the t
ree g
rew
each
mo
nth
. T
he y
-in
terc
ep
t is
28 a
nd
is
the h
eig
ht
of
the t
ree w
hen
h
e b
eg
an
.
2. D
RIV
ING
Ell
en i
s dr
ivin
g to
a f
rien
d’s
hou
se. T
he
grap
h s
how
s th
e di
stan
ce
(in
mil
es)
that
Ell
en w
as f
rom
hom
e t
min
ute
s af
ter
she
left
her
hou
se.
m
tO
5
5
Wri
te a
n e
quat
ion
th
at r
elat
es m
an
d t.
m =
0.8
t
3. C
OST
Car
la h
as a
bu
sin
ess
that
tes
ts
the
air
qual
ity
in a
rtis
t’s s
tudi
os. S
he
had
to
purc
has
e $7
50 w
orth
of
test
ing
equ
ipm
ent
to s
tart
her
bu
sin
ess.
Sh
e ch
arge
s $5
0 to
per
form
th
e te
st. L
et n
be
the
nu
mbe
r of
jobs
sh
e ge
ts a
nd
let
P b
e h
er n
et p
rofi
t. W
rite
an
equ
atio
n t
hat
re
late
s P
an
d n
. How
man
y jo
bs d
oes
she
nee
d to
mak
e $7
50?
P =
50n
- 7
50;
to m
ake $
750,
n =
30
4. P
AIN
T TE
STIN
G A
pai
nt
com
pan
y de
cide
d to
tes
t th
e du
rabi
lity
of
its
wh
ite
pain
t. T
hey
pai
nte
d a
squ
are
all
wh
ite
wit
h t
hei
r pa
int
and
mea
sure
d th
e re
flec
tivi
ty o
f th
e sq
uar
e ea
ch y
ear.
S
even
yea
rs a
fter
bei
ng
pain
ted,
th
e re
flec
tivi
ty w
as 8
5%. T
en y
ears
aft
er
bein
g pa
inte
d, t
he
refl
ecti
vity
dro
pped
to
82.
9%. A
ssu
min
g th
at t
he
refl
ecti
vity
de
crea
ses
stea
dily
wit
h t
ime,
wri
te a
n
equ
atio
n t
hat
giv
es t
he
refl
ecti
vity
R (
as
a pe
rcen
tage
) as
a f
un
ctio
n o
f ti
me
t in
ye
ars.
Wh
at i
s th
e re
flec
tivi
ty o
f a
fres
h
coat
of
thei
r w
hit
e pa
int?
R =
-0.7
t +
89.9
; a f
resh
co
at
is
89.9
% r
efl
ecti
ve.
5. A
RTI
STR
YG
ail
is a
n o
il p
ain
ter.
Sh
e pa
ints
on
can
vase
s m
ade
from
Bel
gian
li
nen
. Bef
ore
she
pain
ts o
n t
he
lin
en, s
he
has
to
prim
e th
e su
rfac
e so
th
at i
t do
es
not
abs
orb
the
oil
from
th
e pa
int
she
use
s. S
he
can
bu
y li
nen
th
at h
as a
lrea
dy
been
pri
med
for
$21
per
yar
d, o
r sh
e ca
n
buy
un
prim
ed l
inen
for
$15
per
yar
d,
but
then
sh
e w
ould
als
o h
ave
to b
uy
a ja
r of
pri
mer
for
$30
.
a. L
et P
be
the
cost
of
Y y
ards
of
prim
ed
Bel
gian
lin
en. W
rite
an
equ
atio
n t
hat
re
late
s P
an
d Y
.
P =
21Y
b.
Let
U b
e th
e co
st o
f bu
yin
g Y
yar
ds
of u
npr
imed
lin
en a
nd
a ja
r of
pri
mer
. W
rite
an
equ
atio
n t
hat
rel
ates
Uan
d Y
.
U =
15Y
+ 3
0
c. F
or h
ow m
any
yard
s w
ould
it
be l
ess
expe
nsi
ve f
or G
ail
to b
uy
the
prim
ed
lin
en?
an
yth
ing
un
der
5 y
d
3-4
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NA
ME
DA
TE
PE
RIO
D
Lesson 3-4
Cha
pte
r 3
29
G
lenc
oe G
eom
etry
Enri
chm
ent
Po
lyg
on
s o
n a
Co
ord
inate
Gri
dW
hen
equ
atio
ns
are
grap
hed
on
a c
oord
inat
e gr
id, t
hei
r li
nes
can
in
ters
ect
in a
w
ay t
hat
th
e se
gmen
ts d
eter
min
ed b
y th
eir
inte
rsec
tion
poi
nts
for
m t
he
side
s of
a
poly
gon
.
1. T
he
foll
owin
g eq
uat
ion
s w
hen
gra
phed
wil
l co
nta
in t
he
side
s of
a p
olyg
on. W
ith
out
grap
hin
g th
e li
nes
, mak
e a
pred
icti
on a
bou
t w
hat
kin
d of
fig
ure
th
e li
nes
wil
l cr
eate
.
y=
1 −
2 x
+ 3
y
= 1 −
2 x
- 2
y
= 2
x +
1
y
= 2
x -
3
Sin
ce t
here
are
tw
o p
air
s o
f p
ara
llel
lin
es,
the l
ines w
ill
form
a
para
llelo
gra
m.
Th
e l
ines a
re n
ot
perp
en
dic
ula
r, s
o t
hey w
ill
no
t fo
rm a
recta
ng
le.
2. G
raph
th
e li
nes
fro
m E
xerc
ise
1 to
det
erm
ine
wh
eth
er y
our
pred
icti
on w
as
corr
ect.
y
xO
3. F
ind
the
equ
atio
ns
of t
he
lin
es t
hat
for
m t
he
side
s to
th
e po
lygo
n s
how
n b
elow
. Wh
at
type
of
poly
gon
is
it?
Exp
lain
you
r re
ason
ing.
y
xO
3-4
y =
2
−
3 x
+ 1
y =
2
−
3 x
- 2
y =
- 3
−
2 x
y =
- 3
−
2 x
+ 3
recta
ng
le,
becau
se c
on
secu
tive
sid
es a
re p
erp
en
dic
ula
r
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Answers (Lesson 3-4)
A01_A20_GEOCRMC03_890512.indd A13A01_A20_GEOCRMC03_890512.indd A13 6/19/08 2:22:06 PM6/19/08 2:22:06 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 3 A14 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
30
G
lenc
oe G
eom
etry
Stud
y G
uide
and
Inte
rven
tion
Pro
vin
g L
ines P
ara
llel
Iden
tify
Par
alle
l Lin
es I
f tw
o li
nes
in
a p
lan
e ar
e cu
t by
a t
ran
sver
sal
and
cert
ain
co
ndi
tion
s ar
e m
et, t
hen
th
e li
nes
mu
st b
e pa
rall
el.
Ifth
en
• co
rre
sp
on
din
g a
ng
les a
re c
on
gru
en
t,
• a
lte
rna
te e
xte
rio
r a
ng
les a
re c
on
gru
en
t,
• co
nse
cu
tive
in
terio
r a
ng
les a
re s
up
ple
me
nta
ry,
• a
lte
rna
te in
terio
r a
ng
les a
re c
on
gru
en
t, o
r
• tw
o lin
es a
re p
erp
en
dic
ula
r to
th
e s
am
e lin
e,
the
lin
es a
re p
ara
llel.
If
m∠
1 =
m∠
2, d
eter
min
e w
hic
h l
ines
, if
any,
are
par
alle
l. S
tate
th
e p
ostu
late
or
theo
rem
th
at j
ust
ifie
s yo
ur
answ
er.
nm
rs
12
∠1
and
∠2
are
corr
espo
ndi
ng
angl
es o
f li
nes
r
and
s. S
ince
∠1
� ∠
2, r
‖ s
by t
he
Con
vers
e of
th
e C
orre
spon
din
g A
ngl
es
Pos
tula
te.
F
ind
m∠
AB
C s
o th
at
m ‖
n.
n
mA
B
C
D
( 3x
+ 1
0)°
( 6x
- 2
0)°
We
can
con
clu
de t
hat
m ‖
n i
f al
tern
ate
inte
rior
an
gles
are
con
gru
ent.
m∠
BA
D =
m∠
AB
C 3
x +
10
= 6
x -
20
10
= 3
x -
20
30
= 3
x
10 =
x m
∠A
BC
= 6
x -
20
=
6(1
0) -
20
or 4
0
Exer
cise
sF
ind
x s
o th
at l
‖ m
. Id
enti
fy t
he
pos
tula
te o
r th
eore
m y
ou u
sed
.
1.
m�
( 5x
- 5
) °
( 6x
- 2
0)°
2.
m
�( 4
x+
20)
°
6x°
3.
m
� ( 3x
+ 1
5)°
1
5;
Alt
. E
xt.
� T
h.
10;
Alt
. In
t. �
Th
. 2
5;
Alt
. In
t. �
Th
.
4.
m�
( 9x
+ 1
) °( 8x
+ 8
) °
5.
m
� ( 3x
- 2
0)°
2x°
6.
m
�
( 5x
+ 2
0)°
70°
7
; A
lt.
Int.
� T
h.
20;
Alt
. E
xt.
� T
h.
10;
Co
rr.
� T
h.
3-5
Exam
ple
1Ex
amp
le 2
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NA
ME
DA
TE
PE
RIO
D
Lesson 3-5
Cha
pte
r 3
31
G
lenc
oe G
eom
etry
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Pro
vin
g L
ines P
ara
llel
Pro
ve L
ines
Par
alle
l Y
ou c
an p
rove
th
at l
ines
are
par
alle
l by
usi
ng
pos
tula
tes
and
theo
rem
s ab
out
pair
s of
an
gles
.
� m
rs
12
43 5
68
7
910 11
12
1413
1516
G
iven
: ∠1
� ∠
2, ∠
1 �
∠3
P
rove
: −
−
A
B ‖
−−
D
C
P
roof
:S
tate
men
tsR
easo
ns
1. ∠
1 �
∠2
∠
1 �
∠3
1. G
iven
2. ∠
2 �
∠3
2. T
ran
siti
ve P
rope
rty
of �
3. −
−
AB
‖ −
−−
DC
3.
If
alt.
in
t. a
ngl
es a
re �
, th
en
the
lin
es a
re ‖
.123
AB
CD
Exer
cise
s1.
Com
ple
te t
he
pro
of.
Giv
en: ∠
1 �
∠5,
∠15
� ∠
5P
rove
: � ‖
m, r
‖ s
Pro
of:
Sta
tem
ents
Rea
son
s
1. ∠
15 �
∠5
1.
2. ∠
13 �
∠15
2.
3. ∠
5 �
∠13
3.
4. r
‖ s
4.
5.
5. G
iven
6.
6. I
f co
rr �
are
�, t
hen
lin
es ‖
.
3-5
Exam
ple
Giv
en
Vert
ical
� a
re �
.
Tra
nsit
ive P
rop
ert
y o
f �
If c
orr
. �
are
�,
then
lin
es ‖
.
∠1 �
∠5
� ‖ m
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Answers (Lesson 3-5)
A01_A20_GEOCRMC03_890512.indd A14A01_A20_GEOCRMC03_890512.indd A14 6/10/08 4:40:50 PM6/10/08 4:40:50 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF 2nd
Chapter 3 A15 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
32
G
lenc
oe G
eom
etry
Giv
en t
he
foll
owin
g in
form
atio
n, d
eter
min
e w
hic
h l
ines
, if
an
y, a
re p
aral
lel.
Sta
te t
he
pos
tula
te o
r th
eore
m t
hat
ju
stif
ies
you
r an
swer
.
1. ∠
3 �
∠7
2. ∠
9 �
∠11
a
‖ b
; A
lt.
Int.
� T
h.
a ‖
b;
Co
rr.
� P
os
t.
3. ∠
2 �
∠16
4.
m∠
5 +
m∠
12 =
180
ℓ
‖ m
; A
lt.
Ext.
� T
h.
ℓ ‖
m;
Co
nsec.
Int.
� T
h.
Fin
d x
so
that
� ‖
m. S
how
you
r w
ork
.
5.
m
k�
( 2x
+ 6
) °
130°
22
6.
mk
�
( 4x
- 1
0)°
( 3x
+ 1
0)°
20
7.
m
k�
( 6x
+ 4
) ° ( 8x
- 8
) °
6
8.
(4x)
°
(x+
6)°
�
k
m
9.
(7x-
5)°
(5x+
19)°
k
m�
10
. (3
x+10
)°
(5x+
18)°
�
km
11. P
RO
OF
Pro
vide
a r
easo
n f
or e
ach
sta
tem
ent
in t
he
proo
f of
Th
eore
m 3
.7.
G
iven
: ∠
1 an
d ∠
2 ar
e co
mpl
emen
tary
.
−−
− B
C ⊥
−−
−
CD
Pro
ve:
−−
BA
‖ −
−−
CD
Pro
of:
Sta
tem
ents
Rea
son
s
1. −−
− B
C ⊥
−−
−
CD
1.
Giv
en
2. m
∠A
BC
= m
∠1
+ m
∠2
2. A
ng
le A
dd
itio
n P
ostu
late
3. ∠
1 a
nd
∠2
are
com
plem
enta
ry.
3. G
iven
4. m
∠1
+ m
∠2
= 9
04
. D
efi
nit
ion
of
co
mp
lem
en
tary
an
gle
s
5. m
∠A
BC
= 9
05
. T
ran
sit
ive P
rop
ert
y o
f E
qu
ality
6. −
−
BA
⊥ −−
− B
C
6. D
efi
nit
ion
of
perp
en
dic
ula
r
7. −
−
BA
‖ −
−−
CD
7.
Skill
s Pr
acti
ceP
rovin
g L
ines P
ara
llel
m�
ab
12
34
87
65
910
1112
1615
1413
1
ADC
B2
3-5
If 2
lin
es a
re ⊥
to
th
e s
am
e l
ine,
then
lin
es a
re ‖
.
14
219
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-5
Cha
pte
r 3
33
G
lenc
oe G
eom
etry
Prac
tice
Pro
vin
g L
ines P
ara
llel
A
BC
D
EF H
K
G J
14
25
36
AB
CD
Giv
en t
he
foll
owin
g in
form
atio
n, d
eter
min
e w
hic
h l
ines
, if
an
y, a
re p
aral
lel.
Sta
te t
he
pos
tula
te o
r th
eore
m t
hat
ju
stif
ies
you
r an
swer
.
1. m
∠B
CG
+ m
∠F
GC
= 1
80
2. ∠
CB
F �
∠G
FH
�
⎯
�
BD
‖ �
⎯
�
EG
;
�
⎯
�
BD
‖ �
⎯
�
EG
;
C
on
vers
e C
on
s.
Int.
�T
h.
Co
nvers
e C
orr
. �
Th
.
3. ∠
EF
B �
∠F
BC
4.
∠A
CD
� ∠
KB
F
�
⎯
�
BD
‖ �
⎯
�
EG
;
�
⎯
�
AJ ‖
�
⎯
�
BH
;
Co
nvers
e A
lt.
Int.
�T
h.
Alt
. E
xt.
� T
h.
Fin
d x
so
that
l ‖
m. I
den
tify
th
e p
ostu
late
or
theo
rem
you
use
d.
5.
( 3x
+ 6
) °
( 4x
- 6
) °
t m
�
6.
( 5
x+
18)
°
( 7x
- 2
4)°
t m
�
7.
( 2x
+ 1
2)° ( 5x
- 1
5)°
t
m �
8. P
RO
OF
Wri
te a
tw
o-co
lum
n p
roof
.
Giv
en:
∠2
and
∠3
are
supp
lem
enta
ry.
P
rove
: −
−
AB
‖ −
−−
CD
9. L
AN
DSC
API
NG
Th
e h
ead
gard
ener
at
a bo
tan
ical
gar
den
wan
ts t
o pl
ant
rose
bush
es i
n
para
llel
row
s on
eit
her
sid
e of
an
exi
stin
g fo
otpa
th. H
ow c
an t
he
gard
ener
en
sure
th
at
the
row
s ar
e pa
rall
el?
S
am
ple
an
sw
er:
If
the g
ard
en
er
dig
s e
ach
ro
w a
t a 9
0 a
ng
le t
o t
he
foo
tpath
, each
ro
w w
ill
be p
erp
en
dic
ula
r to
th
e f
oo
tpath
. If
each
of
the
row
s i
s p
erp
en
dic
ula
r to
th
e f
oo
tpath
, th
en
th
e r
ow
s a
re p
ara
llel.
3-5
12;
co
rr.
�21;
alt
. ext.
�9;
alt
. in
t. �
Pro
of:
Sta
tem
en
tsR
easo
ns
1.
∠2 a
nd
∠3 a
re s
up
ple
men
tary
.1.
Giv
en
2.
�
⎯
�
AB
‖ �
⎯
�
CD
2.
If c
on
sec.
int
� a
re s
up
pl.,
then
lin
es a
re ‖
.
3.
−−
A
B ‖
−−
C
D
3.
Seg
men
ts c
on
tain
ed
in
‖ l
ines
are
‖.
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A01_A20_GEOCRMC03_890512.indd A15A01_A20_GEOCRMC03_890512.indd A15 6/10/08 4:40:55 PM6/10/08 4:40:55 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 3 A16 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
34
G
lenc
oe G
eom
etry
Wor
d Pr
oble
m P
ract
ice
Pro
vin
g L
ines P
ara
llel
1. R
ECTA
NG
LES
Jim
mad
e a
fram
e fo
r a
pain
tin
g. H
e w
ants
to
chec
k to
mak
e su
re t
hat
opp
osit
e si
des
are
para
llel
by
mea
suri
ng
the
angl
es a
t th
e co
rner
s an
d se
ein
g if
th
ey a
re r
igh
t an
gles
. How
m
any
corn
ers
mu
st h
e ch
eck
in o
rder
to
be s
ure
th
at t
he
oppo
site
sid
es a
re
para
llel
? 3
2. B
OO
KS
Th
e tw
o gr
ay b
ooks
on
th
e bo
oksh
elf
each
mak
e a
70°
angl
e w
ith
th
e ba
se o
f th
e sh
elf.
Wh
at m
ore
can
you
say
abo
ut
thes
e tw
o gr
ay b
ooks
? T
hey a
re p
ara
llel.
3. P
ATT
ERN
S A
rec
tan
gle
is c
ut
alon
g th
e sl
ante
d, d
ash
ed l
ine
show
n i
n t
he
figu
re.
Th
e tw
o pi
eces
are
rea
rran
ged
to f
orm
an
oth
er f
igu
re. D
escr
ibe
as p
reci
sely
as
you
can
th
e sh
ape
of t
he
new
fig
ure
. E
xpla
in.
Para
llelo
gra
m;
Th
e t
op
ed
ges
are
perp
en
dic
ula
r to
th
e v
ert
ical
lin
e s
o t
hey a
re a
sin
gle
lin
e.
Th
e
bo
tto
m e
dg
e i
s a
lso
a s
ing
le l
ine
an
d p
erp
en
dic
ula
r to
th
e s
am
e
lin
e a
s t
he t
op
, so
it
is p
ara
llel
to t
he t
op
. T
he t
op
ed
ge i
s
tran
svers
al
to t
he l
eft
an
d r
igh
t
sla
nte
d e
dg
es a
nd
th
e a
ng
les
are
su
pp
lem
en
tary
. S
o,
the l
eft
an
d r
igh
t ed
ges a
re p
ara
llel.
4. F
IREW
OR
KS
A f
irew
orks
dis
play
is
bein
g re
adie
d fo
r a
cele
brat
ion
. Th
e de
sign
ers
wan
t to
hav
e fo
ur
fire
wor
ks
shoo
t ou
t al
ong
para
llel
tra
ject
orie
s.
Th
ey d
ecid
e to
pla
ce t
wo
lau
nch
ers
on a
do
ck a
nd
the
oth
er t
wo
on t
he
roof
of
a bu
ildi
ng.
70˚
30˚
1
To
pull
off
th
is d
ispl
ay, w
hat
sh
ould
th
e m
easu
re o
f an
gle
1 be
? 80
5. S
IGN
SH
arol
d is
mak
ing
a gi
ant
lett
er
“A”
to p
ut
on t
he
roof
top
of t
he
“A i
s fo
r A
pple
” O
rch
ard
Sto
re. T
he
figu
re s
how
s a
sket
ch o
f th
e de
sign
.
21
108 ˚
a. W
hat
sh
ould
th
e m
easu
res
of a
ngl
es
1 an
d 2
be s
o th
at t
he
hor
izon
tal
part
of
th
e “A
” is
tru
ly h
oriz
onta
l? 1
08
b.
Wh
en b
uil
din
g th
e “A
,” H
arol
d m
akes
su
re t
hat
an
gle
1 is
cor
rect
, bu
t w
hen
h
e m
easu
res
angl
e 2,
it
is n
ot c
orre
ct.
Wh
at d
oes
this
im
ply
abou
t th
e “A
”?
Sam
ple
an
sw
er:
On
e s
ide
of
the “
A” i
s l
on
ger
than
th
e o
ther.
3-5
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-5
Cha
pte
r 3
35
G
lenc
oe G
eom
etry
Enri
chm
ent
Scra
mb
led
-Up
Pro
of
Th
e re
ason
s n
eces
sary
to
com
ple
te t
he
foll
owin
g p
roof
are
sc
ram
ble
d u
p b
elow
. To
com
ple
te t
he
pro
of, n
um
ber
th
e re
ason
s to
mat
ch t
he
corr
esp
ond
ing
stat
emen
ts.
Giv
en: ∠
1 an
d ∠
5 ar
e co
mpl
emen
tary
∠2
and
∠6
are
com
plem
enta
ry
∠5
� ∠
6
Pro
ve:
−−
−
AD
‖ −
−
CF
Sta
tem
ents
Rea
son
s
1. ∠
1 an
d ∠
5 ar
e co
mpl
emen
tary
2. ∠
2 an
d ∠
6 ar
e co
mpl
emen
tary
3. ∠
5 �
∠6
4. ∠
1 �
∠2
5. ∠
1 an
d ∠
2 ar
e co
rres
pon
din
g an
gles
for
li
nes
� ��
AD
an
d � �
�
CF
6.
� ��
AD
‖ �
��
CF
7. −
−−
AD
‖ −
−
CF
con
vers
e of
Cor
resp
ondi
ng
An
gles
Th
eore
m G
iven
an
gles
com
plem
enta
ry t
o co
ngr
uen
t an
gles
are
co
ngr
uen
t G
iven
seg
men
ts c
onta
ined
in
par
alle
l li
nes
are
par
alle
l d
efin
itio
n o
f co
rres
pon
din
g an
gles
Giv
en
3-5
5
1
6
2
6 1 4 2 7 5 3
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A01_A20_GEOCRMC03_890512.indd A16A01_A20_GEOCRMC03_890512.indd A16 6/19/08 2:23:53 PM6/19/08 2:23:53 PM
An
swer
s
Co
pyr
ight
© G
lenc
oe/
McG
raw
-Hill
, a d
ivis
ion
of
The
McG
raw
-Hill
Co
mp
anie
s, In
c.
PDF 2nd
Chapter 3 A17 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
36
G
lenc
oe G
eom
etry
Dis
tan
ce F
rom
a P
oin
t to
a L
ine
Wh
en a
poi
nt
is
not
on
a l
ine,
th
e di
stan
ce f
rom
th
e po
int
to t
he
lin
e is
th
e le
ngt
h o
f th
e se
gmen
t th
at c
onta
ins
the
poin
t an
d is
per
pen
dicu
lar
to t
he
lin
e.
C
onst
ruct
th
e se
gmen
t th
at r
epre
sen
ts t
he
dis
tan
ce
from
E t
o �
⎯
�
AF
.
Ext
end
� ��
AF
. D
raw
� ��
EG
⊥ �
��
AF
.−
−−
EG
rep
rese
nts
th
e di
stan
ce f
rom
E t
o � �
�
AF
.
Exer
cise
sC
onst
ruct
th
e se
gmen
t th
at r
epre
sen
ts t
he
dis
tan
ce i
nd
icat
ed.
1. C
to
� ��
AB
2.
D t
o � �
�
AB
A
C
BX
A
CD
BX
3. T
to
� ��
RS
4.
S t
o � �
�
PQ
U
RS
T
X
RT
S
PQ
X
5. S
to
� ���
QR
6.
S t
o � �
�
RT
R
T
S
P
Q
X
XR
T
S
Pdi
stan
ce b
etw
een
M a
nd P
Q�
Q
M
�
AF
BE
AF
G
BE
Stud
y G
uide
and
Inte
rven
tion
Perp
en
dic
ula
rs a
nd
Dis
tan
ce
Exam
ple
3-6
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-6
Cha
pte
r 3
37
G
lenc
oe G
eom
etry
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Perp
en
dic
ula
rs a
nd
Dis
tan
ce
Dis
tan
ce B
etw
een
Par
alle
l Lin
es T
he
dist
ance
bet
wee
n p
aral
lel
lin
es i
s th
e le
ngt
h
of a
seg
men
t th
at h
as a
n e
ndp
oin
t on
eac
h l
ine
and
is p
erpe
ndi
cula
r to
th
em. P
aral
lel
lin
es
are
ever
ywh
ere
equ
idis
tan
t, w
hic
h m
ean
s th
at a
ll s
uch
per
pen
dicu
lar
segm
ents
hav
e th
e sa
me
len
gth
.
F
ind
th
e d
ista
nce
bet
wee
n t
he
par
alle
l li
nes
l a
nd
m w
ith
th
e eq
uat
ion
s y
= 2
x +
1 a
nd
y =
2x
- 4
, res
pec
tive
ly.
x
y O
m�
Dra
w a
lin
e p
thro
ugh
(0,
1)
that
is
perp
endi
cula
r to
� a
nd
m.
x
y O
mp
�
( 0, 1
)
Lin
e p
has
slo
pe -
1 −
2 a
nd
y-in
terc
ept
1. A
n
equ
atio
n o
f p
is y
= -
1 −
2 x
+ 1
. Th
e po
int
of
inte
rsec
tion
for
p a
nd
� i
s (0
, 1).
To
fin
d th
e po
int
of i
nte
rsec
tion
of
p an
d m
, so
lve
a sy
stem
of
equ
atio
ns.
Lin
e m
: y
= 2
x -
4L
ine
p:
y =
- 1 −
2 x
+ 1
Use
su
bsti
tuti
on.
2x
- 4
= -
1 −
2 x
+ 1
4x
- 8
= -
x +
2
5x =
10
x =
2
Su
bsti
tute
2 f
or x
to
fin
d th
e y-
coor
din
ate.
y =
- 1 −
2 x
+ 1
= -
1 −
2 (2
) +
1 =
-1
+ 1
= 0
Th
e po
int
of i
nte
rsec
tion
of
p an
d m
is
(2, 0
).U
se t
he
Dis
tan
ce F
orm
ula
to
fin
d th
e di
stan
ce b
etw
een
(0,
1)
and
(2, 0
).d
= √
(x
2 -
x1)
2 +
(y 2
-y 1)
2
= √
(2
- 0
)2 +
(0
- 1
)2
= √
5
Th
e di
stan
ce b
etw
een
� a
nd
m i
s √
5 u
nit
s.
Exer
cise
sF
ind
th
e d
ista
nce
bet
wee
n e
ach
pai
r of
par
alle
l li
nes
wit
h t
he
give
n e
qu
atio
ns.
1. y
= 8
2.
y =
x +
3
3. y
= -
2xy
= -
3 y
= x
- 1
y
= -
2x -
5
1
1
2 √
�
2
√
�
5
Exam
ple
3-6
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Answers (Lesson 3-6)
A01_A20_GEOCRMC03_890512.indd A17A01_A20_GEOCRMC03_890512.indd A17 6/10/08 4:41:12 PM6/10/08 4:41:12 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF 2nd
Chapter 3 A18 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Cha
pte
r 3
38
G
lenc
oe G
eom
etry
Skill
s Pr
acti
ceP
erp
en
dic
ula
rs a
nd
Dis
tan
ce
Con
stru
ct t
he
segm
ent
that
rep
rese
nts
th
e d
ista
nce
in
dic
ated
.
1. B
to
� ��
AC
2.
G t
o � �
�
EF
3.
Q t
o � �
�
SR
A
B
C
D
EF
G
S
PQ
R
CO
OR
DIN
AT
E G
EO
ME
TR
Y F
ind
th
e d
ista
nce
fro
m P
to
ℓ.
4. L
ine
ℓ co
nta
ins
poin
ts (
0, −
2) a
nd
(6, 6
). P
oin
t P
has
coo
rdin
ates
(−
1, 5
).
5
5. L
ine
ℓ co
nta
ins
poin
ts (
2, 4
) an
d (5
, 1).
Poi
nt
P h
as c
oord
inat
es (
1, 1
).
2
√ �
2
6. L
ine
ℓ co
nta
ins
poin
ts (
−4,
−2)
an
d (2
, 0).
Poi
nt
P h
as c
oord
inat
es (
3, 7
).
2
√ �
�
10
7. L
ine
ℓ co
nta
ins
poin
ts (
−7,
8)
and
(0, 5
). P
oin
t P
has
coo
rdin
ates
(−
5, 3
2).
3
√ �
�
58
Fin
d t
he
dis
tan
ce b
etw
een
eac
h p
air
of p
aral
lel
lin
es w
ith
th
e gi
ven
eq
uat
ion
s.
8. y
= 7
9.
x =
-6
10. y
= 3
xy
= -
1 x
= 5
y
= 3
x +
10
8
1
1
√
��
10
11. y
= -
5x
12. y
= x
+ 9
13
. y =
-2x
+ 5
y =
-5x
+ 2
6 y
= x
+ 3
y
= -
2x -
5
√ �
�
26
3 √
�
2
2 √
�
5
3-6
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 3-6
Cha
pte
r 3
39
G
lenc
oe G
eom
etry
Prac
tice
Perp
en
dic
ula
rs a
nd
Dis
tan
ce
Con
stru
ct t
he
segm
ent
that
rep
rese
nts
th
e d
ista
nce
in
dic
ated
.
1. O
to
� ���
MN
2.
A t
o � �
�
DC
3.
T t
o � �
��
VU
MN
O
AB
CD
T
VW
SU
CO
OR
DIN
AT
E G
EO
ME
TR
Y F
ind
th
e d
ista
nce
fro
m P
to
l.
4. L
ine
l co
nta
ins
poin
ts (
−2,
0)
and
(4, 8
). P
oin
t P
has
coo
rdin
ates
(5,
1).
5
5. L
ine
l co
nta
ins
poin
ts (
3, 5
) an
d (7
, 9).
Poi
nt
P h
as c
oord
inat
es (
2, 1
0).
3
√ �
2
6. L
ine
l co
nta
ins
poin
ts (
5, 1
8) a
nd
(9, 1
0). P
oin
t P
has
coo
rdin
ates
(−
4, 2
6).
2
√ �
5
7. L
ine
l co
nta
ins
poin
ts (
−2,
4)
and
(1, −
9). P
oin
t P
has
coo
rdin
ates
(14
, −6)
.
√ �
�
178
Fin
d t
he
dis
tan
ce b
etw
een
eac
h p
air
of p
aral
lel
lin
es w
ith
th
e gi
ven
eq
uat
ion
.
8. y
= -
x 9.
y =
2x
+ 7
10
. y =
3x
+ 1
2y
= -
x -
4
y =
2x
- 3
y
= 3
x -
18
2
√ �
2
2 √
�
5
3 √
��
10
11. G
raph
th
e li
ne
y =
-x
+ 1
. Con
stru
ct a
per
pen
dicu
lar
segm
ent
thro
ugh
th
e po
int
at (
-2,
-3)
. Th
en f
ind
the
dist
ance
fro
m t
he
poin
t to
th
e li
ne.
3 √
�
2
12. C
AN
OEI
NG
Bro
nso
n a
nd
a fr
ien
d ar
e go
ing
to c
arry
a c
anoe
acr
oss
a fl
at f
ield
to
the
ban
k of
a s
trai
ght
can
al. D
escr
ibe
the
shor
test
pat
h t
hey
can
use
.
S
am
ple
an
sw
er:
Th
e s
ho
rtest
path
wo
uld
be a
perp
en
dic
ula
r seg
men
t fr
om
wh
ere
th
ey a
re t
o t
he b
an
k o
f th
e c
an
al.
x
y
O
y=
-x
+ 1
( –2,
–3)
3-6
023_
042_
GE
OC
RM
C03
_890
512.
indd
394/
11/0
810
:56:
22A
M
Answers (Lesson 3-6)
A01_A20_GEOCRMC03_890512.indd A18A01_A20_GEOCRMC03_890512.indd A18 6/10/08 4:41:16 PM6/10/08 4:41:16 PM
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