C
oor
di
n
a
t
e Ge
om
e
try
—
F
u
ll De
pt
h S
tu
d
y
G
u
ide
(
NC Ma
t
h
2
CDM
)
1
.
COORDINATE PLANE
—
Dee
p
Di
v
e
Di
st
a
n
ce F
ormu
la
Fi
n
d
s
t
he
str
aigh
t
-
li
n
e di
st
a
n
ce be
tw
ee
n
two
po
i
nts
(
x₁
,
y₁
)
a
n
d
(
x₂
,
y₂
).
I
t
c
om
e
s
di
r
ec
t
l
y
f
rom
t
he P
yt
hag
or
ea
n
The
or
e
m
—
t
he h
or
i
zont
al di
st
a
n
ce a
n
d
v
e
rt
ical di
st
a
n
ce be
tw
ee
n
t
he
po
i
nts
f
orm
t
he
two
leg
s
o
f a
r
igh
t
tr
ia
n
gle
,
a
n
d
t
he di
st
a
n
ce be
tw
ee
n
t
he
po
i
nts
i
s
t
he
h
ypot
e
nus
e
.
F
ormu
la
:
d
=
√
[(
x₂
−
x₁
)
²
+
(
y₂
−
y₁
)
²
]
P
ro
ce
ss
:
1
.
S
u
b
tr
ac
t
t
he
x
-
c
oor
di
n
a
t
e
s
,
squ
a
r
e
t
he
r
e
su
l
t
.
2
.
S
u
b
tr
ac
t
t
he
y
-
c
oor
di
n
a
t
e
s
,
squ
a
r
e
t
he
r
e
su
l
t
.
3
.
Add
t
he
two
squ
a
r
e
s
to
ge
t
he
r
.
4
.
Take
t
he
squ
a
r
e
root
.
E
x
a
mp
le
:
Di
st
a
n
ce be
tw
ee
n
(
1
,
2
)
a
n
d
(
4
,
6
):
d
=
√
[(
4
−
1
)
²
+
(
6
−
2
)
²
]
=
√
[
9
+
16
]
=
√
25
=
5
O
r
de
r
d
o
e
sn
'
t
m
a
tt
e
r
f
or
su
b
tr
ac
t
i
on
—
s
i
n
ce
you
squ
a
r
e
t
he di
ff
e
r
e
n
ce
,
(
x₂
−
x₁
)
²
gi
v
e
s
t
he
s
a
m
e
r
e
su
l
t
a
s
(
x₁
−
x₂
)
²
.
J
ust
be c
ons
i
st
e
nt
w
i
t
hi
n
each
p
ai
r
.
Mid
po
i
nt
F
ormu
la
Fi
n
d
s
t
he
po
i
nt
e
x
ac
t
l
y
half
w
a
y
be
tw
ee
n
two
po
i
nts
—
e
ss
e
nt
iall
y
t
he a
v
e
r
age
o
f
t
he
x
-
c
oor
di
n
a
t
e
s
a
n
d
t
he a
v
e
r
age
o
f
t
he
y
-
c
oor
di
n
a
t
e
s
.
F
ormu
la
:
M
=
((
x₁
+
x₂
)/
2
,
(
y₁
+
y₂
)/
2
)
E
x
a
mp
le
:
Mid
po
i
nt
o
f
(
2
,
-
3
)
a
n
d
(
8
,
5
):
M
=
((
2
+
8
)/
2
,
(-
3
+
5
)/
2
)
=
(
10
/
2
,
2
/
2
)
=
(
5
,
1
)
W
or
ki
n
g back
w
a
r
d
(
fin
di
n
g a
n
e
n
d
po
i
nt
gi
v
e
n
t
he
m
id
po
i
nt
a
n
d
on
e e
n
d
po
i
nt
):
If M i
s
t
he
m
id
po
i
nt
o
f
s
eg
m
e
nt
AB
,
a
n
d
you
k
now
A a
n
d M
,
so
l
v
e f
or
B
:
x
_
B
=
2
(
x
_
M
)
−
x
_
A
,
y
_
B
=
2
(
y
_
M
)
−
y
_
A
E
x
a
mp
le
:
M
=
(
4
,
1
),
A
=
(
1
,
-
2
).
Fi
n
d B
.
x
_
B
=
2
(
4
)
−
1
=
7
,
y
_
B
=
2
(
1
)
−
(−
2
)
=
4
→
B
=
(
7
,
4
)
Seg
m
e
nt
Addi
t
i
on
If
po
i
nt
B lie
s
ON
s
eg
m
e
nt
AC
(
be
tw
ee
n
A a
n
d C
),
t
he
n
t
he
two
sm
alle
r
p
iece
s
add
up
to
t
he
w
h
o
le
:
AB
+
BC
=
AC
Thi
s
a
pp
lie
s
di
r
ec
t
l
y
on
t
he c
oor
di
n
a
t
e
p
la
n
e
too
—
if
t
h
r
ee
po
i
nts
a
r
e c
o
lli
n
ea
r
(
on
t
he
s
a
m
e
li
n
e
)
a
n
d B i
s
be
tw
ee
n
A a
n
d C
,
you
ca
n
us
e
t
he di
st
a
n
ce f
ormu
la
on
each
p
iece a
n
d c
onfirm
t
he
y
add
up
c
orr
ec
t
l
y
,
or
us
e
s
eg
m
e
nt
addi
t
i
on
to
so
l
v
e f
or
a
n
un
k
nown
c
oor
di
n
a
t
e
.
C
ommon
us
e
:
you
'
r
e gi
v
e
n
AB a
n
d BC a
s
algeb
r
aic e
xpr
e
ss
i
ons
(
i
n
t
e
rms
o
f
x
)
a
n
d
t
he
tot
al
AC
—
s
e
t
up
t
he e
qu
a
t
i
on
AB
+
BC
=
AC a
n
d
so
l
v
e f
or
x
,
t
he
n
su
b
st
i
tut
e back
to
fin
d ac
tu
al
s
eg
m
e
nt
le
n
g
t
h
s
.
2
.
EQUATIONS OF LINES
—
Dee
p
Di
v
e
Sl
op
e
Mea
sur
e
s
t
he
st
ee
pn
e
ss
/
di
r
ec
t
i
on
o
f a li
n
e
—
"
r
i
s
e
ov
e
r
run
,"
t
he
r
a
t
i
o
o
f
v
e
rt
ical cha
n
ge
to
h
or
i
zont
al cha
n
ge
.
F
ormu
la
(
f
rom
two
po
i
nts
):
m
=
(
y₂
−
y₁
)/(
x₂
−
x₁
)
Readi
n
g
s
l
op
e
typ
e
s
:
P
os
i
t
i
v
e
s
l
op
e
→
li
n
e
r
i
s
e
s
lef
t
to
r
igh
t
Nega
t
i
v
e
s
l
op
e
→
li
n
e fall
s
lef
t
to
r
igh
t
Ze
ro
s
l
op
e
→
h
or
i
zont
al li
n
e
(
y
=
c
onst
a
nt
)
U
n
de
fin
ed
s
l
op
e
→
v
e
rt
ical li
n
e
(
x
=
c
onst
a
nt
);
t
hi
s
ha
pp
e
ns
w
he
n
x₂
−
x₁
=
0
(
ca
n
'
t
di
v
ide b
y
z
e
ro
)
F
rom
a g
r
a
p
h
:
c
ount
r
i
s
e
/
run
be
tw
ee
n
two
clea
r
l
y
m
a
r
ked
po
i
nts
.
F
rom
a
t
able
:
p
ick a
ny
two
(
x
,
y
)
rows
a
n
d a
pp
l
y
t
he
two
-
po
i
nt
f
ormu
la
;
check
t
ha
t
t
he
s
l
op
e i
s
t
he
s
a
m
e be
tw
ee
n
ot
he
r
p
ai
rs
o
f
rows
to
c
onfirm
li
n
ea
r
i
ty
.
Sl
op
e
-
I
nt
e
r
ce
pt
F
orm
y
=
mx
+
b
m
=
s
l
op
e
b
=
y
-
i
nt
e
r
ce
pt
(
w
he
r
e
t
he li
n
e c
ross
e
s
t
he
y
-
a
x
i
s
,
a
t
x
=
0
)
Be
st
f
or
qu
ickl
y
g
r
a
p
hi
n
g a li
n
e
(
p
l
ot
b
on
t
he
y
-
a
x
i
s
,
t
he
n
us
e
t
he
s
l
op
e
to
fin
d a
s
ec
on
d
po
i
nt
)
or
r
eadi
n
g
s
l
op
e
/
y
-
i
nt
e
r
ce
pt
di
r
ec
t
l
y
f
rom
a
n
e
qu
a
t
i
on
.
W
r
i
t
i
n
g
t
he e
qu
a
t
i
on
gi
v
e
n
s
l
op
e a
n
d
y
-
i
nt
e
r
ce
pt
:
p
l
u
g di
r
ec
t
l
y
i
n
.
W
r
i
t
i
n
g
t
he e
qu
a
t
i
on
gi
v
e
n
s
l
op
e a
n
d a
po
i
nt
(
not
t
he
y
-
i
nt
e
r
ce
pt
):
su
b
st
i
tut
e
t
he
po
i
nt
i
nto
y
=
mx
+
b a
n
d
so
l
v
e
f
or
b
,
t
he
n
r
e
wr
i
t
e
t
he f
u
ll e
qu
a
t
i
on
.
P
o
i
nt
-
Sl
op
e F
orm
y
−
y₁
=
m
(
x
−
x₁
)
Be
st
w
he
n
you
ha
v
e a
s
l
op
e a
n
d ANY
po
i
nt
on
t
he li
n
e
(
d
o
e
sn
'
t
n
eed
to
be
t
he
y
-
i
nt
e
r
ce
pt
).
Ve
ry
us
ef
u
l a
s
a
st
a
rt
i
n
g
po
i
nt
,
t
he
n
ca
n
be
s
i
mp
li
fi
ed
/
c
onv
e
rt
ed i
nto
s
l
op
e
-
i
nt
e
r
ce
pt
f
orm
b
y
di
str
ib
ut
i
n
g a
n
d
so
l
v
i
n
g f
or
y
.
W
r
i
t
i
n
g
t
he e
qu
a
t
i
on
gi
v
e
n
two
po
i
nts
:
1
.
Fi
n
d
t
he
s
l
op
e
us
i
n
g
t
he
two
po
i
nts
.
2
.
Pl
u
g
t
he
s
l
op
e a
n
d ONE
o
f
t
he
po
i
nts
i
nto
po
i
nt
-
s
l
op
e f
orm
.
3
.
(
O
pt
i
on
al
)
s
i
mp
lif
y
i
nto
s
l
op
e
-
i
nt
e
r
ce
pt
f
orm
.
E
x
a
mp
le
:
li
n
e
t
h
rou
gh
(
2
,
5
)
a
n
d
(
6
,
13
)
m
=
(
13
−
5
)/(
6
−
2
)
=
8
/
4
=
2
y
−
5
=
2
(
x
−
2
)
→
y
−
5
=
2x
−
4
→
y
=
2x
+
1
Pa
r
allel Li
n
e
s
Pa
r
allel li
n
e
s
ha
v
e
t
he EXACT SAME
s
l
op
e
,
b
ut
di
ff
e
r
e
nt
y
-
i
nt
e
r
ce
pts
(
so
t
he
y
n
e
v
e
r
i
nt
e
rs
ec
t
).
W
r
i
t
i
n
g a
p
a
r
allel li
n
e
'
s
e
qu
a
t
i
on
:
us
e
t
he
s
a
m
e
s
l
op
e a
s
t
he gi
v
e
n
li
n
e
,
p
l
u
g i
n
t
he
n
e
w
po
i
nt
us
i
n
g
po
i
nt
-
s
l
op
e f
orm
.
Pe
rp
e
n
dic
u
la
r
Li
n
e
s
Pe
rp
e
n
dic
u
la
r
li
n
e
s
i
nt
e
rs
ec
t
a
t
a
90
°
a
n
gle
.
Thei
r
s
l
op
e
s
a
r
e NEGATIVE RECIPROCALS
o
f
each
ot
he
r
—
fl
i
p
t
he f
r
ac
t
i
on
AND cha
n
ge
t
he
s
ig
n
.
R
u
le
:
m₁
×
m₂
=
−
1
E
x
a
mp
le
:
a li
n
e
w
i
t
h
s
l
op
e
3
/
4
→
p
e
rp
e
n
dic
u
la
r
s
l
op
e
=
−
4
/
3
E
x
a
mp
le
:
a li
n
e
w
i
t
h
s
l
op
e
−
2
(
i
.
e
.,
−
2
/
1
)
→
p
e
rp
e
n
dic
u
la
r
s
l
op
e
=
1
/
2
E
x
a
mp
le
:
a h
or
i
zont
al li
n
e
(
s
l
op
e
0
)
→
p
e
rp
e
n
dic
u
la
r
li
n
e i
s
v
e
rt
ical
(
un
de
fin
ed
s
l
op
e
)
W
r
i
t
i
n
g a
p
e
rp
e
n
dic
u
la
r
li
n
e
'
s
e
qu
a
t
i
on
:
fin
d
t
he
n
ega
t
i
v
e
r
eci
pro
cal
o
f
t
he gi
v
e
n
s
l
op
e
,
p
l
u
g i
nto
po
i
nt
-
s
l
op
e f
orm
w
i
t
h
t
he
n
e
w
po
i
nt
.
15
PRACTICE PROBLEMS
1
.
Fi
n
d
t
he di
st
a
n
ce be
tw
ee
n
(−
2
,
3
)
a
n
d
(
4
,
−
5
).
2
.
Fi
n
d
t
he di
st
a
n
ce be
tw
ee
n
(
0
,
0
)
a
n
d
(
7
,
24
).
3
.
Fi
n
d
t
he
m
id
po
i
nt
o
f
t
he
s
eg
m
e
nt
c
onn
ec
t
i
n
g
(−
6
,
2
)
a
n
d
(
10
,
-
8
).
End of Document
331
