Math (2)_text.pdf

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09
INTRODUCTION
TO
COORDINATE
GEOMETRY
In
the
right
triangle
Define
Coordinate
Geometr
The
study
of
geometrical
shapes
in
a
plane
is
called
plane
geometry.
of
geometrical
shapes
in
the
Cartesian
plane
is
PNQ,
lx~2
IQNI
=
y
2
~
y\
I
I
and
IPNI
=
-
xJ.
Coordinate
geometry
Using
Pythagoras
Theorem
1
2
2
the
study
{PQ)
=(pn)
=(qn)
=>
=>
(coordinate
plane).
We
We
known
that
a
plane
is
divided
d
d
2
=
I
x
2
-X[
2
I
+1
y
2
2
I
-
>’[
2
I
into
four
quadrants
lines
called
the
by
two
perpendicular
axes
intersccting
at
origin.
hâve
also
seen
that
there
is
onc
to
one
=
±^\
x
2
-X|
\j\x
2
+1
y
2
-
y
2
I
{
correspondence
between
the points
of
the
plane
and
the
ordered
pairs
in
d
=
since
-
x,
2
I
+1
y
2
-
y!
2
t
R
x
R.
d >
0
always.
Finding
Distance
between
two
points
Let
P(x
y,
)
and
Q(x
2
,y
2
)
be
{
,
two
points
is
in
the
coordinate
plane
where
d
Using
the
distance
formula,
find
the
distance
between
the
points.
(i)
the
length
of
the
line
segment PQ.
i.e.
P(1
,
2)
and
Q(0,3)
!PQI
=
d.
The
line
segments
MQ
(ü)
S(—
1,
3)
and
R(3,
-2)
and
LP
parallel
to
y-axis
mect
x-axis
at
points
and
L,
rcspeclively
with
coordinates
M(x
2
0)
and
L(x
,,()).
,
M
ISolutions
US
=
7(0-l)
+(3-2)
2
V(-l)
+(l)
2
2
(i)IPQI
2
—
The
line-segment
PN
is
parallel
to
x-axis.
=
N
/m=
V2
(ii)ISRI
=
x/(3
—
(
—
l))
2
+
(—2
—
3)
2
\[4Ï
2
2
=
V(3+l)
+(-5)
=-y/l6+25=
Collinear
or
Non-collinear
Points
in
the
Two
which
lie
more
than
two
points
on
the
same
straight
line
are
or
they
are
called
called
collinear
points
with
respect
to that
line;
otherwise
non-
collinear.
Let
PQ
be
a
line,
then
ail
the
points
on
line
m
are
collinear.
6
)
2
In
the
given
figure
the
points
P
and
Q
are
colîinear
with
respect
to
the
line
m
Sol
lPSI=V(-2-l)
+(-l
+
l)
2
2
and
the
points
P
and
R
are
not
colîinear
with
respect
to
il.
=
Since
I
n
/(-
3)
2
+0
=3
1
R
4
se
of
Distance
—
—
R
=
QS
=
X
/(I
-
0
2
+
(-
-
3)
2
I
>
»
and
=
/l+Ï6
=
I
n/17,
I,
PQ
l+l
QS
l#l
PS
Fo
Therefore
Ihe
points,
P,Q
and
S
are
IIUÜ
I
ollineàritÿ
dit
Thgeè
or
niore
Pointe
and
be
three
points
in
not
colîinear
and
hence,
the
points
P,
Q,
R
and
S
are
also
not
Let
P,
the
plane.
colîinear.
Q
+
They
are called colîinear
A
closed
figure
in
a
plane
obtaincd
is
If
IPQI
IQRI
IPRI,
otherwise
by
joining
three
non-collinear
points
called
a
triangle.
In
they
are
non-collinear.
Exàmple
Using
distance
formula
show
the
points.
(i)
the
triangle
ABC"
the
non-
that
collinear
points
A,
vcrtices
of
the
B
and
C
arc the
three
triangle
ABC.
The
line
and
P(—
2,—
1),
Q(0,
3)
and
R(l,
5)
are
colîinear.
segments
AB,
of
the
triangle.
BC
CA
are
called
sides
(ii)
The
above P,Q,R
and
S
not
colîinear
(1,-1) are
Sol.
11
By
using
the
distance
formula,
we
nd
|PQ|
—
-y/tü+2)
2
+(3
+
I)
2
=
J
4
+
1
=
n
/20
=
2
75
If
the
lenglhs
of
ail
the
three
sides
is
|QR|
2
=
V(1-0)
h-(+5-3)
2
of
a
triangle
are
saine,
then
the
triangle
called
an
équilatéral
triangle.
=
n/
Ï
+4
=
\PR\
2
S
2
=
J(\+2)
+
(5
+
})
The
=3n/5
triangle
OPQ
is
an
équilatéral
triangle
f
1
=
n/9
+
36=V45
Since
PQ|
+
|QR|
j
'
,0
since
the
points
0(0,0),
P
and
=
2^5
+
JK
i
V
=
3
Js
=|PR|
points
P,
Q,
(ii)
S
’
A
are
not
colîinear
,
where
R
are
colîinear.
(1,-1)
2V2
IOPI
2n/2
I
The
above
points
P,Q,R
and
S
are
not
colîinear
lî
,
IQOI
=
..VL'
2/i
The
triangle
triangle
PQR
is
an
isoscelcs
points
in
the
2V2
as
for
the
non-collinear
1)
14
8
8
P(—
1,0),
Q(l,
0)
and
R(0,
following
figure.
shown
Q
I
(
PQI=
V3_
-0
2V2
2
1-2
v
V2
/
'
2V2
+
V3
ÎZ
^
^U>/2
1
.
2v/2
V'
+
1
-
/£
8
V
8
8
i.e.,
IOPI
=
IQOI
=
!PQI=—
a
real
number
V2
and
the
points
0(0.0),
1
Q
2V2
collinear.
’
VL
2V2
andpf
-~
the
I
PQI
=
t
0
n
/(
1-(-1))
2
+
(0
-
O)
=
v
'(l
+
2
2
l
)
+0
=
-J
4
=
2
i
are
not
I
2
2
QRI=n/(0-1)
+(!-())'
=yj(-
l)
+
2
l
=Vl
+
=yjï
sfî
Hence
triangle
OPQ
is
I
PR
=
\J(Q
•-
(•-
I
I
équilatéral.
Sinee
QRM
-
0)
2
=
Jî+Ï
=
PRI
=
>/2
and PQI
=
2*
1
2
))
h-
(
1
I
so
the
non-collinear
points
isoscelcs
triangle
P,
Q,
R
form
an
PQR.
A
triangle
in
which
one
of
the
angles
has
to
90"
is
mcasure
equal
triangle.
callcd
a
right
angle
Lel
0(0,
0),
P(—
3,
0)
and
Q(0,
2)
be
three
non-collinear
points.
Verify
lhat
triangle
OPQ
is
right-angled.
I
An
triangle
isoscelcs
triangle
its
PQR
OQI
=
yj(Q
—
O)
2
+
(2-
O)
2
2
2
=V?=2
is
a
which
has
two
of
sides
with
IOPI=7(-3)
+0
2
I
=y/9
2
=3
equal
length
while
the
third
sidc
has
a
different
length.
PQI
=
V(-3)
+
(-2)
=
s/Ô+4
=
VÏ3
3
Hence
PQI=
VÎÔ,I
QRI=4and!
PRI
=
sfï
I
The
points
P,
I
Q
and
R
are
non-collinear
since,
PQ!+I
QRI>!
PRI
Thus
the
given
points
form
a
scalene
triancrlp
f
OQI
2
+1
OP!
=(2)
2
2
2
2
+
(3)
=1
3
and!
PQ!
=1
Since
i
OQ!
2
+1
OPI
2
=
PQ!
2
I
,
therefore
ZPOQ
=90°
If
A(2,
2),
B(2,
-2),
Hence
the
given
non-collinear
points
form
a
right
triangle.
C(—
2,
-2)
and
in
Scalene
Triangle
D(-2,
2)
be
four
non-collinear
points
is
A
triangle
if
triangle
called
ail
a
scalene
the
plane,
then
verify
thaï
they
form
a
measures
of
the
three
sides
square
ABCD.
ABI=^/(2-2)
+(-2-2)
2
2
are
different.
Show
and
R(2,
triangle.
Since
that
the
points
P(l,
2),
1)
in
the
plane
!
Q(-2,
1)
2
2
=
/o
+
M)
=
716=4
form
a
scalene
a
!PQ!=7i
(—
2-1)
+
(1—
2)
=>/(-
3)
+(-l)
2
:
I
BC\=yj
(—2
-
2)
+
(—2
+
2)
2
2
2
2
2
2
—
V
(
—
4)
+0
=
\f\6
=
4
2
=>
/9
+
l=Vîô
I
CDI
=
^/(-2
-
(—2))
+
(2
-
(-2))
2
2
1
QRI=
^
f(2+2)
2
+(l-l)
2
=
a
/
(-
2
+
2)
2
+(
2
+
2)
2
=
7
0+16
=
7Ï6=4
I
=
a/4
2
+0
2
and
1
= V4
2
=
4
2
2
2
DAI=V(2+2)
+(2-2)
2
PRI=V(
2—
l)
+(l-2)
2
=
a
/(+4)
+0=VÏ6 =4
2
2
2
2
=
V
l
+
(-l)
=
A
/l
+l
=
72
Hence
!ABf
=
IBCI
=
ICDI
=
IDA!
=
4
Aiso
!
ACI=7(-2
-
2)
2
+
(-2
-
2)
2
Solution:
=
VÏ6
+
16=>/32=4V2
Now
2
I
ABI
2
+1
BCI
=(4)
+(4)
2
=32,
and
2
2
s
s
H
!
»
i
as
|AC|
=
(A\fl)
2
=
32
ABI
2
+1
BC!
2
=1
ACI
2
,
i
a
M
3a
S
mm
m
s
|
i
m
By
Sincel
îherefore
ZABC
=
90°
Hence
the
given
four
non-collinear
points
l'orm
a
square.
rr
D(
te
F
p
——
Y
y
TC™
m
i
m
1
II
1
:
m
11
B:
fl
j
m
a
m
m
IS
s
5
m
a
K
11
fl
H
bm
B
K
H
m
a
flfl
a
mm
m
II
B
m
i
1
Ss5
1
RI
i
m
11
1
1
fl
1
l
im
«1
ifl
fl:
IBB
n
1
+
distance
formula.
—
)
I
ABI
=
1
=
7^
+0
=\/Ï6=
"û"
><
4
3
2
r
r
1_
“i
>
—
ICDI=7(3+l)
2
+
(3-3)
2
2
=
V
4
+0
=
VÎ6-4
L_
t—
bç
2,-5
-â
Çf
?L
__
A
points
.
H
j3
ram
r
I
ADI=7(-l
+
2)
2
+(3—
l)
2
L_
-7^
+
2
2
=^+4=>/5
IBCI=V(3-2)
+
(3-l)
2
2
ëfinë
fiarallelb
figure
in
fonned
by
four
non-collinear
the
if
plane
is
called
a
parallelogram
(i)
(ii)
Since
its
its
opposite
sûtes
are
of
equal
length
opposite
sides
are
parallel
in
90°
IABI=ICDI
=
4
and
IADI=IBCI
=
\/5
So
opposite
sides
of
the
quadrilatéral
(iii)
measure
of
none
of
the
angles
ABCD
are
equal.
Aiso
IACI
=
/(3
+
2)
2
Show
that
the
points
A(-2,
1),
B(2,
+
(3
-l)
2
1),
C(3,
3)
and
D(-l,
3)
forai
a
parallelogram.
—
7(5)”
+2‘
=
,{25+4
=
V29
Now
I
2
2
ABI
+IBCI
=
6
+
5
=
2
and
IACI
2
=
29
1
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