CGc10.pdf

1. Cartesian Coordinate
system and Quadrants
2. Distance formula
3. Area of a triangle
4. Section formula
Objectives

• The use of algebra to study
geometric properties i.e.
operates on symbols is defined
as the coordinate system.
What is co-ordinate geometry ?
IntroductIon

 A system of geometry where the position of points on the plane is described
using an ordered pair of numbers.
 The method of describing the location of points in this way was
 proposed by the French mathematician René Descartes .
 He proposed further that curves and lines could be described by
equations using this technique, thus being the first to link
algebra and geometry.
 In honor of his work, the coordinates of a point are often referred to as
its Cartesian coordinates, and the coordinate plane as the Cartesian
Coordinate Plane.
Coordinate Geometry

SoME BASIc PoIntS
To locate the position of a point on a plane,we
require a pair of coordinate axes.
The distance of a point from the y-axis is called its
x-coordinate,OR abscissa.
The distance of a point from the x-axis is called its
y-coordinate,OR ordinate.
The coordinates of a point on the x-axis are
of the form (x, 0) and of a point on the y-axis
are of the form (0, y).

RECAP
Coordinate Plane
X
X’
Y
Y’
O
Origin
1 2 3 4
+ve direction
-1
-2
-3
-4
-ve direction
-1
-2
-3
-ve
direction
1
2
3
+ve
direction
X-axis : X’OX
Y-axis : Y’OY

Quadrants
X
X’ O
Y
Y’
I
II
III IV
(+,+)
(-,+)
(-,-) (+,-)

Coordinates
X
X’
Y
Y’
O
1 2 3 4
-1
-2
-3
-4
-1
-2
-3
1
2
3
(2,1)
(-3,-2)
Ordinate
Abcissa
(?,?)

X
X’ O
Y
Y’
I
II
III IV
(+,+)
(-,+)
(-,-) (+,-)
Q : (1,0) lies in which Quadrant?
Ist? IInd?
A : None. Points which lie on the axes do not lie in any
quadrant.

Distance Formula
0
x1
x2
y2
y1
P(x1,y1)
Q(x2,y2)
Y-axis
X-axis

 Let us now find the distance between any two points P(x1, y1) and
Q(x1, y2)
 Draw PR and QS ⊥ x-axis. A
perpendicular from the point P on
 QS is draw to meet it at the point T
So, OR = x1 , OS = x2 ,
PR = PS = y1 , QS = y2
Then , PT = x2 – x1 ,
QT = y2 – y1
x
Y
P (x1 , y1)
Q(x2 , y2)
T
R S
O
Distance Formula

Now, applying the Pythagoras theorem in ΔPTQ, we get
Therefore
2
2
2
QT
PT
PQ +
=
( ) ( )2
1
2
2
1
2 y
y
x
x
PQ −
+
−
=
which is called the distance formula.

Example 1: Find the distance between P(1,-3) and Q(5,7).
The exact distance between A(1, -3) and B(5, 7) is

Distance From Origin
Distance of P(x, y) from the origin is
( ) ( )
2 2
x 0 y 0
= − + −
2 2
x y
= +

Applications of Distance Formula
To check which type of triangle is
formed by given 3 coordinates.
and
To check which type of quadrilateral is
formed by given 4 coordinates.

Parallelogram
Prove opposite sides are equal or diagonals bisect
each other

Rhombus
Prove all 4 sides are equal

Rectangle
Prove opposite sides are equal and diagonals are
equal.

Square
Prove all 4 sides are equal and diagonals are equal.

Collinearity of Three Points
Use Distance Formula
a b
c
Show that a+b = c

Section Formula
2
1
m
m
PB
PA
=
 Consider any two points A(x1 , y1) and B(x1 ,y2) and assume that P (x, y)
divides AB internally in the ratio m1: m2 i.e.
 Draw AR, PS and BT ⊥ x-axis.
 Draw AQ and PC parallel to the x-axis.
Then,
by the AA similarity criterion,
x
Y
A (x1 , y1)
B(x2 , y2)
P (x , y)
R S
O T
m1
m2
Q
C

Section Formula
ΔPAQ ~ ΔBPC
---------------- (1)
Now,
AQ = RS = OS – OR = x– x1
PC = ST = OT – OS = x2– x
PQ = PS – QS = PS – AR = y– y1
BC = BT– CT = BT – PS = y2– y
Substituting these values in (1), we get
BC
PQ
PC
AQ
BP
PA
=
=
( )
( )
( )
( )
y
y
y
y
x
x
x
x
m
m
−
−
=
−
−
=
2
1
2
1
2
1 ( )
( )
( )
( )
y
y
y
y
x
x
x
x
m
m
−
−
=
−
−
=
2
1
2
1
2
1

Section Formula
For x - coordinate
Taking
or
( )
( )
x
x
x
x
m
m
−
−
=
2
1
2
1
( ) ( )
1
2
2
1 x
x
m
x
x
m −
=
−
1
2
2
1
2
1 x
m
x
m
x
m
x
m −
=
−
or
or ( )
1
2
1
2
2
1 m
m
x
x
m
x
m +
=
+
1
2
1
2
2
1
m
m
x
m
x
m
x
+
+
=

Section Formula
For y – coordinate
Taking ( )
( )
y
y
y
y
m
m
−
−
=
2
1
2
1
( ) ( )
1
2
2
1 y
y
m
y
y
m −
=
−
1
2
2
1
2
1 y
m
y
m
y
m
y
m −
=
−
( )
1
2
1
2
2
1 m
m
y
y
m
y
m +
=
+
1
2
1
2
2
1
m
m
y
m
y
m
y
+
+
=
or
or
or

Midpoint
Midpoint of A(x1, y1) and B(x2,y2)
m:n ≡ 1:1
1 2 1 2
x x y y
P ,
2 2
+ +
 
∴ ≡  
 
Find the Mid-Point of P(1,-3) and Q(5,7).

Area of a Triangle
 Let ABC be any triangle whose vertices are A(x1, y1), B(x2 , y2) and
C(x3 , y3).
 Draw AP, BQ and CR
 perpendiculars from A,B and C,
 respectively, to the x-axis.
Clearly ABQP, APRC and
BQRC are all trapezium,
Now, from figure
QP = (x2 – x1)
PR = (x3 – x1)
QR = (x3 – x2)
x
Y
A (x1 , y1)
B(x2 , y2)
C (x3 , y3)
P Q
O R

Area of a Triangle
X
X’
Y’
O
Y A(x1, y1)
C(x3, y3)
B(x
2
,
y
2
)
M L N
Area of ∆ ABC =
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC

Area of a Triangle
Area of ABC
Δ = Area of trapezium ABQP + Area of
trapezium BQRC– Area of trapezium APRC.
We also know that ,
Area of trapezium =
Therefore,
Area of ABC
Δ =
( )( )
em
between th
distance
sides
parallel
of
sum
2
1
( ) ( ) ( )PR
CR
+
AP
2
1
CR
+
BQ
2
1
QP
AP
+
BQ
2
1
−
+ QR
( )( ) ( )( ) ( )( )
1
3
3
1
2
3
3
2
1
2
1
2
2
1
2
1
2
1
x
x
y
y
x
x
y
y
x
x
y
y −
+
−
−
+
+
−
+
=
( ) ( ) ( )
[ ]
1
3
3
3
1
1
3
1
2
3
3
3
2
2
3
2
1
1
2
1
1
2
2
2
2
1
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y −
+
−
−
−
+
−
+
−
+
−
=
( ) ( ) ( )
[ ]
1
2
3
3
1
2
2
3
1
2
1
y
y
x
y
y
x
y
y
x −
+
−
+
−
=
Area of Δ ABC

PROJECT
Mark coordinate axes on your city map and
find distances between important landmarks-
bus stand, railway station, airport, hospital, school,
Your house, Any river etc.

CGc10.pdf

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