CLASS X MATHS Coordinate geometry

CLASS X MATHS Coordinate geometry

1. Coordinate Geometry
Class X

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

3. •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

4. 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

5. 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).

6. RECAP
Coordinate Plane
XX’
Y
Y’
O
Origin
1234
+ve direction
-1-2-3-4
-ve direction
-1
-2
-3 -
v
e

d
i
r
e
c
t
i
o
n

1
2
3
+
v
e

d
i
r
e
c
t
i
o
n

X-axis : X’OX
Y-axis : Y’OY

7. Quadrants
XX’ O
Y
Y’
III
IIIIV
(+,+)(-,+)
(-,-) (+,-)

8. Coordinates
XX’
Y
Y’
O
1234-1-2-3-4
-1
-2
-3
1
2
3
(2,1)
(-3,-2)
Ordinate
Abcissa
(?,?)

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

10. Distance Formula
0
x
1
x
2
y
2
y
1
P(x
1
,y
1
)
Q(x
2
,y
2
)
Y-axis
X-axis

11. Let us now find the distance between any two points P(x
1
, y
1
) and
Q(x
1
, y
2
)
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 = x
1
, OS = x
2
,
PR = PS = y
1
, QS = y
2

Then , PT = x
2
– x
1
,
QT = y
2
– y
1
x
Y
P (x
1
, y
1
)
Q(x
2
, y
2
)
T
R SO
Distance Formula

12. Now, applying the Pythagoras theorem in ΔPTQ, we get
Therefore

222
QTPTPQ +=
( ) ( )
2
12
2
12
yyxxPQ -+-=
which is called the distance formula.

13. 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

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

15. 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.

16. Applications of Distance Formula
Parallelogram
Prove opposite sides are equal or diagonals bisect
each other

17. Applications of Distance Formula
Rhombus
Prove all 4 sides are equal

18. Applications of Distance Formula
Rectangle
Prove opposite sides are equal and diagonals are
equal.

19. Applications of Distance Formula
Square
Prove all 4 sides are equal and diagonals are equal.

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

21. Section Formula
2
1
m
m
PB
PA
=
Consider any two points A(x
1
, y
1
) and B(x
1
,y
2
) and assume that P (x, y)
divides AB internally in the ratio m
1
: m
2
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 (x
1
, y
1
)
B(x
2
, y
2
)
P (x , y)
R SO T
m
1
m
2
Q
C

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

23. Section Formula
For x - coordinate
Taking
or
( )
( )xx
xx
m
m
-
-
=
2
1
2
1
( ) ( )
1221
xxmxxm -=-
122121 xmxmxmxm -=-or
or ( )
121221
mmxxmxm +=+
12
1221
mm
xmxm
x
+
+
=

24. Section Formula
For y – coordinate
Taking ( )
( )yy
yy
m
m
-
-
=
2
1
2
1
( ) ( )
1221
yymyym -=-
122121 ymymymym -=-
( )
121221
mmyymym +=+
12
1221
mm
ymym
y
+
+
=
or
or
or

25. Midpoint
Midpoint of A(x
1
, y
1
) and B(x
2
,y
2
)
m:n º 1:1
1 2 1 2
x x y y
P ,
2 2
+ +æ ö
\ º
ç ÷
è ø
Find the Mid-Point of P(1,-3) and Q(5,7).

26. Area of a Triangle
Let ABC be any triangle whose vertices are A(x
1
, y
1
), B(x
2
, y
2
) and
C(x
3
, y
3
).
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 = (x
2
– x
1
)
PR = (x
3
– x
1
)
QR = (x
3
– x
2
)
x
Y
A (x
1
, y
1
)
B(x
2
, y
2
)
C (x
3
, y
3
)
P QO R

27. Area of a Triangle
XX’
Y’
O
Y
A(x
1
, y
1
)
C(x
3
, y
3
)
B
(
x
2
,

y
2
)
M L N
Area of D ABC =
Area of trapezium ABML+ Area of trapezium ALNC
- Area of trapezium BMNC

28. 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
=
( )( )embetween th distancesides parallel of sum
2
1
( ) ( ) ( )PRCR + AP
2
1
CR + BQ
2
1
QPAP + BQ
2
1
-+ QR
( )( ) ( )( ) ( )( )
133123321212
2
1
2
1
2
1
xxyyxxyyxxyy -+--++-+=
( )( )( )[ ]
133311312333223211211222
2
1
xyxyxyxyxyxyxyxyxyxyxyxy -+---+-+-+-=
( ) ( ) ( )[ ]
123312231
2
1
yyxyyxyyx -+-+-=
Area of Δ ABC

29. 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.

31. THANK YOU