This document provides an overview of coordinate geometry. It defines key concepts like the Cartesian coordinate system, quadrants, and using an ordered pair (x,y) to locate points on a plane. It then explains how to calculate the distance between two points using the distance formula. Other topics covered include finding the area of a triangle using the coordinates of its vertices, using the section formula to divide a line segment internally, and finding the midpoint of two points.

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
1 2 3 4
+ve direction
-1-2-3-4
-ve direction
-1
-2
-3
-vedirection
1
2
3
+vedirection
X-axis : X’OX
Y-axis : Y’OY

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

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

9. XX’ O
Y
Y’
III
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.

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

11.  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 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(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 SO T
m1
m2
Q
C

22. 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
==
( )
( )
( )
( )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(x1, y1) and B(x2,y2)
m:n ≡ 1:1
1 2 1 2x x y y
P ,
2 2
+ + 
∴ ≡  
 
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(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 QO R

27. Area of a Triangle
XX’
Y’
O
Y A(x1, y1)
C(x3, y3)
B(x2,y2)
M L N
Area of ∆ ABC =
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC

28. Area of a Triangle
Area of ABCΔ = Area of trapezium ABQP + Area of
trapezium BQRC– Area of trapezium APRC.
We also know that ,
Area of trapezium =
Therefore,
Area of ABCΔ =
( )( )embetween thdistancesidesparallelofsum
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