cg class10

1. Class X
Prepared By:
Sharda Chauhan
TGT Mathematics

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.
INTRODUCTION
What is co-ordinate geometry ?

4. ▪ Asystemof geometrywherethepositionof points onthe plane isdescribed
usinganorderedpairof numbers.
▪ The method of describing the location of points in this waywas
▪ proposed bythe French mathematician René Descartes .
▪ He proposed further that curvesand lines could be described by
equations usingthis technique, thusbeing the first to link
algebraand geometry
.
▪ In honor of his work, the coordinates of apoint are often referred to as
its Cartesiancoordinates, andthe coordinate planeasthe Cartesian
Coordinate Plane.
What is Coordinate Geometry?

5. IMPORTANTPOINTS
● 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
X
X’
Y’
O
-4 -3 -2 -1
Origin
1 2 3 4
-1
+ve direction
-2
-ve direction
-3
-ve
direction
1
+ve
direction
X-axis : X’OX
Y 3
Y-axis : Y’OY
2

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

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

9. X
X’ O
Y
Y’
(-,+) (+,+)
II I
III IV
(-,-) (+,-)
Ist? IInd?
Q : (1,0) lies in which Quadrant?
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. P(x1, y1) and
▪ Draw PRandQS x-axis.
perpendicular fromthe pointPon
▪ QSisdraw tomeet it at thepointT
1 2
So, OR = x , OS= x ,
Then,
PR= PS= y1 , QS= y2
PT= x2 –x1 ,
QT = y2 –y1
x
P (x1 , y1)
Q(xA
2 , y2)
T
S
O R
Distance Formula
▪ Letusnowfindthedistancebetweenanytwo points
Q(x1, y2)
Y

12. ▪Now, applying the Pythagoras theorem in ΔPTQ, we get
Therefore
 QT 2
PQ2
 PT 2
2
2
2 1  y2  y1 
PQ  x  x 
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
 x  02
 y  02
√x
2
 y
2

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. 2
PB m
PA

m1
Section Formula
▪ Consider any two points A(x1, y1) and B(x1 ,y2)and assume that P(x, y)
dividesABinternallyin the ratio
x
m
1:m
2 i.e.
Y
A (x1 , y1)
B(x2 , y2)
S
▪ DrawAR, PSand BT  x-axis.
▪DrawAQ and PC parallel to the x-axis.
Then,
by theAAsimilarity criterion, O R T
m1
m2
P (x , y) Q
C

22. Section Formula
Now, BP PC BC
ΔPAQ ~ ΔBPC
PA

AQ

PQ---------------- (1)
AQ = RS= OS – OR = x–
x1PC = 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
m1

x  x1 
y  y1 
m2 x2  x y2  y

23. Section Formula
For x - coordinate
Taking
or
m

x  x 
1
m2 x2  x
1
or
or
m1x2  x m2x  x1
m1x2  m1x  m2x  m2x1
m1x2  m2 x1  xm2  m1 
x 
m1x2  m2 x1
m2  m1

24. Section Formula
For y –coordinate
Taking
m2 y2  y
m1

y  y1 
m1y2  y m2y  y1
m1y2  m1y  m2 y  m2 y1
m1 y2  m2 y1  ym2  m1 
y 
m1 y2  m2 y1
m2  m1
or
or
or

25. Midpoint
Midpoint of A(x1, y1) and B(x2,y2)
m:n  1:1
Find the Mid-Point of P(1,-3) and Q(5,7).

26. Area of a Triangle
▪ Let ABC be any triangle whose verticesare A(x1, y1
), B(x2 , y2
) and
C(x3 , y3).
▪
▪
▪
Draw AP, BQ and CR
perpendicularsfrom A,B and C,
respectively, to the x-axis.
Clearly ABQP, APRCand
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)
O P Q R

27. Area of a Triangle
X
Y’
X’ 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

28. Areaof a Triangle
AreaofΔ ABC= Areaoftrapezium ABQP+ Areaof
trapezium BQRC– Areaof trapezium APRC.
We also know that ,
Areaoftrapezium =
Therefore,
AreaofΔ ABC =
sum of parallelsidesdistance between them
2
1

  
1

1
 1

BQ + AP QP  BQ + CR QR  AP + CR PR
2 2 2
1 3 3 1
2 1 2 1 2 3 3 2
2
2 2

1
y  y x x 
1
y  y x  x 
1
y  y x  x 
2 2
2
y2x1  y1x2 y1x1y2x3 y2x2  y3x3 y3x2y1x3 y1x1  y3x3 y3x1

1
y x
1 3 2 2 1 3 3 2 1
2

1
x y  y  x y  y  x y  y 
Area of ΔABC

29. Thankyou!
Any
Questions?