Mathematics

Session

Session Objectives
1. Cartesian Coordinate system and
Quadrants
2. Distance formula
3. Area of a triangle
4. Collinearity of three points
5. Section formula
6. Special points in a triangle
7. Locus and equation to a locus
8. Translation of axes - shift of origin
9. Translation of axes - rotation of axes

René Descartes

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

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

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

Coordinates
Y
3
(-3,-2)
2
1
(2,1)
Abcissa
X’
-4 -3 -2 -1
Ordinate
O X
Y’
-1
1 2 3 4
-2
-3 (4,-2.5)

Quadrants
X’
(-,+)
O
Y
(+,+)
(-,-)
Y’
(+,-)
X

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

Distance Formula
Y
N x
1
X’ O
Y’
 PQ  x
2
(x
2
-x
1
)
 x
2
 x
1
X
PQN is a right angled  .
 PQ 2 = PN 2 + QN 2
 PQ 2 = (x
2
-x
1
) 2 +(y
2
-y
1
) 2
 y
2
 y
1
 2

Distance From Origin
Distance of P(x, y) from the origin is

 x 0
  y 0
 2
 x 2  y 2

Applications of Distance Formula

Applications of Distance Formula

Applications of Distance Formula

Applications of Distance Formula

Area of a Triangle
Y
A(x
1
, y
1
)
C(x
3
, y
3
)
X’ O M L N X
Y’
Area of  ABC =
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC

Area of a Triangle
Y
A(x
1
, y
1
)
C(x
3
, y
3
)
X’ O
Y’
M L N
X


Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC

1
2
1
2
1
2

 y x x
2
2
3
 y y y
1
2
3

  x x
1 y
1
1
1
1
1
 x

2

1
2


1
2
 y
1
 y
 
3
 x
3

1
2
 x

1
 
1
2

Points clockwise y
2
 

Sign of Area : Points anticlockwise y
3


 x
3
+ve
-ve
 x
2


Area of Polygons
Area of polygon with points A i
 (x i
, y i
) where i = 1 to n

1
2

 x
1 y
1 x
2 y
2
 x
2 y
2 x
3 y
3
 . . .
 x x n y y n
Can be used to calculate area of Quadrilateral,
Pentagon, Hexagon etc.
 x n y n x
1 y
1 

Collinearity of Three Points
Method I :
Use Distance Formula a b c

Collinearity of Three Points
Method II :
Use Area of Triangle
A  (x
1
, y
1
)
B  (x
2
, y
2
)
C  (x
3
, y
3
)
Show that x x
1
2 y y
1
2
1
1 x
3 y
3
1
 0

Section Formula – Internal Division
Y
K
H
X’ O
Y’
L N M

 mx
2
 nx
1 , my
2
 ny
1


X
Clearly  AHP ~  PKB


AP
BP

AH
PK

PH
BK m n
 x x x
2

1 x
 y
2


1 y

Midpoint
Midpoint of A(x
1
, y
1
) and B(x
2
,y
2
) m:n  1:1

 x
1

2 x
2 , y
1

2 y
2




Section Formula – External Division
Y
P divides AB externally in ratio m:n
X’ O
Y’
K
L N M
H
X
Clearly  PAH ~  PBK


AP
BP

AH
BK

PH
PK m n
 x x

1
2



1
2

 mx
2
 nx
1 , my
2
 ny
1



Centroid
Intersection of medians of a triangle is called the centroid.
A(x
1
, y
1
)
F
G
E
B(x
2
, y
2
) D
D
E
 x
2

2 x
3 , y
2

2 y
3


 x
1

2 x
3 , y
1

2 y
3


C(x
3
, y
3
) Centroid is always denoted by G.
F
 x
1

2 x
2 , y
1

2 y
2



Centroid
A(x
1
, y
1
)
F G E
D
 x
2

2 x
3 , y
2

2 y
3


B(x
2
, y
2
) D C(x
3
, y
3
)
E  

 x
1

2 x
3 , y
1

2 y
3


F
 x
1

2 x
2 , y
1

2 y
2
Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1


L  


 x
1
 2 x
2

2 x
3
, y
1
 2 y
2

2 y
3





Centroid
A(x
1
, y
1
)
F G E
D
 x
2

2 x
3 , y
2

2 y
3


B(x
2
, y
2
) D C(x
3
, y
3
)
E  

 x
1

2 x
3 , y
1

2 y
3


F
 x
1

2 x
2 , y
1

2 y
2
Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1


M  


 x
2
 2 x
1

2 x
3
, y
2
 2 y
1

2 y
3





Centroid
A(x
1
, y
1
)
F G E
D
 x
2

2 x
3 , y
2

2 y
3


B(x
2
, y
2
) D C(x
3
, y
3
)
E  

 x
1

2 x
3 , y
1

2 y
3


F
 x
1

2 x
2 , y
1

2 y
2
Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1


N  


 x
3
 2 x
1

2 x
2
, y
3
 2 y
1

2 y
2





Centroid
A(x
1
, y
1
)
F G E
D
 x
2

2 x
3 , y
2

2 y
3


B(x
2
, y
2
) D C(x
3
, y
3
)
E
L
M
N
 


 x
1

2 x
3 , y
1

2 y
3 x
1
 x
3
2
 x
3 ,

 y
1
 y
F
3
2

 x
1

2 x
2 , y
1

2 y
2 y
3



 x
1
 x
3
2
 x
3 , y
1
 y
3
2
 y
3 x
1
 x
2
3
 x
3 , y
1
 y
2
3
 y
3






Medians are concurrent at the centroid, centroid divides medians in ratio 2:1
We see that L  M  N  G

Centroid
A(x
1
, y
1
)
F G E
D
 x
2

2 x
3 , y
2

2 y
3


B(x
2
, y
2
)
E
L
D C(x
3
, y
3
)
 


 x
1

2 x
3 , y
1

2 y
3 x
1
 x
3
2
 x
3 ,

 y
1
 y
F
3
2

 x
1

2 x
2 , y
1

2 y
2 y
3




M
N
 x
1
 x
3
2
 x
3 , y
1
 y
3
2
 y
3


 x
1
 x
2
3
 x
3 , y
1
 y
2
3
 y
3


G  


Centroid

3
2

,

3
2
 y
3



We see that L  M  N  G

Incentre
Intersection of angle bisectors of a triangle is called the incentre
A(x
1
, y
1
)
F
I
E
B(x
2
, y
2
) D C(x
3
Let BC = a, AC = b, AB = c
, y
3
AD, BE and CF are the angle bisectors of A, B and C respectively.

BD
DC

AB
AC
 b c
 D


) Incentre is the centre of the incircle bx
2
 cx
3 , by
2
 cy
3



Incentre
F
A(x
1
, y
1
)
I
E
B(x
2

, y
2
)
BD
DC

AB
AC
 b c
Now,
  



AI
ID ax
1


AB
BD

D a 



C(x
 D


3
, y bx
2
3
)
 cx
3 , by
2

 cy
3
AC
DC

AB AC
BD DC bx
2
 cx
3




,
 ay
1 c b
 a
 a 




 by
2
 cy
3







 ax
1
 bx
2
 cx
3



Similarly I can be derived using E and F also

Incentre
F
A(x
1
, y
1
)
I
E
B(x
2

, y
2
)
BD
DC

AB
AC
 b c
Now,
  



AI
ID ax
1


AB
BD

D a 



C(x
 D


3
, y bx
2
3
)
 cx
3 , by
2

 cy
3
AC
DC

AB AC
BD DC bx
2
 cx
3




,
 ay
1 c b
 a
 a 




 by
2
 cy
3







 ax
1
 bx
2
 cx
3



Angle bisectors are concurrent at the incentre

Excentre
Intersection of external angle bisectors of a triangle is called the excentre
E
A(x
1
, y
1
)
F E
B(x
2
, y
2
)
E
A
D C(x
3
, y
3
)
= Excentre opposite A
E
A

 ax
1
 bx
2
 cx
3 ,
 ay
1
 by
2
 cy
3


Excentre is the centre of the excircle

Excentre
Intersection of external angle bisectors of a triangle is called the excentre
E
A(x
1
, y
1
)
F E
B(x
2
, y
2
)
E
B
D C(x
3
, y
3
)
= Excentre opposite B
E
B
 ax
1
 bx
2
 cx
3 , ay
1
 by
2
 cy
3


Excentre is the centre of the excircle

Excentre
Intersection of external angle bisectors of a triangle is called the excentre
E
A(x
1
, y
1
)
F E
B(x
2
, y
2
)
E
C
D C(x
3
, y
3
)
= Excentre opposite C
E
C
 ax
1
 bx
2
 cx
3 , ay
1
 by
2
 cy
3


Excentre is the centre of the excircle

Cirumcentre
Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre.
A
C
O
OA = OB = OC
= circumradius
B
The above relation gives two simultaneous linear equations. Their solution gives the coordinates of O.

Orthocentre
Intersection of altitudes of a triangle is called the orthocentre.
A
H
Orthocentre is always denoted by H
B C
We will learn to find coordinates of Orthocentre after we learn straight lines and their equations

Cirumcentre, Centroid and
Orthocentre
The circumcentre O, Centroid G and
Orthocentre H of a triangle are collinear.
H
G
O
G divides OH in the ratio 1:2

Locus – a Definition
The curve described by a point which moves under a given condition or conditions is called its locus e.g. locus of a point having a constant distance from a fixed point
:

Locus – a Definition
The curve described by a point which moves under a given condition or conditions is called its locus e.g. locus of a point equidistant from two fixed points :
Perpendicular bisector!!

Equation to a Locus
The equation to the locus of a point is that relation which is satisfied by the coordinates of every point on the locus of that point
Important :
A Locus is NOT an equation. But it is associated with an equation

Equation to a Locus
Algorithm to find the equation to a locus :
Step I : Assume the coordinates of the point whose locus is to be found to be (h,k)
Step II : Write the given conditions in mathematical form using h, k
Step III : Eliminate the variables, if any
Step IV : Replace h by x and k by y in Step III. The equation thus obtained is the required equation to locus

Illustrative Example
Find the equation to the locus of the point equidistant from
A(1, 3) and B(-2, 1)
Solution :
Let the point be P(h,k)
PA = PB (given)
 PA 2 = PB 2
 (h-1) 2 +(k-3) 2 = (h+2) 2 +(k-1) 2
 6h+4k = 5
 equation of locus of (h,k) is 6x+4y = 5

Illustrative Example
A rod of length l slides with its ends on perpendicular lines. Find the locus of its midpoint.
Solution :
Let the point be P(h,k)
Let the  lines be the axes
Let the rod meet the axes at
A(a,0) and B(0,b)
 h = a/2, k = b/2
Also, a 2 +b 2 = l 2
 4h 2 +4k 2 = l 2 O
 equation of locus of (h,k) is 4x 2 +4y 2 = l 2
B(0,b)
P(h,k)
A(a,0)

Shift of Origin
Y
P(x,y)
X Y y
O’(h,k)
Consider a point P(x, y)
X’ O X
Let the origin be shifted to
O’ with coordinates (h, k) relative to old axes x
Y’ Let new P  (X, Y)
 x = X + h, y = Y + k
 X = x - h, Y = y - k
O  (-h, -k) with reference to new axes

Illustrative Problem
Show that the distance between two points is invariant under translation of the axes
Solution :
Let the points have vertices
A(x
1
, y
1
), B(x
2
, y
2
)
Let the origin be shifted to (h, k) new coordinates : A(x
1
-h, y
1
-k), B(x
2
-h, y
2
-k)
 Old dist.
 (x
1
 x ) 2  (y
1
 y ) 2
& New dist.
 (x
1
 
2
 h) 2  (y
1
 
2
 h) 2
= Old dist.

Rotation of Axes
Y
P(x,y)


X’ O
Y’ x
 cos   x
R
, sin   y
Consider a point P(x, y) y
R
,
X
Let the axes be rotated through an angle  .
Let new P  (X, Y) make an angle  with the new x-axis sin  
Y
R
, cos  
X
R

Rotation of Axes


 cos
  cos cos sin cos
X
R cos
 
  x
R
,
 
Y
R sin sin sin sin
 

  x
R y
R
, sin  
 
  x
R y
R
Y
R
, cos  
X
R

X
R sin  
Y
R cos   y
R x  X cos   Y sin  y  X sin   Y cos 
 X  x cos   y sin 
Y  y cos   x sin 

Class Exercise

Class Exercise - 1
If the segments joining the points
A(a,b) and B(c,d) subtend an angle  at the origin, prove that cos  
 a 2  b 2
 c 2  d 2


Solution
Let O be the origin.
 OA 2 = a 2 +b 2 , OB 2 = c 2 +d 2 , AB 2 = (c-a) 2 +(d-b) 2
Using Cosine formula in  OAB, we have
AB 2 = OA 2 +OB 2 -2OA.OBcos

  c a     2
 a 2  b 2  c 2  d 2 

2  b 2
 c 2 


On simplifying, cos  
 a 2  b 2
 c 2  d 2


Class Exercise - 2
Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given such that


DBC
ABC

1
2
Find x.
Solution :
Given that  ABC = 2  DBC

6 3 1 x 3x 1
  3 5 1  
4  2 1 4  2 1






 










 

 2 28x 14  49
 28x 14  
49
2
11
8 or x  
3
8


Class Exercise - 3
If a  b  c, prove that (a,a 2 ), (b,b 2 ) and (c,c 2 ) can never be collinear.
Solution :
Let, if possible, the three points be collinear.

1
2 a a 2 1 b b 2 1  0 c c 2 1
R
2
 R
2
-R
1
, R
3 a a 2 1
 R
3
- R
2
  2  a 2 0  0 a a 2 1

 


  0
 2  b 2 0

Solution Cont.
R
2
 R
2
-R
3 a a 2 1

 


  0

 




 0
This is possible only if a = b or b = c or c = a.
But a  b  c. Thus the points can never be collinear.
Q.E.D.

Class Exercise - 4
Three vertices of a parallelogram taken in order are (a+b,a-b),
(2a+b,2a-b) and (a-b,a+b). Find the fourth vertex.
Solution :
Let the fourth vertex be (x,y).
Diagonals bisect each other.
 a b a b
2

 
2 and a b a b
2

 
2
 the required vertex is (-b,b)

Class Exercise - 5
If G be the centroid of  ABC and P be any point in the plane, prove that
PA 2 +PB 2 +PC 2 =GA 2 +GB 2 +GC 2 +3GP 2 .
Solution :
Choose a coordinate system such that G is the origin and P lies along the X-axis.
Let A  (x
1
,y
1
), B  (x
2
,y
2
), C  (x
3
,y
3
), P  (p,0)
 LHS = (x
1
-p) 2 +y
1
2 +(x
2
-p) 2 +y
2
2 +(x
3
-p) 2 +y
3
2
= (x
1
2 +y
1
2 )+(x
2
2 +y
2
2 )+(x
3
2 +y
3
2 )+3p 2 -2p(x
1
+x
2
+x
3
)
=GA 2 +GB 2 +GC 2 +3GP 2
Q.E.D.
=RHS

Class Exercise - 6
The locus of the midpoint of the portion intercepted between the axes by the line xcos  +ysin  = p, where p is a constant, is
(a)x 2  y 2  4p (b)
1 x 2

1 y 2

4 p 2
(c)x 2  y 2 
4 p 2
(d)
1 x 2

1 y 2

2 p 2

Solution
Let the line intercept at the axes at A and B. Let R(h,k) be the midpoint of AB.

 p
2 cos 
, p
2 sin 


 sin   p
2k
, cos   p
2h
 p 2 p 2
4k 2

4h 2
 1  Locus 
1 x 2

1 y 2

4 p 2
 Ans : (b)

Class Exercise - 7
A point moves so that the ratio of its distance from (-a,0) to (a,0) is 2:3.
Find the equation of its locus.
Solution :
Let the point be P(h,k). Given that




 2
 k

2
 k
2
2

2
3






 2
2

 k k
2
2

4
9
 h h
2
2




2
2

 k k
2
2

4
9
 5h 2   2  5a 2  0
 the required locus is
5x 2   2  5a 2  0

Class Exercise - 8
Find the locus of the point such that the line segments having end points
(2,0) and (-2,0) subtend a right angle at that point.
Solution :
Let A  (2,0), B  (-2,0)
Let the point be P(h,k). Given that
PA 2  PB 2  AB 2

 h 2
 2
 k 2 
 h 2
 2
 k 2 

2 2
 2
 2h 2  2k 2  
 the required locus is x 2  y 2  4

Class Exercise - 9
Find the coordinates of a point where the origin should be shifted so that the equation x 2 +y 2 -6x+8y-9 = 0 will not contain terms in x and y. Find the transformed equation.
Solution :
Let the origin be shifted to (h,k). The given equation becomes
(X+h) 2 +(Y+k) 2 -6(X+h)+8(Y+k)-9 = 0
Or, X 2 +Y 2 +(2h-6)X+(2k+8)Y+(h 2 +k 2 -6h+8k-9) = 0
 2h-6 = 0; 2k+8 = 0  h = 3, k = -4.
Thus the origin is shifted to (3,-4).
Transformed equation is X 2 +Y 2 +(9+16-18-32-9) = 0
Or, X 2 +Y 2 = 34

Class Exercise - 10
Through what angle should the axes be rotated so that the equation
11x 2 +4xy+14y 2 = 5 will not have terms in xy?
Solution :
Let the axes be rotated through an angle  . Thus equation becomes

  Y sin


  Y cos
 
  Y sin 

X sin   Y cos 


5

Solution Cont.



2

4 cos 2

11sin 2
 
 
     14 sin 2 

X
   4 sin 2 

XY
   14 cos 2 5
 2cos 2      2sin 2   0
  cos   2 sin   2 cos   sin 
 tan   
1
2 or tan   2
0
Therefore, the required angle is
 tan  1
1
2 or tan 2

Thank you