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Mathematics
Session
Cartesian Coordinate Geometry
And
Straight Lines
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
Y
Y-axis : Y’OY
3
2
X-axis : X’OX
O
-4
-3
-2
-1
X
1
-1
Origin
-2
-ve direction
Y’
-3
2
-ve direction
X’
1
+ve direction
Coordinates
3
4
+ve direction
Coordinates
Y
3
2
Abcissa
1
X’
(2,1)
O
-4
-3
-2
-1
X
1
-1
Ordinate
2
3
4
-2
(-3,-2)
Y’
-3
(?,?)
Coordinates
Y
3
2
Abcissa
1
X’
(2,1)
O
-4
-3
-2
-1
X
1
-1
Ordinate
2
3
4
-2
(-3,-2)
Y’
-3
(4,?)
Coordinates
Y
3
2
Abcissa
1
X’
(2,1)
O
-4
-3
-2
-1
X
1
-1
Ordinate
2
3
4
-2
(-3,-2)
Y’
-3
(4,-2.5)
Quadrants
Y
(-,+)
X’
II
(+,+)
O
IV
III
(-,-)
I
Y’
(+,-)
X
Quadrants
Y
(-,+) (+,+)
X’
II
O
X
IV
III
(-,-)
I
Y’
(+,-)
Ist? IInd?
Q : (1,0) lies in which Quadrant?
A : None. Points which lie on the axes do not lie in any
quadrant.
Distance Formula
x1
X’ O
Y’
(x2-x1)
x2
 PQ 
PQN is a right angled .
y1
N
y2
y2-y1
Y
 PQ2 = PN2 + QN2
 PQ2 = (x2-x1)2+(y2-y1)2
X
2
 x2  x1 
2
  y2  y1 
Distance From Origin
Distance of P(x, y) from the
origin is
2
2

 x  0   y  0

x2  y2
Applications of Distance Formula
Parallelogram
Applications of Distance Formula
Rhombus
Applications of Distance Formula
Rectangle
Applications of Distance Formula
Square
Area of a Triangle
A(x1, y1)
B(x2, y2)
Y
X’ O
M
C(x3, y3)
L
N
X
Y’
Area of  ABC =
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC
Area of a Triangle
X’O
Y’
A(x1, y1)
B(x2, y2)
Y
M
C(x3, y3)
L
N
X
Area of trapezium ABML + Area of trapezium ALNC
- Area of trapezium BMNC
1
1
1
 BM  AL  ML    AL  CN LN  BM  CN MN
2
2
2
1
1
1
  y2  y1   x1  x2    y1  y3   x3  x1    y2  y3   x3  x2 
2
2
2
x1 y1 1 Sign of Area : Points anticlockwise  +ve
1
 x2 y2 1
Points clockwise
 -ve
2
x3 y3 1
Area of Polygons
Area of polygon with points Ai  (xi, yi)
where i = 1 to n
1  x1
 
2  x2
y1
y2

x2
y2
x3
y3
 ... 
xn 1
yn 1
xn
yn

Can be used to calculate
area of Quadrilateral,
Pentagon, Hexagon etc.
xn
x1
yn 

y1 
Collinearity of Three Points
Method I :
Use Distance Formula
a
b
c
Show that a+b = c
Collinearity of Three Points
Method II :
Use Area of Triangle
A  (x1, y1)
B  (x2, y2)
C  (x3, y3)
Show that
x1
y1 1
x2
y2 1  0
x3
y3 1
Section Formula – Internal Division
Y
K
H
X’ O
L
N
M
Y’
 mx2  nx1 my2  ny1 
P  
,

m

n
m

n


Clearly AHP ~ PKB
AP AH PH



BP PK BK
y  y1
X
m x  x1



n
x2  x y2  y
Midpoint
Midpoint of A(x1, y1) and B(x2,y2)
m:n  1:1
 x1  x2 y1  y2 
P  
,

2
2


Section Formula – External Division
Y
P divides AB externally in ratio m:n
K
H
X’ O
Y’
L
N
M
Clearly PAH ~ PBK
AP AH PH



BP BK PK
y  y1
X
m x  x1



n
x  x2 y  y2
 mx2  nx1 my2  ny1 
P  
,

m

n
m

n


Centroid
Intersection of medians of a
triangle is called the centroid.
A(x1, y1)
F
B(x2, y2)
E
G
D
 x  x3 y2  y3 
D 2
,

2
2


 x  x3 y1  y3 
E 1
,

2
2


C(x3, y3)
Centroid is
always denoted
by G.
 x  x2 y1  y2 
F 1
,

2
2


Centroid
A(x1, y1)
F
B(x2, y2)
E
G
D
 x  x3 y1  y3 
E 1
,

2
2


 x  x3 y2  y3 
D 2
,

2
2


C(x3, y3)
 x  x2 y1  y2 
F 1
,

2
2


Consider points L, M, N dividing AD, BE
and CF respectively in the ratio 2:1
x2  x3
y2  y3

x

2
y

2
1
 1
2
2
L 
,
12
12







Centroid
A(x1, y1)
F
B(x2, y2)
E
G
D
 x  x3 y1  y3 
E 1
,

2
2


 x  x3 y2  y3 
D 2
,

2
2


C(x3, y3)
 x  x2 y1  y2 
F 1
,

2
2


Consider points L, M, N dividing AD, BE
and CF respectively in the ratio 2:1
x1  x3
y1  y3

x

2
y

2
2
 2
2
2
M
,
12
12







Centroid
A(x1, y1)
F
B(x2, y2)
E
G
D
 x  x3 y1  y3 
E 1
,

2
2


 x  x3 y2  y3 
D 2
,

2
2


C(x3, y3)
 x  x2 y1  y2 
F 1
,

2
2


Consider points L, M, N dividing AD, BE
and CF respectively in the ratio 2:1
x1  x2
y1  y2

x

2
y

2
3
 3
2
2
N
,
12
12







Centroid
A(x1, y1)
F
B(x2, y2)
E
G
D
 x  x3 y1  y3 
E 1
,

2
2


 x  x3 y2  y3 
D 2
,

2
2


C(x3, y3)
 x  x2 y1  y2 
F 1
,

2
2


 x  x2  x3 y1  y2  y3 
L  1
,

3
3


 x  x2  x3 y1  y2  y3 
M 1
,

3
3


 x  x2  x3 y1  y2  y3 
N 1
,

3
3


We see that L  M  N  G
Medians are
concurrent at the
centroid, centroid
divides medians in
ratio 2:1
Centroid
A(x1, y1)
F
B(x2, y2)
E
G
D
 x  x3 y1  y3 
E 1
,

2
2


 x  x3 y2  y3 
D 2
,

2
2


C(x3, y3)
 x  x2 y1  y2 
F 1
,

2
2


 x  x2  x3 y1  y2  y3 
L  1
,

3
3


 x  x2  x3 y1  y2  y3 
M 1
,

3
3


 x  x2  x3 y1  y2  y3 
N 1
,

3
3


We see that L  M  N  G
Centroid
 x1  x2  x3 y1  y2  y3 
G
,

3
3


Incentre
Intersection of angle bisectors of a
triangle is called the incentre
A(x1, y1)
F
B(x2, y2)
E
I
D
C(x3, y3)
Let BC = a, AC = b, AB = c
Incentre is
the centre of
the incircle
AD, BE and CF are the angle
bisectors of A, B and C
respectively.
BD AB b  D   bx2  cx3 , by2  cy3 



 bc

b

c


DC AC c
Incentre
A(x1, y1)
F
B(x2, y2)
E
I
D
BD AB b



DC AC c
C(x3, y3)
 bx  cx3 by2  cy3 
D   2
,
b

c
b  c 

AI AB AC AB  AC c  b




ID BD DC BD  DC
a
bx2  cx3
by2  cy3 

ay1  b  c 
 ax1  b  c  b  c

b

c
I  
,

a  b  c 
a  b  c 




Similarly I can be derived
 ax1  bx2  cx3 
I  

using E and F also
abc


Now,
Incentre
A(x1, y1)
F
B(x2, y2)
E
I
D
BD AB b



DC AC c
C(x3, y3)
 bx  cx3 by2  cy3 
D   2
,
b

c
b  c 

AI AB AC AB  AC c  b




ID BD DC BD  DC
a
bx2  cx3
by2  cy3 

ay1  b  c 
 ax1  b  c  b  c

b

c
I  
,

a  b  c 
a  b  c 




Angle bisectors are
 ax1  bx2  cx3 
I  

concurrent at the incentre
abc


Now,
Excentre
Intersection of external angle
bisectors of a triangle is called
the excentre
E
A(x1, y1)
F
B(x2, y2)
E
D
C(x3, y3)
EA = Excentre opposite A
Excentre is
the centre of
the excircle
 ax1  bx2  cx3 ay1  by2  cy3 
EA  
,


a

b

c

a

b

c


Excentre
Intersection of external angle
bisectors of a triangle is called
the excentre
E
A(x1, y1)
F
B(x2, y2)
E
D
C(x3, y3)
EB = Excentre opposite B
 ax1  bx2  cx3 ay1  by2  cy3 
EB  
,

a

b

c
a

b

c


Excentre is
the centre of
the excircle
Excentre
Intersection of external angle
bisectors of a triangle is called
the excentre
E
A(x1, y1)
F
B(x2, y2)
E
D
C(x3, y3)
EC = Excentre opposite C
 ax1  bx2  cx3 ay1  by2  cy3 
EC  
,

a

b

c
a

b

c


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
B
Orthocentre
is always
denoted by H
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
O
G
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
:
Circle!!
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)
 PA2 = PB2
 (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
B(0,b)
A(a,0) and B(0,b)
 h = a/2, k = b/2
P(h,k)
Also, a2+b2 = l2
 4h2+4k2 = l2
O
 equation of locus of (h,k) is 4x2+4y2 = l2
A(a,0)
Shift of Origin
Y
P(x,y)
X
O’(h,k)
X’ O
Y’
x
Y
y
Consider a point P(x, y)
Let the origin be shifted to
O’ with coordinates (h, k)
X
relative to old axes
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(x1, y1), B(x2, y2)
Let the origin be shifted to (h, k)
new coordinates : A(x1-h, y1-k), B(x2-h, y2-k)
 Old dist. 
& New dist. 
(x1  x2 )2  (y1  y2 )2
(x1  h  x2  h)2  (y1  h  y2  h)2
= Old dist.
Rotation of Axes
Y
P(x,y)
y

Consider a point P(x, y)
Let the axes be rotated
through an angle .
X’
O
X
Let new P  (X, Y) make
x
an angle  with the new
Y’
x-axis
x
y
Y
X
 cos       , sin       , sin   , cos  
R
R
R
R

Rotation of Axes
 cos      
x
y
Y
X
, sin       , sin   , cos  
R
R
R
R
x
R
y
sin  cos   cos  sin  
R
X
Y
x
 cos   sin  
R
R
R
 cos  cos   sin  sin  
X
Y
y
sin   cos  
R
R
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
ac  bd

 b2 c2  d2

Solution
Let O be the origin.
 OA2 = a2+b2, OB2 = c2+d2, AB2 = (c-a)2+(d-b)2
Using Cosine formula in OAB, we have
AB2 = OA2+OB2-2OA.OBcos
  c  a   d  b   a2  b2  c2  d2  2
2
2
On simplifying, cos  
a
2
a
ac  bd
 b2
 c
2
 d2
2



 b2 c2  d2 cos 
Class Exercise - 2
Four points A(6,3), B(-3,5), C(4,-2)
and D(x,3x) are given such that
DBC 1 Find x.

ABC 2
Solution :
Given that ABC = 2DBC
6
 3
4
3
1
5
1  2 3
2 1
x
4
3x 1
5
1
2 1
 6 5  2  3  4  3  1 6  20  2 x 5  2  3x  4  3  1 6  20 
 2 28x  14  49
49
 28x  14  
2
x 
11
3
or x  
8
8
Class Exercise - 3
If a  b  c, prove that (a,a2), (b,b2)
and (c,c2) can never be collinear.
Solution :
Let, if possible, the three points be
collinear.
a a2 1

1
b b2 1  0
2
c c2 1
R2  R2-R1, R3  R3- R2
a
2
a
1
 b  a b2  a2 0  0
c  b c2  b2
0
a
a2
1
 b  a  c  b  1 b  a 0  0
1 c b 0
Solution Cont.
R2  R2-R3
a
a2
1
 b  a  c  b  0 a  c 0  0
1 cb 0
 b  a  c  b   c  a  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 2a  b  x
a  b  a  b 2a  b  y

and

2
2
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
PA2+PB2+PC2=GA2+GB2+GC2+3GP2.
Solution :
Choose a coordinate system such that G is
the origin and P lies along the X-axis.
Let A  (x1,y1), B  (x2,y2), C  (x3,y3), P  (p,0)
 LHS = (x1-p)2+y12+(x2-p)2+y22+(x3-p)2+y32
= (x12+y12)+(x22+y22)+(x32+y32)+3p2-2p(x1+x2+x3)
=GA2+GB2+GC2+3GP2
=RHS
Q.E.D.
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
1
1
4
(a) x2  y2  4p2 (b) 2  2  2
x
y
p
4
1
1
2
(c) x2  y2  2 (d) 2  2  2
p
x
y
p
Solution
Let the line intercept at the
axes at A and B. Let R(h,k) be
the midpoint of AB.
p 
 p
 R h,k   
,

 2 cos  2 sin  
 sin  
p
p
, cos  
2k
2h
1
1
4
p2
p2
 2  2  1  Locus  2  2  2
x
y
p
4k
4h
 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
 h  a  k 2
2
 h  a  k 2
2

2
3
h  a  k 2


2
h

a

  k2
2
h2  2ah  a2  k2 4
 2

2
2
9
h  2ah  a  k
 5h2  26ah  5k2  5a2  0
 the required locus is
5x2  26ax  5y2  5a2  0

4
9
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
PA2  PB2  AB2
2
2
2
  h  2   k 2  h  2   k 2   2  2 
 2h2  2k2  8  16
 the required locus is
x2  y2  4
Class Exercise - 9
Find the coordinates of a point where the
origin should be shifted so that the equation
x2+y2-6x+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, X2+Y2+(2h-6)X+(2k+8)Y+(h2+k2-6h+8k-9) = 0
 2h-6 = 0; 2k+8 = 0  h = 3, k = -4.
Thus the origin is shifted to (3,-4).
Transformed equation is X2+Y2+(9+16-18-32-9) = 0
Or, X2+Y2 = 34
Class Exercise - 10
Through what angle should the axes
be rotated so that the equation
11x2+4xy+14y2 = 5 will not have
terms in xy?
Solution :
Let the axes be rotated through an
angle . Thus equation becomes
11  X cos   Y sin    4  X cos   Y sin    X sin   Y cos  
2
14  X sin   Y cos    5
2
Solution Cont.


Or, 11cos2   4 sin  cos   14 sin2  X

 11sin

 4 cos2   6 sin  cos   4 sin2  XY
2

  4 sin  cos   14 cos2   5
 2 cos2   3 sin  cos   2 sin2   0
  cos   2 sin   2 cos   sin    0
 tan   
1
or tan   2
2
Therefore, the required angle is
 tan1
1
or tan1 2
2
Thank you