jee maths

The great philospher and mathematician of France Rane Descartes(1596-1665) published a book ‘La Geometric’ in 1637. Descartes gave a new idea i e. . , Each point in a plane is expressed by an ordered pair of algebraic realnumbers like ( , ), ( , )x y r θ etc., called co-ordinates of the point. The point ( , )x y iscalled cartesian co-ordinates and ( , )r θ is called polar co-ordinates of the point.Then represents different forms of equations which are developed for all types ofstraight lines and curves.

Thus the Co-ordinate Geometry (or Analytical Geometry) is that branch ofmathematics in which geometrical problems are solved with the help of Algebra.

The position of a point in a plane is determined with reference to twointersecting straight lines called the co-ordinate axes and their point ofintersection is called the origin of co-ordinates. If these two axes of reference(generally we call them x and y axes) cut each other at right angle, they are called rectangular axes otherwise they are called oblique axes. The axes divide the co-ordinate plane in four quadrants.

(a) Rectangular axes (b) Oblique axes

INTRODUCTION1.1

CO-ORDINATE AXES1.2

1CO-OR D I N AT E SY ST EM S AN D

CO-OR D I N AT ES

IIquadrant

IIIquadrant

Iquadrant

IVquadrant

Y

XX'

Y'

90°y-a

xis

x-axisO

IIquadrant

IIIquadrant

Iquadrant

IVquadrant

Y

XX'

Y'

αx-axisO

obliq

ue a

xis

(α ≠ 90°)

Let X OX′ and Y OY′ be two perpendicular axes inthe plane of paper intersecting at O. Let P be any point in the plane of the paper. Draw PM perpendicular toOX. Then the lengths OM and PM are called therectangular cartesian co-ordinates or briefly theco-ordinates of P.

Let OM x= and MP y=Then the position of the point P in the plane with

respect to the co-ordinate axes is represented by theordered pair (x, y). The ordered pair (x, y) is called the co-ordinates of point P.

i e. . , OM x= -co-ordinate or abscissa of the point P

and MP y= -co-ordinate or ordinate of the point P.

1. The ordinate of every point on x-axis is 0.2. The abscissa of every point on y-axis is 0.3. The abscissa and ordinate of the origin O(0, 0) are both zero.4. The abscissa and ordinate of a point are at perpendicular distance from y-axis

and x-axis respectively.5. Table for conversion sign of co-ordinates :

Quadrants XOY(I)

X OY′(II)

X OY′ ′(III)

XOY ′(IV)

Sign of x co-ordinates + − − +Sign of y co-ordinates + + − −Sign of ( , )x y ( , )+ + ( , )− + ( , )− − ( , )+ −

6. Equation of x-axis, y = 0 and equation of y-axis, x = 0.

If OP r= (radius vector)

and ∠ =XOP θ (vectorial angle)

Then the ordered pair of real numbers ( , )r θcalled the polar co-ordinates of the point P.

1. r may be positive or negative according as θis measured in anticlockwise or clockwisedirection. θ lies between −π to π i e. . , − < ≤π θ π. If it is greater than π, thenwe subtract 2π from it and if it is less than −π, then we add 2π to it. It is also known as principal value of P.

2. Always taken θ in radian.

2 CO-ORDINATE GEOMETRY

RECTANGULAR CARTESIAN CO-ORDINATES OF A POINT1.3

Y

XX'

Y'

O Mx

y

P x y ( , )

NOTE

POLAR CO-ORDINATES OF A POINT1.4

X

Q (r, – θ)

P (r, θ)

r

Oθ

(Pole) Initial line

NOTE

EXAMPLE 1. Draw the polar co-ordi nates 23

,π

, −

23

,π

, − −

23

,π

and 23

, −

π on the

plane.

Solution :

EXAMPLE 2.

Draw the polar co-ordi nate 35

4,

π

on the plane.

Solution : Here θ π π= >5

4

then θ π π π π− = − = −25

42

3

4

∴ 35

4,

π

is same 3

3

4, −

π

Let P x y( , ) be the cartesian co-ordinates with respect to axes OX and OY and ( , )r θ be its polar co-ordinates with respect to pole O and initial line OX.

It is clear from figure

OM x r= = cosθ …(1)

and MP y r= = sinθ …(2)

Squaring and adding (1) and (2), we get

x y r2 2 2+ = or r x y= +( )2 2

Dividing (2) by (1), then

tanθ =

y

x

or θ =

−tan 1 y

x

i e. . , ( cos , sin )r rθ θ ⇒ ( , )x y …(3)

and ( ), tanx yy

x

2 2 1+

− ⇒ ( , )r θ …(4)

If r and θ are known then we can find ( , )x y from (3) and if x and y are

known then we can find ( , )r θ from (4).

CO-ORDINATE SYSTEMS AND CO-ORDINATES 3

X

P

3

O

3π4

5π4

3, or 3, 3π4

–

RELATION BETWEEN THE POLAR AND CARTESIAN CO-ORDINATES1.5

Y

X

P x, y( )

MO

θ

r y

x

X

2 2

2 2

O

π3

–2, – Sπ3

2, P

π3

–2, Qπ3

2, R –

Q θ =

−tan 1 y

x

If α =

−tan 1 y

x

Then values of θ in four quadrants

Quadrant I II III IV

θ α π α− − +π α −α

EXAMPLE 1. Find the carte sian co-ordi nates of the points whose polar co-ordi nates are

(i) 54

31, tanπ −

− (ii) 5 24

,π

Solution : (i) Given r = 5, θ π= −

−tan 1 4

3

Now, x r= = −

−cos cos tanθ π54

31

= −

−54

31cos tan

= −

= − × = −−53

55

3

531cos cos

and y r= = −

−sin sin tanθ π54

31

=

−54

31sin tan

=

= × =−54

55

4

541sin sin

Hence cartesian co-ordinates of the given point, will be ( , )−3 4 .

(ii) Given, r = 5 2, θ π=4

Now, x r= =

cos cosθ π5 2

4 = × =5 2

1

25

and y r= =

sin sinθ π5 2

4 = × =5 2

1

25

Hence cartesian co-ordinates of the given point, will be ( , )5 5 .

EXAMPLE 2. Find the polar co-ordi nates of the points whose carte sian co-ordi nates are(i) ( , )− −2 2 (ii) ( , )−3 4

Solution : (i) Given x = − 2 , y = − 2

∴ r x y= + = + =( ) ( )2 2 4 4 2 2

and α π=

= −

−

= =− − −tan tan tan1 1 12

21

4

y

x

Since, point ( , )− −2 2 lies in 3rd quadrant.


∴ θ π α π π π= − + = − + = −4

3

4

Hence polar co-ordinates of the given points will be 2 23

4, .−

π

Remark : If we find θ, from the equation,

tan θ = = −−

=y

x

2

21

then θ π=4

and then ( , ) ( cos , sin ) ,x y r r= = × ×

θ θ 2 21

22 2

1

2

= ≠ − −( , ) ( , )2 2 2 2

(ii) Given x = − 3, y = 4

∴ r x y= + = + =( ) ( )2 2 9 16 5

and α =

=

−

=

− − −tan tan tan1 1 14

3

4

3

y

x

Since, point ( , )−3 4 lies in 2nd quadrant

∴ θ π α π= − = −

−tan 1 4

3

Hence polar co-ordinates of the given points, will be 54

31, tan .π −

−

EXAMPLE 3. Trans form the equa tion r a2 2 2= cos θ into carte sian form.

Solution : Q r x y= +( )2 2 and θ =

−tan 1 y

x

or r x y2 2 2= +( ) and tan θ = y

x

Given r a a2 2 22

22

1

1= = −

+

cos

tan

tanθ θ

θ

or ( )x y a

y

xy

x

2 2 2

2

2

2

2

1

1

+ =−

+

or ( ) ( )x y a x y2 2 2 2 2 2+ = −

This is the required equation in cartesian form.

Alternative Method :

r a2 2 2= cos θ

or r a2 2 2 2= −(cos sin )θ θ

Q x r= cosθ and y r= sin θand r x y2 2 2= +

then r ax

r

y

r

2 22

2

2

2= −

or r a x y4 2 2 2= −( )

or ( ) ( ).x y a x y2 2 2 2 2 2+ = −


EXAMPLE 4. Trans form the equa tion x y ax2 2+ = into polar form.

Solution : Q x r= cosθ, y r= sin θGiven, x y ax2 2+ =

⇒ r a r2 = ( cos )θ

or r a= cosθThis is the required equation in polar form.

1. The polar co-ordinates of the point whose cartesian co-ordinates are ( , )− −1 1 is :

(a) 24

,π

(b) 2

3

4,

π

(c) 24

, −

π (d) 2

3

4, −

π

2. The cartesian co-ordinates of the point whose polar co-ordinates are

135

121, tanπ −

− is :

(a) ( , )12 5 (b) ( , )−12 5(c) ( , )− −12 5 (d) ( , )12 5−

3. The transform equation of r a2 2 2 2cos cosθ θ= to cartesian form is

( )x y x a2 2 2 2+ = λ, then value of λ is :

(a) y x2 2− (b) x y2 2−(c) xy (d) x y2 2

4. The co-ordinates of P′ in the figure is :

(a) 33

,π

(b) 33

, −

π

(c) − −

33

,π

(d) −

33

,π

5. The cartesian co-ordinates of the point Q in thefigure is :

(a) ( , )3 1

(b) ( , )− 3 1

(c) ( , )− −3 1

(d) ( , )3 1−


1.1INTRODUCTORY EXERCISE

A. Objective

XO

P'

π3

–

–3, π3

P –

X

Q2 2

2

–π/6

π/3

O

2, π3P

2, π6S –

–2, π3

R

2

1. A point lies on x-axis at a distance 5 units from y-axis. What are its co-ordinates ?

2. A point lies on y-axis at a distance 4 units from x-axis. What are its co-ordinates ?

3. A point lies on negative direction of x-axis at a distance 6 units from y-axis. What are its co-ordinates ?

4. Transform the equation y x= tan α to polar form.

5. Transform the equation r a= 2 cosθ to cartesian form.

Theorem : The distance between two points P x y( , )1 1 and Q x y( , )2 2 is given by

| | ( ) ( )PQ x x y y= − + −2 12

2 12

Proof : Let P x y( , )1 1 and Q x y( , )2 2 be anytwo points in the plane. Let us assume that thepoints P and Q are both in 1st quadrant (for thesake of exactness).

From P and Q draw PL and QMperpendiculars to x-axis. From P draw PRperpendicular to QM and join PQ. Then

OL x= 1, OM x= 2, PL y= 1, QM y= 2

∴ PR LM OM OL x x= = − = −2 1

and QR QM RM QM PL= − = − = −y y2 1

Since PRQ is a right angled triangle, therefore by Pythagoreas theorem. ( ) ( ) ( )PQ PR QR2 2 2= +

∴ | | ( ) ( )PQ PR QR= +2 2 (Q PQ is always positive)

= − + −( ) ( )x x y y2 12

2 12

∴ The distance PQ between the points P x y( , )1 1 and Q x y( , )2 2 is

given by ( ) ( )x x y y2 12

2 12− + −

or (difference in co-ordinates) (difference in 2x y+ co -ordinates)2

or (difference of abscissaes) (difference of ordina2 + tes)2

Notations : We shall denote the distance between two points P and Q of the co-ordinate plane, either by | |PQ or by PQ.

Corollary 1 : The above formula is true for all positions of the points(i.e., either point or both points are not in the 1st quadrant) keeping in mind, theproper signs of their co-ordinates.

Corollary 2 : The distance of the point P x y( , ) from the origin O( , )0 0 is given by

| | ( ) ( ) ( )OP x y x y= − + − = +0 02 2 2 2

Corollary 3 : The above formula can also be used as

( ) ( )x x y y1 22

1 22− + −


DISTANCE BETWEEN TWO POINTS1.6

O

Q (x , y )2 2

y –y2 1

y2RP

y1

(x , y )1 1

x – x2 1

x1x2

Y'

X' X

Y

ML

B. Subjective

Corollary 4 : (i) If PQ is parallel to x-axis, then y y1 2= and so

| | ( ) | |PQ x x x x= − = −2 12

2 1

(ii) If PQ is parallel to y-axis, then x x1 2= and so

| | ( ) | |PQ y y y y= − = −2 12

2 1

Corollary 5 : If distance between two points is given, then use ± sign.

1. If three points A x y( , )1 1 , B x y( , )2 2 and C x y( , )3 3 are collinear, then

| | | | | |AB BC AC± =

2. When three points are given and it is required to : (i) an Isosceles triangle, show that two of its sides (or two angles) are

equal. (ii) an Equilateral triangle, show that its all sides are equal or each angle is

of 60°. (iii) a Right angle triangle, show that the sum of the squares of two sides is

equal to the square of the third side. (iv) an Isosceles right angled triangle, show that two of its sides are equal

and the sum of the squares of two equal sides is equal to the square of thethird side.

(v) a Scalene triangle, show that its all sides are unequal.

3. When four points are given and it is required to

(i) a Square, show that the four sides are equal and the diagonals are also

equal.

(ii) a Rhombus, show that the four sides are equal and the diagonals are not

equal.

(iii) a Rectangle, show that the opposite sides are equal and the diagonals are

also equal.

(iv) a Parallelogram, show that the opposite sides are equal and the diagonals

are not equal.

(v) a Trapezium, show that the two sides are parallel and the other two sides

are not parallel.4. If A, B, C be the vertices of a triangle and we have to find the co-ordinates of the

circumcentre then, let the circumcentre be P x y( , ) and use PA PB2 2= and

PA PC2 2= this will give two equations in x and y then solve these two equations

and ( , )x y .

Impor tant Note for Objec tive Ques tions

If two vertices of an equilateral triangle are ( , )x y1 1 and ( , )x y2 2 then co-ordinates of

the third vertex are

x x y y y y x x1 2 2 1 1 2 2 13

2

3

2

+ − + ± −

m ( ),

( )


A

B

C

NOTE

A

B

C

EXAMPLE 1. Prove that the distance of the point ( cos , sin )a aα α from the origin isinde pend ent of α.

Solution : Let P a a≡ ( cos , sin )α α and O ≡ ( , )0 0

then | | ( cos ) ( sin )OP a a= − + −α α0 02 2

= +( cos sin )a a2 2 2 2α α

= +a2 2 2(cos sin )α α

= a2

=| |a , which is independent of α.

EXAMPLE 2. Find the distance between the points ( cos , sin )a aα α and ( cos , sin )a aβ βwhere a > 0.

Solution : Let P a a≡ ( cos , sin )α α and Q a a≡ ( cos , sin )β β

then | | ( cos cos ) ( sin sin )PQ a a a a= − + −α β α β2 2

= − + −a2 2 2{(cos cos ) (sin sin ) }α β α β

= + − + + −a2 2 2 2 22 2{cos cos cos cos sin sin sin sin }α β α β α β α β

= + − +a2 1 1 2{ (cos cos sin sin )}α β α β

= − −a2 2 2( cos( ))α β

= − −2 12a ( cos( ))α β

= ⋅ −

2 22

2 2a sinα β

= −

42

2 2a sinα β

= −

2

2a sin

α β

= −

2

2a sin

α β ( )Q a > 0

EXAMPLE 3. If the point ( , )x y be equi dis tant from the points ( , )6 1− and ( , )2 3 , prove that x y− = 3 .

Solution : Let P x y≡ ( , ) , A ≡ −( , )6 1 and B ≡ ( , )2 3

By the given condition, PA PB=

⇒ ( ) ( ) ( ) ( )x y x y− + + = − + −6 1 2 32 2 2 2

or ( ) ( ) ( ) ( )x y x y− + + = − + −6 1 2 32 2 2 2

or x x y y2 212 36 2 1− + + + + = − + + − +x x y y2 24 4 6 9

or 8 8 24x y− =

or x y− = 3.


EXAMPLE 4. Using distance formula, show that the points ( , ), ( , )1 5 2 4 and ( , )3 3 arecollin ear.

Solution : Let A ≡ ( , )1 5 , B ≡ ( , )2 4 and C ≡ ( , )3 3 be the given points, then

| | ( ) ( )AB = − + − =1 2 5 4 22 2

| | ( ) ( )BC = − + − =2 3 4 3 22 2

and | | ( ) ( )AC = − + − =1 3 5 3 2 22 2

Clearly, | | | | | |AB BC AC+ = + = =2 2 2 2 .

Hence A, B, C are collinear.

EXAMPLE 5. An equi lat eral trian gle has one vertex at the point ( , )0 0 and another at ( , )3 3 . Find the co-ordi nates of the third vertex.

Solution : Let O ≡ ( , )0 0 and A ≡ ( , )3 3 be the givenpoints and let B x y≡ ( , ) be the required point. Then

OA OB AB= =

⇒ ( ) `( ) ( )OA OB AB2 2 2= =

⇒ ( ) ( ) ( ) ( )3 0 3 0 0 02 2 2 2− + − = − + −x y

= − + −( ) ( )x y3 32 2

⇒ 12 6 2 3 122 2 2 2= + = + − − +x y x y x y

Taking first two members then

x y2 2 12+ = …(1)

and taking last two members then

6 2 3 12x y+ = or y x= −3 2( ) …(2)

From (1) and (2), we get

x x2 23 2 12+ − =( ) or 4 12 02x x− =

⇒ x = 0 3,

Putting x = 0 3, in (2), we get y = −2 3 3,Hence, the co-ordinates of the third vertex B are ( , )0 2 3 or ( , )3 3− .

Short Cut Method : According to important note :

x x y y y y x x1 2 2 1 1 2 2 13

2

3

2

+ − + ± −

m ( ),

( )

i.e.,0 3 3 3 0

2

0 3 3 3 0

2

+ − + ± −

m ( ),

( )

or3 3

2

3 3 3

2

m,

±

⇒ ( , )0 2 2 or ( , )3 3−

EXAMPLE 6. Show that four points ( , )1 2− , ( , )3 6 , ( , )5 10 and ( , )3 2 are the verti ces of aparal lel o gram.

Solution : Let A ≡ −( , )1 2 , B ≡ ( , )3 6 , C ≡ ( , )5 10 and D ≡ ( , )3 2 be the given points. Then

| | ( ) ( )AB = − + − − = + =1 3 2 6 4 64 2 172 2

| | ( ) ( )BC = − + − = + =3 5 6 10 4 16 2 52 2


A

B

2

2

C

X' X

Y'

Y

O(0, 0)

B(x, y)

(x, y)B

A(3, 3)

| | ( ) ( )CD = − + − = + =5 3 10 2 4 64 2 172 2

| | ( ) ( )AD = − + − − = + =1 3 2 2 4 16 2 52 2

| | ( ) ( )AC = − + − − = + =1 5 2 10 16 144 4 102 2

and | | ( ) ( )BD = − + − =3 3 6 2 42 2

Clearly, | | | |AB CD= , | | | |BC AD= and | | | |AC BD≠Hence ABCD is a parallelogram.

EXAMPLE 7. Let the oppo site angu lar points of a square be ( , )3 4 and ( , )1 1− . Find theco-ordi nates of the remain ing angu lar points.

Solution : Let A( , )3 4 and C( , )1 1− be the given angular points of a square ABCD and let B x y( , ) be the unknown vertex. Then

AB BC=⇒ ( ) ( )AB BC2 2=

⇒ ( ) ( ) ( ) ( )x y x y− + − = − + +3 4 1 12 2 2 2

⇒ 4 10 23 0x y+ − =

⇒ xy= −

23 10

4 …(1)

Also in ∆ ABC,

( ) ( ) ( )AB BC AC2 2 2+ =

⇒ ( ) ( ) ( ) ( )x y x y− + − + − + +3 4 1 12 2 2 2 = − + +( ) ( )3 1 4 12 2

⇒ x y x y2 2 4 3 1 0+ − − − = …(2)

Substituting the value of x from (1) into (2), we get

23 10

44

23 10

43 1 0

22−

+ − −

− − =yy

yy

⇒ 4 12 5 02y y− + = or ( )( )2 1 2 5 0y y− − =

∴ y = 1

2 or

5

2

Putting y = 1

2 in (1), we get x = 9

2,

and putting y = 5

2 in (1), we get x = − 1

2

Hence the required vertices of the square are 9

2

1

2,

and −

1

2

5

2, .

EXAMPLE 8 Find the circumcentre of the trian gle whose verti ces are ( , )− −2 3 , ( , )−1 0 and ( , )7 6− . Also find the radius of the circumcircle.

Solution : Let A ≡ − −( , )2 3 , B ≡ −( , )1 0 and C ≡ −( , )7 6 . Let P x y≡ ( , ) be the circumcentre of ∆ABC.

Since P is the circumcentre

∴ | | | | | |PA PB PC= =

⇒ ( ) ( ) ( )PA PB PC2 2 2= =


D C

A B

A(3, 4)

B x y( , )

D

C(1, –1)

O

(–1, 0)B

(–2, –3)A

C(7, –6)

P ( x, y )

( ) ( ) ( ) ( )x y x y+ + + = + + −2 3 1 02 2 2 2

= − + +( ) ( )x y7 62 2

⇒ x y x y x y x2 2 2 24 6 13 2 1+ + + + = + + +

= + − + +x y x y2 2 14 12 85

Taking first two members, we get

x y+ + =3 6 0 …(1)

and taking 1st and last member then, we get

3 12 0x y− − = …(2)

Solving (1) and (2), we get

x = 3, y = − 3

Hence circumcentre is ( , )3 3− .

Radius of the circumcircle = = + + − −PB ( ) ( )3 1 3 02 2

= +16 9 = 5 units

EXAMPLE 9. If the line segment join ing the points A a b( , ) and B c d( , ) subtends an angle θ atthe origin O, prove that

cos( )( )

θ = +

+ +

ac bd

a b c d2 2 2 2

or OA OB ac bd. cosθ = +

Solution : Let OA r= 1 and OB r= 2

Now r OA a b12 2= = +| | ( ) …(1)

and r OB c d22 2= = +| | ( ) …(2)

Also | | ( ) ( )AB a c b d= − + −2 2

= + + + − −a b c d ac bd2 2 2 2 2 2

= + − +r r ac bd12

22 2( ) (from (1) and (2))

By using Cosine formula in ∆ AOB

cos( ) ( ) ( )

.θ = + −

⋅OA OB AB

OA OB

2 2 2

2

= + − + − +r r r r ac bd

r r12

22

12

22

1 2

2

2

( ( ))

= + = +2

2 1 2 1 2

( ) ( )ac bd

r r

ac bd

r r …(3)

= +

+ +

( )

( ) ( )

ac bd

a b c d2 2 2 2(from (1) and (2))

= +

+ +

( )

( )( )

ac bd

a b c d2 2 2 2

Also from (3),

r r ac bd1 2 cosθ = + or OA OB ac bd. cosθ = +


X' X

Y'

Y

O

A(a, b)

B(c, d)r1

r2θ

For simplification we carefully choose the axes or the origin. Some situationsare given below :

(i) If two lines are perpendicular then point of intersection is taken as origin and these lines must be taken as the co-ordinate axes.

(ii) If two fixed points A and B are given then we take | |AB a= 2 and the mid

point of AB as origin ‘O’, line AOB as x-axis and the line perpendicular toAB through O is taken as y-axis then the co-ordinates of the fixed pointsare ( , )±a 0 . Similarly if AOB as y-axis and the line perpendicular to AB

through O is taken as x-axis then the co-ordinates of the fixed points are ( , )0 ± a .

(iii) If there is a symmetry of any kind then take the co-ordinates of thepoints in a general way i e. . , ( , )x yi i , i = …1 2 3, , , etc.

EXAMPLE 1. Show that the trian gle, the co-ordi nates of whose verti ces are given by inte -gers, can never be an equi lat eral trian gle.

Solution : Let A ≡ ( , )0 0 , B a≡ ( , )0 and C b c≡ ( , ) be the vertices of equilateral triangle ABC where a b c, , are integers then,

| | | | | |AB BC CA= =⇒ ( ) ( ) ( )AB BC CA2 2 2= =

⇒ a a b c b c2 2 2 2 2= − + = +( )

From first two members we get

b c ab2 2 2+ = …(1)

and taking first and third members then

b c a2 2 2+ = …(2)

From (1) and (2), we get a b= 2 ( )Q a ≠ 0

From (2), b c b2 2 22+ = ( )

or c b2 23=

or c b= ± 3

which is impossible, since b and c are integers.

EXAMPLE 2. In any trian gle ABC, show that

AB AC AD BD2 2 2 22+ = +( )

where D is the middle point of BC.

Solution : Let D as the origin and DC and DYas the x and y axes respectively. Let BC a= 2 , then

B a≡ −( , )0 , C a≡ ( , )0 and let A b c≡ ( , )

Now,

LHS = +AB AC2 2

= + + − + − + −( ) ( ) ( ) ( )b a c b a c2 2 2 20 0

= + +2 2 2 2( )a b c …(1)


CHOICE OF AXES1.7

Y

Y '

X' XD (0, 0) C a( , 0)B a(– , 0)

A b c( , )

and RHS = +2 2 2( )AD BD

= − + − +2 0 02 2 2{( ) ( ) }b c a

= + +2 2 2 2( )a b c …(2)

From (1) and (2), we get

AB AC AD BD2 2 2 22+ = +( )

EXAMPLE 3. Let ABCD be a rect an gle and P be any point in its plane. Show that

PA PC PB PD2 2 2 2+ = +

Solution : Let A as the origin and AB and AD as the xand y axes respectively. Let AB a= and AD b= then

B a≡ ( , )0 , D b≡ ( , )0 and C a b≡ ( , )

Let P ≡ ( , )α βNow, LHS = +PA PC2 2

= − + − + − + −( ) ( ) ( ) ( )α β α β0 02 2 2 2a b

= + − − + +2 2 2 22 2 2 2α β α βa b a b …(1)

and RHS = +PB PD2 2

= − + − + − + −( ) ( ) ( ) ( )α β α βa b2 2 2 20 0

= + − − + +2 2 2 22 2 2 2α β α βa b a b …(2)

From (1) and (2), we getPA PC PB PD2 2 2 2+ = +

Let O be the pole and OX be the initial line.

Let P and Q be two given points whose polar

co-ordinates are ( , )r1 1θ and ( , )r2 2θrespectively.

Then OP r= 1, OQ r= 2

and ∠ =POX θ1, ∠ =QOX θ2

∴ ∠ = −POQ θ θ1 2

By using Cosine formula in ∆ POQ,

cos( )( ) ( ) ( )

( )( )∠ = + −

POQOP OQ PQ

OP OQ

2 2 2

2

or cos( )( )θ θ1 2

12

22 2

1 22− = + −r r PQ

r r

∴ | | ( cos( ))PQ r r r r= + − −12

22

1 2 1 22 θ θ

Always taking θ1 and θ2 in radians.


DISTANCE BETWEEN TWO POINTS IN POLAR CO-ORDINATES1.8

Y'

X' X

Y'

(0, b)D C(a, b)

P(α, β)

B(a, 0)A(0, 0)

NOTE

P r( , )1 1θ

Q r( , )2 2θr1

r2

XO

(–

)

1

2

θθ

θ1θ2

(Pole)Initial line

EXAMPLE 1. Prove that the points ( , )0 0 , 32

,π

and 3

6,

π

are the verti ces of an equi lat -

eral trian gle.

Solution : Let A ≡ ( , )0 0 , B ≡

32

,π

and C ≡

36

,π

Here given co-ordinates are in polar form

∴ | | ( cosAB = + − ⋅ ⋅ −

=0 3 2 0 32

0 32 2 πunits

| | cosBC = + − ⋅ ⋅ −

3 3 2 3 32 6

2 2 π π

= −

= − =18 186

18 9 3sin ( )π

units

and | | cosCA = + − ⋅ ⋅ −

=3 0 2 3 06

0 32 2 πunits

∴ | | | | | |AB BC CA= =Hence points A B C, , are the vertices of an equilateral triangle.


Q ∠ =BAXπ2

and ∠ =CAXπ6

∴ ∠ = − =BACπ π π2 6 3

Q In ∆ ABC, AB AC=∴ ∠ = ∠ =ACB ABC α ( )say

∴ α α π π+ + =3

or α π=3

Hence | | | | | |AB BC CA= = .

1. If the distance between the points ( , )a 2 and ( , )3 4 be 8, then a =(a) 2 3 15+ (b) 2 3 15−(c) 2 3 15± (d) 3 2 15±

2. The three points ( , )−2 2 , ( , )8 2− and ( , )− −4 3 are the vertices of :(a) an isosceles triangle (b) an equilateral triangle(c) a right angled triangle (d) none of these


XA(0, 0)

3

3π/3

π/6

3, π6

C

3, π2

B

A. Objective

3. The distance between the points 34

,π

and 7

5

4,

π

is :

(a) 8 (b) 10(c) 12 (d) 14

4. Let A( , )6 1− , B ( , )1 3 and C x( , )8 be three points such that AB BC= then the value of x are :

(a) 3 5, (b) −3 5,(c) 3 5, − (d) − −3 5,

5. The points ( , )a + 1 1 , ( , )2 1 3a + and ( , )2 2 2a a+ are collinear, if :

(a) a = −1 2, (b) a = 1

22,

(c) a = 2 1, (d) a = − 1

22,

6. If A ≡ ( , )3 4 and B is a variable point on the lines | |x = 6. If AB ≤ 4 then the number of positions of B with integral co-ordinates is :(a) 5 (b) 6(c) 10 (d) 12

7. The number of points on x-axis which are at a distance c units ( )c < 3 from ( , )2 3 is :

(a) 1 (b) 2(c) 0 (d) 3

8. The point on the axis of y which its equidistant from ( , )−1 2 and ( , )3 4 , is :

(a) ( , )0 3 (b) ( , )0 4(c) ( , )0 5 (d) ( , )0 6−

1. Find the distance between the points ( , )at at12

12 and ( , )at at22

22 , where t1 and t 2

are the roots of the equation x x2 2 3 2 0− + = and a > 0.

2. If P at at( , )2 2 , Qa

t

a

t2

2, −

and S a( , )0 be any three points, show that

1 1

SP SQ+ is

independent of t .

3. Prove that the points ( , )3 4 , ( , )8 6− and ( , )13 9 are the vertices of a right angledtriangle.

4. Show that the points ( , )0 1− , ( , )6 7 , ( , )−2 3 and ( , )8 3 are the vertices of a rectangle.

5. Find the circumcentre and circumradius of the triangle whose vertices are ( , )−2 3 , ( , )2 1− and ( , )4 0 .

6. The vertices of a triangle are A( , )1 1 , B( , )4 5 and C( , )6 13 . Find cos A.

7. Two opposite vertices of a square are ( , )2 6 and ( , )0 2− . Find the co-ordinates of theother vertices.

8. If the point ( , )x y is equidistant from the points ( , )a b b a+ − and ( , )a b a b− + , provethat bx ay= .

9. If a and b are real numbers between 0 and 1 such that the points ( , )a 1 , ( , )1 b and

( , )0 0 form an equilateral triangle, find a and b.

10. An equilateral triangle has one vertex at ( , )3 4 and another at ( , )−2 3 . Find theco-ordinates of the third vertex.

11. If P be any point in the plane of square ABCD, prove that

PA PC PB PD2 2 2 2+ = +


B. Subjective

Definition : If P be any point on the line ABbetween A and B then we say that P divides segment ABAB internally in the ratio AP PB: .

Also, if P be any point on the line AB but notbetween A and B ( P may be to the right or the left ofthe points A B, ) then P divides AB externally in the ratio

AP PB:

AP

PB=

Positive, in internally division

Negative, in externally division

(1) Formula for Inter nal Divi sion

Theorem : If the point P x y( , ) divides the line segment joining the points A x y( , )1 1 and B x y( , )2 2 internally in the ratio m n: , then prove that

xmx nx

m ny

my ny

m n= +

+= +

+2 1 2 1,

Proof : The given points are A x y( , )1 1 and B x y( , )2 2 . Let us assume that the points A and Bare both in 1st quadrant (for the sake ofexactness). Since P x y( , ) divides AB internally in the ratio m n: i e. . , AP PB m n: := . From A B,and P draw AL, BM and PN perpendiculars tox-axis. From A and P draw AH and PJperpendiculars to PN and BM respectively then

OL x ON x OM x AL y= = = =1 2 1, , , , PN y=and BM y= 2

∴ AH LN ON OL x x= = − = − 1

PJ NM OM ON x x= = − = −2

PH PN HN PN AL y y= − = − = − 1

and BJ BM JM BM PN y y= − = − = −2

Clearly, the ∆ s AHP and PJB are similar and therefore, their sides areproportional

∴ AH

PJ

PH

BJ

AP

PB= =

or x x

x x

y y

y y

m

n

−−

= −−

=1

2

1

2

(i) (ii) (iii)

From (i) and (iii), we havex x

x x

m

n

−−

=1

2

⇒ nx nx mx mx− = −1 2


SECTION FORMULAE1.9

NOTE

O

Y'

X' X

Y

P

B(x , y )2 2

A (x ,y )1 1

J

H

L N M

(x, y

)

m

n

AP

B

AB

P

BA

P

⇒ ( )m n x mx nx+ = +2 1

∴ xmx nx

m n= +

+2 1

and from (ii) and (iii), we havey y

y y

m

n

−−

=1

2

⇒ ny ny my my− = −1 2

⇒ ( )m n y my ny+ = +2 1

∴ ymy ny

m n= +

+2 1

Thus, the co-ordinates of P are mx nx

m n

my ny

m n2 1 2 1+

+++

, .

Corollary 1 : The above section formula is true for all positions of the points ( . . ,i e either point or both points are not in the 1st quadrant), keeping in mind, the proper signs of their co-ordinates.

Corollary 2 : If P is the mid point of AB then m n= , the co-ordinates of themiddle point of AB are

x x y y1 2 1 2

2 2

+ +

,

1. If P ( , )α β be the mid point of AB and if co-ordinates of A are ( , )λ µ then theco-ordinates of B are ( , )2 2α β µ− λ − , i e. . , (Double the x-co-ordinate of midpoint – x-co-ordinate of given point, Double the y-co-ordinate of mid point –y-co-ordinate of given point).

2. The following diagram will help to remember the section formula.

3. For finding ratio, use ratio λ : 1 then co-ordinates of P are x x y y1 2 1 2

1 1

+ λ+ λ

+ λ+ λ

, .

If λ is positive then divides internally and if λ is negative then divides externally.4. The straight line ax by c+ + = 0 divides the

joint of points A x y( , )1 1 and B x y( , )2 2 inthe ratio

AP

PB= λ

1

= − + ++ +

( )

( )

ax by c

ax by c1 1

2 2

If ratio is positive then divides internallyand if ratio is negative then dividesexternally.


NOTE

A(x , y )1 1

P (x, y) B(x , y )2 2

m n

A(x , y )1 1

B(x , y )2 2

P

ax + b

y + c =

0

λ 1

Proof : Co-ordinates of P are x x y y1 2 1 2

1 1

+ λ+ λ

+ λ+ λ

,

Q P lies on the line ax by c+ + = 0, then

ax x

by y

c1 2 1 2

1 10

+ λ+ λ

+ + λ

+ λ

+ =

or ( ) ( )ax by c ax by c1 1 2 2 0+ + + λ + + =

or λ = − + +

+ +11 1

2 2

( )

( )

ax by c

ax by c

5. The line joining the points ( , )x y1 1 and ( , )x y2 2 is divided by the x-axis in the

ratio − y

y1

2

and by y-axis in the ratio − x

x1

2

.

6. In square, rhombus, rectangle and parallelogram diagonals bisect to each other.

EXAMPLE 1. Find the co-ordi nates of the point which divides the line segment join ing thepoints ( , )5 2− and ( , )9 6 in the ratio 3 : 1.

Solution : Let the required point be (x y, ), then

x = × + ×+

=3 9 1 5

3 18 and y = × + × −

+

=3 6 1 2

3 14

( )

Thus the co-ordinates of the required point are ( , )8 4 .

EXAMPLE 2. Find the length of median through A of a trian gle whose verti ces are A B( , ), ( , )− −1 3 1 1 and C ( , )5 1 .

Solution : Let D be the mid point of BC, then

co-ordinates of D are 1 5

2

1 1

2

+ − +

, i e. . , ( , )3 0

∴ Median AD = + + −( ) ( )3 1 0 32 2

= +16 9 = 25

= 5 units

EXAMPLE 3. Deter mine the ratio in which y x− + =2 0 divides the line join ing ( , )3 1− and ( , )8 9 .

Solution : Suppose the line y x− + =2 0 divides the line segment joining A ( , )3 1− and

B ( , )8 9 in the ratio λ : 1 at point P, then the co-ordinates of the point P are

8 3

1

9 1

1

λ +λ +

λ −λ +

, .

But P lies on y x− + =2 0 therefore

9 1

1

8 3

12 0

λ −λ +

− λ +

λ +

+ =

⇒ 9 1 8 3 2 2 0λ − − λ − + λ + =

⇒ 3 2 0λ − = or λ = 2

3

So, the required ratio is 2

31: , i e. . , 2 : 3 (internally) since here λ is positive.


A(–1, 3)

B(1, –1)

D(3, 0)

C(5, 1)

Short Cut Method : According to Note 4 :

λ = − − − +− +

1 3 2

9 8 2 = 2

3

or λ =: :1 2 3

EXAMPLE 4. The co-ordi nates of three consec u tive verti ces of a paral lel o gram are ( , ), ( , )1 3 1 2− and ( , ).2 5 Then find the co-ordi nates of the fourth vertex.

Solution : Let the fourth vertex be D ( , )α β . Since ABCD is a parallelogram, the diagonalsbisect to each other.

i e. . , mid point of BD = mid point of AC

∴ α β− +

= + +

1

2

2

2

2 1

2

5 3

2, ,

or α β− +

=

1

2

2

2

3

24, ,

On equating abscissaes and ordinates, we getα − =1

2

3

2 or α − =1 3 or α = 4

and β + =2

24 or β + =2 8 or β = 6

Hence the co-ordinates of the fourth vertex D ( , )α β is ( , )4 6 .

EXAMPLE 5. In what ratio does x-axis divide the line segment join ing ( , )2 3− and ( , )5 6 ?

Solution : Let the given points be A ( , )2 3− and B ( , )5 6 . Let AB be divided by thex-axis at P x( , )0 in the ratio λ : 1 internally. Considering the ordinate of P, then

06 1 3

1= λ × + × −

λ +( )

or λ = 1

2

∴ The ratio is 1

21: i e. . , 1 : 2 (Internally)

Short Cut Method :

According to Note 5 :

λ = − = − − =1

3

6

1

21

2

y

y

( )

∴ The ratio is 1

21: i e. . , 1 : 2 (internally)

EXAMPLE 6. The mid points of the sides of a trian gle are ( , ), ( , )1 2 0 1− and ( , ).2 1− Findthe co-ordi nates of the verti ces of a trian gle with the help of two unknowns.

Solution : Let D E( , ), ( , )1 2 0 1− and F ( , )2 1− be the mid points of BC CA, and AB

respectively.

Let the co-ordinates of A be ( , )α β then co-ordinates of B and C are ( , )4 2− − −α β and

( , )− − −α β2 respectively (see note 1)


OX

B(5, 6)

A(2, –3)

P(x, 0)

1

λ

A(1, 3)

C(2, 5)

B(–1, 2)

D(1, 2)

E (0, –1)

F (2, –1)

Q D is the mid point of B and C

then 14

2= − −α α

⇒ 1 2= − α or α = 1

and 22 2

2= − − − −β β

⇒ 2 2= − −β or β = − 4

Hence co-ordinates of A B, and C are ( , ), ( , )1 4 3 2− and ( , )−1 2 respectively.

EXAMPLE 7. Prove that in a right angled trian gle the mid point of the hypot e nuse isequi dis tant from its verti ces.

Solution : Let the given right angled triangle be ABC,with right angled at B. We take B as the origin and BAand BC as the x and y axes respectively.

Let BA a= and BC b=then A a≡ ( , )0 and C b≡ ( , )0

Let M to be the mid point of the hypotenuse AC,

then co-ordinates of M are a b

2 2,

∴ | |( )

AM aa b a b

= −

+ −

=+

20

2 2

2 2 2 2

... (1)

| |( )

BMa b a b

= −

+ −

=+

02

02 2

2 2 2 2

…(2)

and | |CMa

bb= −

+ −

02 2

2 2

=+( )a b2 2

2 …(3)

From (1), (2) and (3), we get

| | | | | |AM BM CM= =

EXAMPLE 8 Show that the line join ing the mid points of any two sides of a trian gle is halfthe third side.

Solution : We take O as the origin and OC and OY as the x and y axes respectively.

Let BC a= 2 then B a C a≡ − ≡( , ), ( , )0 0

Let A b c≡ ( , ), if E and F are the mid points of

sides AC and AB respectively.

Then Ea b c≡ +

2 2

, and Fb a c≡ −

2 2

,


Y'

X' X

Y

B(0, 0) A(a, 0)

C(0, b)

Ma2

b2

Y

Y'

X' XO C a( , 0)B a(– , 0)

A b c( , )

F Ea b+

2c2

b a–2

c2

Now FEa b b a c c

a= + − −

+ −

=2 2 2 2

2 2

= =1

22

1

2( ) ( )a BG

Hence the line joining the mid points of any two sides of a triangle is half the thirdside.

(2) Formula for Exter nal Divi sion

Theorem : If the point P x y( , ) divides the line joining the points A x y( , )1 1

and B x y( , )2 2 externally in the ratio m n: then prove that

xmx nx

m ny

my ny

m n= −

−= −

−2 1 2 1,

Proof : The given points are A x y( , )1 1 and B x y( , )2 2 . Let us assume that the points A and Bare both in the 1st quadrant (for the sake ofexactness). Let P x y( , ) be the point whichdivides AB externally in the ratio m n: , so that AP

BP

m

n= .

From A B, and P draw AL BM, and PNperpendiculars on x-axis. Also from A and B draw AR and BS perpendiculars on PN,

then AR LN ON OL x x= = − = − 1

BS MN ON OM x x= = − = − 2

PR PN RN PN AL y y= − = − = − 1

and PS PN SN PN BM y y= − = − = − 2

Clearly, the ∆s APR and BPS are similar and therefore their sides areproportional.

∴ AP

PB

AR

BS

PR

PS= =

or m

n

x x

x x

y y

y y= −

−= −

−1

2

1

2

(i) (ii) (iii)From (i) and (ii), we have

m

n

x x

x x= −

−1

2

⇒ mx mx nx nx− = −2 1

⇒ ( )m n x mx nx− = −2 1

or xmx nx

m n= −

−2 1

Also from (i) and (iii), we have

m

n

y y

y y= −

−1

2


O

Y'

X' X

Y

B

P(x, y)

A (x ,y )1 1

S

R

L NM

(x, y

)

2

2

⇒ my my ny ny− = −2 1

⇒ ( )m n y my ny− = −2 1

or ymy ny

m n= −

−2 1

Thus the co-ordinates of P are mx nx

m n

my ny

m n2 1 2 1−

−−−

, . (Here m n≠ )

Corollary 1 : The above formula is true for all positions of the points,keeping in mind, the proper signs of their co-ordinates.

Corollary 2 : The above co-ordinates can also be expressed as

mx n x

m n

my n y

m n2 1 2 1+ −

+ −+ −+ −

( )

( ),

( )

( )

and this can be thought of as the co-ordi nates of the point divid ing AB inter nallyin the ratio m n: −

Corollary 3 : Q AP

PB

m

n=

or AP

PB

m

n− = −1 1

or AP PB

PB

m n

n

− = −

or AB

PB

m n

n= −

Now we can say that B divides AP in the ratio m n n− : internally.

i e. . , xm n x nx

m n n2

1= − +− +

( )

( ) ⇒ x

mx nx

m n= −

−2 1

and ym n y ny

m n n2

1= − +− +

( )

( ) ⇒ y

my ny

m n= −

−2 1

Corollary 4 : (for proving A, B and C are collinear)

If A B C, , three points are collinear then let C divides AB in the ratio λ : 1

internally.

If λ = + ve rational then divide internally

and if λ = − ve rational then divide externally.

1. The following diagram will help to remember the section formula

2. Let m

n= λ then

mx nx

m n

my ny

m n2 1 2 1−

−−−

,


Ax y( , )1 1

Bx y( , )2 2

P( )x, y

NOTE

A x y( , )1 1

B x y( , )2 2m

n

–+

or

m

nx x

m

n

m

ny y

m

n

2 1 2 1

1 1

−

−

−

−

, or λ −

λ −λ −

λ −

x x y y2 1 2 1

1 1,

EXAMPLE 1. Find the co-ordi nates of a point which divides exter nally the line join ing ( , )1 3− and ( , )−3 9 in the ratio 1 : 3.

Solution : Let the co-ordinates of the required point be P x y( , ).

Then x = × − − ×−

1 3 3 1

1 3

( ) and y = × − × −

−

1 9 3 3

1 3

( )

i e. . , x = 3 and y = − 9

Hence the required point is ( , )3 9− .

EXAMPLE 2. The line segment join ing A ( , )6 3 to B ( , )− −1 4 is doubled in length by havingits length added to each end. Find the co-ordi nates of the new ends.

Solution : Let P and Q be the required new ends

Let the co-ordinates of P be ( , )x y1 1

Given AB AP= 2

⇒ AB

AP= 2

1

i e. . , A divides BP internally in the ratio 2 : 1.

Then 62 1 1

2 11= × + × −

+x ( )

⇒ 19 2 1= x or x119

2=

and 32 1 4

2 11= × + × −

+y ( )

⇒ 13 2 1= y or y113

2=

∴ Co-ordinates of P are 19

2

13

2,

.

Also let co-ordinates of Q be ( , )x y2 2

Given AB BQ= 2

⇒ AB

BQ= 2

1

i e. . , B divides AQ internally in the ratio 2 : 1

Then − = × + ×+

12 1 6

2 12x

⇒ − =9 2 2x or x29

2= −

and − = × + ×+

42 1 3

2 12y

⇒ − =15 2 2y or y215

2= −

∴ Co-ordinates of Q are − −

9

2

15

2,


Q AB AP= 2

⇒ AB

AP= 2

1

⇒ AB

AP+ = +1

2

11


Q x y( , )2 2

P x y( , )1 1

A(6, 3)

B(–1, –4)

⇒ AB AP

AP

+ = 3

1 ⇒

BP

AP= 3

1

∴ P divides AB externally in the ratio 1 : 3

Then x11 1 3 6

1 3

19

2= × − − ×

−=( )

and y11 4 3 3

1 3

13

2= × − − ×

−=( )

∴ Co-ordinates of P are 19

2

13

2,

Also AB BQ= 2

⇒ AB

BQ= 2

1

⇒ AB

BQ+ = +1

2

11

⇒ AB BQ

BQ

+ = 3

1

⇒ AQ

BQ= 3

1

∴ Q divides AB externally in the ratio 3 : 1

then x23 1 1 6

3 1

9

2= × − − ×

−= −( )

and y23 4 1 3

3 1

15

2= × − − ×

−= −( )

∴ Co-ordinates of Q are − −

9

2

15

2, .

EXAMPLE 3. Using section formula show that the points ( , ), ( , )1 1 2 1− and ( , )4 5 arecollin ear.

Solution : Let A B≡ − ≡( , ), ( , )1 1 2 1 and C ≡ ( , )4 5

Suppose C divides AB in the ratio λ : 1 internally then

42 1 1

1= λ × + ×

λ +

⇒ 4 4 2 1λ + = λ +

or λ = − 3

2

i e. . , C divides AB in the ratio 3 : 2 (externally).

Hence A B C, , are collinear.

EXAMPLE 4. Find the ratio in which the point ( , )2 y divides the line segment join ing (4, 3)and ( , )6 3 and hence find the value of y.

Solution : Let A B≡ ≡( , ), ( , )4 3 6 3 and P y≡ ( , )2

Let P divides AB internally in the ratio λ : 1

then 26 4

1= λ +

λ +


A(1, –1)

C(4, 5)

B(2, 1)

⇒ 2 2 6 4λ + = λ +⇒ − λ =4 2

or λ = − 1

2

∴ P divides AB externally in the ratio 1 : 2 (Qλ is negative)

Now y = × − ×−

=1 3 2 3

1 23

(3) Harmonic Conju gates :

If four points in a line, then the system is said to form a range. Let four pointssay P Q R S, , , .

If the range ( , )PQ RS has a cross ratio equal to −1, then it is called harmonic.

i e. . , PR

RQ

SQ

SP. = − 1

⇒ PR

RQ

SP

SQ= − = λ (say)

∴ PR

RQ= λ

1 ⇒ PR RQ: := λ 1 (internally)

and SP

SQ= − λ

1 ⇒ PS SQ: := λ 1 (externally)

Hence R and S are called the harmonic conjugates to each other withrespect to the points P and Q.

EXAMPLE 1 Find the harmonic conju gates of the point R ( , )5 1 with respect to the points P ( , )2 10 and Q ( , ).6 2−

Solution : Let S ( , )α β be the harmonic conjugates of the point R( , )5 1 . Suppose Rdivides PQ in the ratio λ : 1 internally then S divides PQ in the ratio λ : 1externally then

56 2

1= λ +

λ +

⇒ 5 5 6 2λ + = λ +

∴ λ = 3

Also, 12 10

1= − λ +

λ +

λ + = − λ +1 2 10

⇒ 3 9λ =∴ λ = 3

Now, α = × − ×−

=3 6 1 2

3 18

and β = × − − ×−

= −3 2 1 10

3 18

( )

Hence harmonic conjugates of R( , )5 1 is S( , ).8 8−


P

S

QR

S(α, β)

P(2, 10)

R(5, 1)

Q(6, –2)

A(2, 3) B(4, 3) C(6, 3)

Definition : The point of intersection of the medians of a triangle is calledthe centroid of the triangle and it divides the median internally in the ratio 2 : 1.

Theorem : Prove that the co-ordinates of the centroid of the triangle whosevertices are ( , ),( , )x y x y1 1 2 2 and ( , )x y3 3 are

x x x y y y1 2 3 1 2 3

3 3

+ + + +

,

Also deduce that the medians of a triangle are concurrent.Proof : Let A x y B x y≡ ≡( , ), ( , )1 1 2 2 and C x y≡ ( , )3 3 be the vertices of the

triangle ABC. Let us assume that the points A B, and C are in the 1st quadrant (for the sake of exactness) whosemedians are AD BE, and CFrespectively so D E, and F arerespectively the mid points of BC CA, and AB then theco-ordinates of D E F, , are

Dx x y y≡ + +

2 3 2 3

2 2,

Ex x y y≡ + +

3 1 3 1

2 2,

and Fx x y y≡ + +

1 2 1 2

2 2,

The co-ordinates of a point dividing AD in the ratio 2 1: are

22

1

2 1

22

1

2 1

2 31

2 31⋅ +

+ ⋅

+

⋅ +

+ ⋅

+

x xx

y yy

,

or x x x y y y1 2 3 1 2 3

3 3

+ + + +

,

and the co-ordinates of a point dividing BE in the ratio 2 : 1 are

22

1

2 1

22

1

2 1

3 12

3 12⋅ +

+ ⋅

+

⋅ +

+ ⋅

+

x xx

y yy

,

or x x x y y y1 2 3 1 2 3

3 3

+ + + +

,

Similarly the co-ordinates of a point dividing CF in the ratio 2 1: are

x x x y y y1 2 3 1 2 3

3 3

+ + + +

,


CENTROID OF A TRIANGLE1.10

O

Y'

X' X

Y

A(x , y )1 1

E

D

C(x , y )3 3

x +x3 1

2

y +y3 1

2

x +x2 3

2

y +y2 3

2

x +x1 2

2

y +y1 2

2 F

B(x , y )2 2

G

∴ The common point which divides AD BE, and CF in the ratio 2 1: is

x x x y y y1 2 3 1 2 3

3 3

+ + + +

,

Hence medians of a triangle are concurrent and the co-ordinates of the

centroid are

x x x y y y1 2 3 1 2 3

3 3

+ + + +

,

Important theorem : Centroid of the triangle obtained by joining the

middle points of the sides of a triangle is the same as the centroid of the original

triangle.

Or

If ( , ), ( , )a b a b1 1 2 2 and ( , )a b3 3 are the mid points of the sides of a triangle,

then its centroid is given by

a a a b b b1 2 3 1 2 3

3 3

+ + + +

,

Proof : Let D E F, , are the mid points of BC CA, and AB respectively now let

co-ordinates of A are ( , )α β then co-ordinates of B and C are ( , )2 23 3a b− −α βand ( , )2 22 2a b− −α β are respectively.

Q D a b( , )1 1 is the mid point of B and C, then

2 2 21 3 2a a a= − + −α α ⇒ α = + −a a a2 3 1

and 2 2 21 3 2b b b= − + −β β

⇒ β = + −b b b2 3 1

Now, co-ordinates of B are ( , )2 23 3a b− −α βor ( , )a a a b b b3 1 2 3 1 2+ − + −

and co-ordinates of C are ( , )2 22 2a b− −α β

or ( , )a a a b b b2 1 3 2 1 3+ − + −Hence co-ordinates of A B C, and are

A a a a b b b≡ + − + −( , )2 3 1 2 3 1 ,

B a a a b b b≡ + − + −( , )3 1 2 3 1 2

and C a a a b b b≡ + − + −( , )2 1 3 2 1 3

∴ Co-ordinates of centroid of triangle ABC are a a a b b b1 2 3 1 2 3

3 3

+ + + +

,

which is same as the centroid of triangle DEF.

Corollary 1 : If mid points of the sides of a triangle are ( , )x y1 1 , ( , )x y2 2

and ( , )x y3 3 then co-ordinates of the original triangle are

( , )x x x y y y2 3 1 2 3 1+ − + − , ( , )x x x y y y3 1 2 3 1 2+ − + −and ( , ).x x x y y y1 2 3 1 2 3+ − + −


A(α, β)

E(a , b )2 2

D(a , b )1 1

(a , b )F3 3

B C

Corollary 2 : If two vertices of a triangle are ( , )x y1 1 and ( , )x y2 2 and theco-ordinates of centroid are ( , )α β then co-ordinates of the third vertex are

( , )3 31 2 1 2α β− − − −x x y y

Corol lary 3 : Accord ing to impor tant theo rem ∆s ABC and DEF are simi lar

∴ Area of

Area of

∆∆

ABC

DEF

BC

EF= ( )

( )

2

2

= − + −− + −

=442 3

22 3

2

2 32

2 32

{ ( ) ( ) }

{( ) ( ) }

a a b b

a a b b

∴ Area of ∆ ABC = 4 × Area of ∆DEF

i e. . , Area of a triangle is four times the area of the triangle formed by joiningthe mid points of its sides.

EXAMPLE 1. Two verti ces of a trian gle are ( , )−1 4 and ( , )5 2 . If its centroid is ( , )0 3− , find the third vertex.

Solution : Let the third vertex be ( , )x y then the co-ordinates of the centroid of triangleare

− + + + +

1 5

3

4 2

3

x y, i e. . ,

4

3

6

3

+ +

x y,

Now, 4

3

6

30 3

+ +

= −x y, ( , )

⇒ 4

30

+ =x and

6

3

+ y = − 3

⇒ 4 0+ =x and y + = −6 9

or x = − 4 and y = −15

Hence the third vertex is ( , ).− −4 15

Short Cut Method : According to corollary 2 :

( , ) ( ( ) , ( ) )x y = × − − − × − − −3 0 1 5 3 3 4 2 = − −( , )4 15

EXAMPLE 2. The verti ces of a trian gle are ( , ), ( , ) ( , ).1 2 3 4h k− −and Find the value of

{( ) ( ) }h k h k+ + +2 23 . If the centroid of the trian gle be at the point ( , )5 1− .

Solution : Here 1 4

35

+ − =h and

2 3

31

− + = −k

then we get h k= = −18 2,

∴ ( ) ( ) ( ) ( )h k h k+ + + = − + −2 2 2 23 18 2 18 6

= + =( )16 12 202 2

EXAMPLE 3. If D E( , ), ( , )− −2 3 4 3 and F ( , )4 5 are the mid points of the sides BC CA, and

AB of trian gle ABC, then find (| | | | | | )AG BG CG2 2 2+ − where G is the centroid of ∆ ABC.

Solution : Let the co-ordinates of A be ( , )α βthen co-ordinates of B are ( , )8 10− −α βand co-ordinates of C are ( , )8 6− − −α β


Q D is the mid point of BC then

8 8

22

− + − = −α α and

10 6

23

− − − =β β

i e. . , α = 10 and β = −1

∴ Co-ordinates of A B C, , are ( , ), ( , )10 1 2 11− − and ( , )− −2 5 respectively.

Now co-ordinates of centroid

G ≡ − − − + −

10 2 2

3

1 11 5

3,

i e. . , G ≡

25

3,

∴ AG = − + − −

( )10 2 15

32

2

= +

=6464

9

8

310( )

BG = − − + −

( )2 2 115

32

2

= + =1628

9

4

358

2( )( )

and CG = − − + − −

( )2 2 55

32

2

= +

=16400

9

4

334( )

Hence (| | | | | | )AG BG CG2 2 2+ − = × + × − ×

64

910

16

958

16

934

= + −32

920 29 17( )

= ×

=32

932

32

3

EXAMPLE 4. If G be the centroid of the ∆ABC and O be any other point in the plane of thetrian gle ABC, then show that

OA OB OC GA GB GC GO2 2 2 2 2 2 23+ + = + + + .

Solution : Let G be the origin and GO be x-axis.

O a A x y B x y≡ ≡ ≡( , ), ( , ), ( , )0 1 1 2 2 and C x y≡ ( , )3 3

Now, LHS = + +OA OB OC2 2 2

= − + +( )x a y12

12 ( ) ( )x a y x a y2

222

32

32− + + − +

= + + + + +( ) ( )x x x y y y12

22

32

12

22

32 − + + +2 31 2 3

2a x x x a( )

= + − +∑∑ x y a12

12 20 3

Qx x x

i e x x x

1 2 3

1 2 3

30

0

+ + =

+ + =

. . ,

= + +∑∑ x y a12

12 23 ...(1)


B(–2, 11)

A(10, –1)

C(–2, –5)

F(4, 5)

(–2, 3)D

E(4, –3)

Documents

jee maths