AriHant MATHEMATICS

IntroductionCoordinate geometry is the branch of Mathematics

which includes the study the different curves and figures by representing points in a plane by ordered pairs of real number called cartesian coordinates representing lines and curves by algebraic equations. It is the branch of Mathematics in which methods of algebra are used to solve geometrical problems which is known as analytic geometry. Since, to study geometrical figure, we deal all things with coordinate that's why it is called coordinate geometry.

Coordinate SystemCartesian Coordinates System

Let XOX' and YOY' be two perpendicular straight lines drawn through any fix point O in the plane of the paper. Then,

Y

X' X

Y'

O (0, 0)

1

Cartesian System of Rectangular Coordinates

Chapter

Chapter in a SnapshotIntroductio nCoordinate Syste mDistance Formul aApplication of Distance Formul aSection Formula eArea of Triangl eArea of Quadrilatera lCollinearity of Three Given Points Some Standard Points of a Triangl eLocu sTransformation of Axe sPoints to Remembe r

2 Objective Approach To Mathematics Vol.2

Axis of x The horizontal line XOX' is called axis of x.Axis of y The vertical line YOY' is called axis of y.Coordinate axes x-axis and y-axis together are called

axes of coordinate or axes of reference.Origin The point O is called the Origin of ordinates

or the Origin.Oblique axes If both the axes are not perpendicular,

then they are called as Oblique axes as shown in the figure

Y

X' X

Y'

QO (0, 0)

Cartesian coordinates The ordered pair of perpendicular distance from both axes of a point P lying in the plane is called

Cartesian coordinates of P. If the Cartesian coordinates of a point P are (x, y), then x is called the abscissa or X-coordinate of point P and y is called the ordinate or Y-coordinate of point P.

Y

X' X

Y'

O (0, 0)

xy P (x, y)

Important !Coordinates of the origin are (0, 0). y -Coordinate of a point on x-axis is zero.x -Coordinate of a point on y-axis is zero.

Polar Coordinates SystemLet P be a point whose distance be r from origin and

makes an angle with x-axis. cos and sinx y

r r= =

or x = r cos and y = r sin Thus, to represent (x, y) in (r, ) is called polar

representation, as shown in the figure

Q

r y

P (x, y)

xO X

Y

Relation between Polar and Cartesian Coordinates System

Let P is any point in a plane whose cartesian coordinates are P (x, y) and let same point when polar system is used have coordinates (r, ).

OA = x = r cos ...(i)OB = y = r sin ...(ii)

Y

X' X

Y'

rQ

x

y

P (x, y)B

AO

On squaring and adding Eqs. (i) and (ii), we get2 2OP r x y= = +

Again, on dividing Eq. (ii) by Eq. (i), we get

1tanyx

=

Illustration 1 The polar coordinates of points x = 3 and y = 1, are

(a) 2,3 (b) 2, 6

(c) 2,3 (d) 2, 6

Solution. Let 3 cosr= and 1 = r sin 3 1 2r = + =

1

1

tan

1tan

63

yx

= = =

Polar coordinates are 2,6

.

Hence, (b) is the correct answer.

Illustration 2 If polar coordinates of any point are

2,3

, then its cartesian coordinates are

(a) ( 3, 1) (b) (1, 3)(c) (1, 3) (d) None of these

Solution. Let 2 cos , 2 sin3 3

x y = =

x = 1, 3y = Cartesian coordinates of P are (1, 3).Hence, (b) is the correct answer.

Chapter 1 : Cartesian System of Rectangular Coordinates 3

Illustration 3 The polar coordinates of a point having cartesian coordinates ( 1, 1) are

(a) 2,4

(b) 3

2,4

(c) 5

2,4

(d) 2, 4

Solution. The cartesian coordinates of P are ( 1, 1).So, x = 1 = r cos and y = 1 = r sin 1 1 2r = + =

1 11tan tan (1)1

= = Since, the point lie in the third quadrant.

5 3

or4 4 =

The polar coordinate of 5( 1, 1) are 2, .4

P P

Hence, (c) is the correct answer.

Illustration 4 Point P (5, 60) having polar coordinates is equivalent to

(a) ( 5, 60) (b) (5, 240) (c) (5, 420) (d) (5, 300)

Solution. Since, point having coordinates (r, ) is equivalent to (r, 2n + ) where .n IHence, (5, 60) = (5, 2n + 60) = (5, 420) [take n = 1]Hence, (c) is the correct answer.

Important !On adding 360 (or any multiple of 360) to the vectorial angle does not alter the final position of revolving lines so that (r, ) is always the same point as (r, + 2n), where n I .On adding 180 (or any odd multiple of 180) to the vectorial angle and changing the sign of radius vector gives the same point as before.Thus, the point + + [ , (2 1) ]r n is same as + [ , ].r

Distance Formula(i) When coordinates of two points are given in

cartesian formLet 1 1( , )P x y and 2 2( , )Q x y be the two points.

2 22 1 2 1( ) ( )PQ d x x y y= = +

x x1 2

y y2 1

Q (x , y )2 2

P (x ,

y )1

1

M

Y

Y'

X' X

Hence, the distance between P and Q will be given by

2 2(diff. of abscissa) (diff. of ordinate)PQ = +

2 22 1 2 1( ) ( )d x x y y= + (ii) When coordinate axes are inclined at an angle If the coordinate system is oblique means if the

coordinate axes are inclined at an angle . In this case the distance between two points P and Q will be given by

Y

Y'

X' X

180 W W(x , y )2 2

Q

WWO (0, 0)

d

P (x , y )1 1

2 21 2 1 2 1 2 1 2( ) ( ) 2( ) ( ) cosPQ d x x y y x x y y= = + +

(iii) When coordinates are in polar formIf (r1, 1) and (r2, 2) are given points, then distance

between them is 2 2 12 1 21 2 2 cos ( )r r r r+ .

Proof Here,OA = r1, OB = r2 and AB = d

Q

Y

Y'

X' XQ

Q Q

r2

r1

B (r , )Q2 2

A(r , )Q1 1

O

d

By cosine law2 22

12 1 21 2 2 cos ( )d r r r r= +

2 2 12 1 21 2| | 2 cos ( )AB r r r r= +

Illustration 5 The distance between points P (3, 2)and Q ( 6, 7), the axis being inclined at 60, is

(a) 3 7 (b) 7

(c) 2 7 (d) None of these


Solution. 2 2(3) (9) 2 3 ( 9) cos 60PQ = + +

1

9 81 542

63 3 7

PQ = +

= =Hence, (a) is the correct answer.

Illustration 6 If the distance between the points (x, 2)and (3, 4) is 2, then the value of x is (a) 2 (b) 1

(c) 3 (d) 4

Solution. Given that, 2 22 ( 3) (2 4)x= + 4 = (x 3)2 + 4 x 3 = 0

x = 3Hence, (c) is the correct answer.

Illustration 7 The point whose abscissa is equal to its ordinate and which is equidistant from A (5, 0) and B (0, 3) is

(a) (1, 1) (b) (2, 2) (c) (3, 3) (d) (4, 4)

Solution. Let the point be P(h, h). Given that, PA = PB

2 2 2 2( 5) ( 3)h h h h + = + 4h = 16 h = 4 Coordinates of P be (4, 4).Hence, (d) is the correct answer.

Illustration 8 The cartesian form of r2 = a2 cos 2 is (a) (x2 + y2 = a(x2 y2)(b) (x2 + y2 = a2(x2 y2)2

(c) (x2 + y2)2 = a2(x2 y2)(d) None of the above

Solution. Since, x = r cos , y = r sin x2 + y2 = r2

Also, 2 2cos 2 cos sin =

2 2

2 2x yr r

=

2 2

2cos 2x y

r =

2 2

2 2cos2 ( )

x yx y

=+

2 2 cos 2r a=

2 2

2 2 22 2

( ).( )

x yx y a

x y+ =+

2 2 2 2 2 2( ) ( )x y a x y+ = Which is cartesian form of r2 = a2 cos 2.Hence, (c) is the correct answer.

Application of Distance Formula(a) Collinearity of three given points The three given

points A, B, C are collinear ie, lie on the same straight line, if any of the three points (say B) lie on the straight line joining the other two points.

AB + BC = AC

Important !The other conditions for collinearity is

Area of ABC is zero. It means

= =1 1

2 2

3 3

11

1 02

1

x y

x y

x y

+ + =1 2 3 2 3 1 3 1 21[ ( ) ( ) ( )] 0

2x y y x y y x y y

Slope of AB = Slope of BC

(b) Slope formula If is the angle at which a straight line is inclined to the positive direction of the x-axis and 0 < 180, q 90,then the slope of the line, denoted by m and is defined by m = tan . If is 90, m does not exist, but the line is parallel to the y-axis.

If = 0, then m = 0 and the line is parallel to the x-axis. If A (x1, y1) and B (x2, y2), x1 x2, are points on a straight line, then the slope m of the line is given by

1 2

1 2

( )( )y y

mx x=

or 2 12 1

y yx x

(c) Types of triangles If A, B and C are vertices of triangle, then it would be (i) equilateral triangle, when AB = BC = CA. (ii) isosceles triangle, when any two sides are equal. (iii) right angle triangle, when sum of square of any

two sides is equal to square of the third side.(d) Position of four points Let A, B, C and D be

the four given point in a plane. By joining these points following figure are formed. (i) Square, if AB = BC = CD = DA and AC = BD (ii) Rhombus, if AB = BC = CD = DA and AC BD (iii) Parallelogram, if AB = DC, BC = AD, AC BD (iv) Rectangle, if AB = CD, BC = DA, AC = BD

Quadrilateral Diagonals Angles between Diagonals

(i) Parallelogram Not equal2

(ii) Rectangle Equal2

(iii) Rhombus Not equal2 =

(iv) Square Equal2 =


FamilyBrief description of Euclidian figures Convex Polygon

of Quadrilateral(Convex figures)

Parallelogram Trapezium

Rectangle Rhombus

Square

Isosceles/EquilateralTrapezium

Kite

(Diagonals butonly one diagonaldivide the figureinto two congruenttriangles)

Cyclic Quad.

Oppositeangles aresupplementry

Parallelogram 4 ways to prove quadrilateral a parallelogram (A = p1 p2 cosec ) also, area = bh

Rectangle Diagonals equalRhombus Diagonals perpendicular

= 1 2Area2

d dwhere d1 and d2

are the length of diagonalsSquare Area = d2/2 (where d is the length

of diagonal)Trapezium Exactly one pair of parallel sides.

Important !If the distance between the pair of opposite sides of a parallelogram is equal it is rhombus.

Diagonals of square, rhombus, rectangle and parallelogram always bisect each other.Diagonals of rhombus and square bisect each other at right angle.

Illustration 9 The points (1, 1), (2, 7) and (3, 3)(a) form a right angled triangle(b) form an isosceles triangle(c) are collinear(d) None of the aboveSolution. Let A (1, 1), B (2, 7) and C (3, 3) be the given points, then

= + = + =

2 2( 2 1) (7 1)

9 36 3 5

AB

= + + = + =

2 2(3 2) ( 3 7)

25 100 5 5

BC

2 2(3 1) ( 3 1) 4 16 2 5AC = + = + =Clearly, BC = AB + AC A, B, C are collinear.Hence, (c) is the correct answer.

Illustration 10 The points ( , 2 2 ), ( 1, 2 )k k k k + and( 4 , 6 2 )k k are collinear for

(a) all value of k (b) 1

12

k or= (c) k = 1 (d) no value of kSolution. The given points are collinear, ifArea of = 0

2 2 1

1 2 1 04 6 2 1

k kk k

k k

+ =

Applying R2 R2 R1, R3 R3 R1

2 2 1

2 1 4 2 0 04 2 4 0

k kk k

k

+ =

2 1 4 2

04 2 4

k kk

+ =

(1 2k) (k + 1) = 0

k = 1 or 1

,2

k = neglecting 1

,2

as when 1

,2

k = points are same.Hence, (c) is the correct answer.

Illustration 11 The points A (2a, 4a), B (2a, 6a) andC (2 3 , 5 )a a a+ (when a > 0) are vertices of(a) an obtuse angled triangle(b) an equilateral triangle(c) an isosceles obtuse angled triangle(d) a right angled triangle

Solution. 2 2(2 2 ) (4 6 ) 2AB a a a a a= + =2 2 2 2(2 3 2 ) (5 6 ) ( 3 ) 2BC a a a a a a a a= + + = + =

and 2 2 2 2(2 2 3 ) (5 4 ) ( 3 ) 2CA a a a a a a a a= + = + =Since, AB = BC = CA, the triangle is an equilateral.Hence, (b) is the correct answer.

Illustration 12 The triangle formed by P (2, 3), Q (3, 2) and R (0, 0) is(a) an equilateral (b) an isosceles(c) a right angled (d) None of these

Solution. 25 25 5 2PQ = + =9 4 13RQ = + =4 9 13PR = + =

RQ = PRHence, PQR is an isosceles.Hence, (b) is the correct answer.

Section FormulaeCoordinates of a point which divides the line segment

joining two points P(x1, y1)and Q(x2, y2) in the ratio m1 : m2 are

(i) 1 2 2 1 1 2 2 11 2 1 2

,m x m x m y m y

m m m m + + + +

(internal division)

(ii) 1 2 2 1 1 2 2 11 2 1 2

,m x m x m y m y

m m m m

(external division)

When m1, m2 are of opposite sign, the division is external.


Important !Coordinates of any point on one line segment which divide the line segment joining two points P(x1, y1)

and Q(x2, y2) in the ratio : 1 is given by

+ + + +1 2 1 2, , ( 1)

1 1x x y y

Lines formed by joining ( x1, y1) and (x2, y2) is divided by

(a) x-axis in the ratio 12

yy

(b) y-axis in the ratio 12

xx

If ratio is positive, then the axis divides it internally and if ratio is negative, then the axis divides externally.

Line Ax + By + C = 0 divides the line joining the points (x1, y1) and (x2, y2) in the ratio : 1, then

+ + = + + 1 1

2 2

Ax By CAx By C

If is positive it divides internally and if is negative, then it divides externally.

Illustration 13 The coordinates of point which divides the line segment joining points A (0, 0) and B (9, 12) in the ratio 1 : 2, are

(a) ( 3, 4) (b) (3, 4) (c) (3, 4) (d) None of these

Solution. Since, P divides the line segment joining points A (0, 0) and B (9, 12) internally in 1 : 2 ratio.

9 1 0 2 9 33 3

x + = = =

12 1 2 0 124

3 3y

+ = = =

The coordinates of point P are (3, 4).Hence, (b) is the correct answer.

Illustration 14 The ratio in which the line 3x + y 9 = 0divides the segment joining the points (1, 3) and (2, 7), is(a) 1 : 2 (b) 4 : 3(c) 3 : 4 (d) None of theseSolution. Let the line 3x + y 9 = 0 divides the line segment joining A (1, 3) and B (2, 7) in the ratio k : 1 at point

C, then the coordinates of C are + +

+ +2 1 7 3

, .1 1

k kk k

C lie on line 3x + y 9 = 0, so it satisfies the equation of line.

2 1 7 33 9 0

1 1k kk k+ + + = + +

34

k =

So, the required ratio is 3 : 4 (internally).Hence, (c) is the correct answer.

Aliter

Required ratio 1 12 2

( ) (3 3 9) 3( ) (6 7 9) 4

Ax By CAx By C

+ + + = = =+ + +


Illustration 15 The points of trisection of line segment joining the point A (2, 1) and B (5, 3), are

(a) 7 5

3, , , 43 3

(b)

5 74, , 3,

3 3

(c) 5 7

3, , 4,3 3

(d)

7 74, , 3,

3 3

Solution. Let P1 and P2 are the points which trisect the line segment joining the points A (2, 1) and B (5, 3).

11 5 2 2 1 3 2 1 5

( , ) , 3,1 2 1 2 3

P x y + + = = + +

22 5 1 2 2 3 1 1 7

( , ) , 4,1 2 1 2 3

P x y + + = = + +


Illustration 16 The ratio in which the line segment joining the points (3, 4) and (5, 6) is divided by the x-axis, is(a) 2 : 3 (b) 3 : 2

(c) 6 : 4 (d) None of theseSolution. Let P is the point on x-axis which divides the line segment joining points in the ratio : 1.

Then, 5 3

1x

+= + ,

6 41

y = +

But as it is know for point on x axis, then its y-coordinate = 0

6 4 201 3

= = +

Hence, (a) is the correct answer.

Area of TriangleLet (x1, y1), (x2, y2) and (x3, y3) respectively be the

coordinates of the vertices A, B, C of a ABC. Then, the area of ABC, is

1 2 3 2 3 1 3 1 21

[ ( ) ( ) ( )]2

x y y x y y x y y+ + ...(i)

= 1 1

2 2

2 3

11

12

1

x yx yx y

...(ii)

While using formula (i) or (ii), order of the points (x1, y1), (x2, y2) and (x3, y3) has not been taken into account. If we plot the points A(x1, y1), B(x2, y2) and C(x3, y3), then the area of the triangle as obtained by using formula (i) or (ii) will be positive or negative as the point A, B, C are in anti-clockwise or clockwise directions.

C

A B(Anti-clockwise)

B

A C(Clockwise)

So, while finding the area of ABC, we take modulus.


Important !If area of triangle is zero. Then, the points will be collinear.

Area of triangle = 0 or =1 1

2 2

3 3

1

1 0

1

x y

x y

x y

In case of polygon with vertices ( x1, y1), (x2, y2),.......(xn, yn) in order, then area of polygon is given by

1

2|(x1y2 y1x2) + (x2y3 y2x3) + ....+ (xn 1 yn yn 1 xn)

+ (xn y1 yn x1)|

Illustration 17 If the vertices of a triangle are (1, 2), (4, 6) and (3, 5), then its area is

(a) 23

2

sq unit (b) 25

2

sq unit

(c) 12 sq unit (d) None of theseSolution. Area of triangle whose vertices are given

1 2 3 2 3 1 3 1 21

[ ( ) ( ) ( )]2

x y y x y y x y y = + +

1[1( 6 5) 4(5 2) 3(2 6)]

2 = + + +

252

= sq unit


Illustration 18 If A (6, 3), B (3, 5), C (4, 2) and D ( x, 3x)

are four points, if ( )

1: 2, ( )

ar DBCar ABC

=

then x is equal to

(a) 118

(b)811

(c) 3 (d) None of these

Solution. Since, ar ( ) 1ar ( ) 2

DBCABC

=

2 ar ( DBC) = ar ( ABC)

3 1 6 3 1

112 3 5 1 3 5 1

84 2 1 4 2 1

x xx = =


Illustration 19 The area of a triangle is 5 sq unit. Two of its vertices are (2, 1) and (3, 2). The third vertex lying on y = x + 3. The coordinates of the third vertex are given by

(a) 3 3 7 13

, , ,4 4 4 2 (b)

1 1 3 3, , ,

4 2 4 2

(c) 2 2 7 3

, , ,3 3 2 2

(d) 3 3 7 13

, , ,2 2 2 2

Solution. As the vertex lies on the line y = x + 3, its coordinates are of the form (x, x + 3). The area of the triangle those vertices are (2, 1), (3, 2) and (x, x + 3), is

2 1 11 1

3 2 1 |4 4| |2 2|2 2

3 1x x

x x = = +

Accordingly to the given condition 2x 2 = 5

3 7

,2 2

x =

Thus, the coordinates of the third vertex can be3 3 7 13

, or ,2 2 2 2

Hence, (d) is the correct answer.

Illustration 20 If the points (a, 1), (2, 1) and (4, 5) are collinear. Then, the value of a is (a) 1 (b) 2

(c) 3 (d) 4Solution. The essential condition of collinearity is = 0

=1 1

2 1 1 04 5 1

a

a( ) ( ) ( )1 5 1 2 4 110 4 0 + + = a = 1Hence, (a) is the correct answer.

Area of QuadrilateralIf 1 1 2 2 3 3( , ), ( , ), ( , )x y x y x y and (x4, y4) are vertices of

a quadrilateral, then its area will be given byy

x

D C

AB

L NO R M

y'

x'

Area of quadrilateral = Area of trapezium ALND+ Area of trapezium DNRC + Area of trapezium CRMB Area of trapezium ALBM

On substituting the value of area of trapezium

1(sum of parallel sides) distance between them

2=

Area of quadrilateral

1 2 2 1 2 3 3 2 3 4 4 31

[( ) ( ) ( )2

x y x y x y x y x y x y= + +

4 1 1 4( )]x y x y+


Important !Area can also be calculated as,

(i) Row method When vertices of polygon are given, then area of polygon will be

+ + +

1 2 2 3 1

1 2 2 3 1

1.......

2n

n

x x x x x x

y y y y y y

(ii) Stair method Again, when vertices of polygon are given. Say (x1, y1), (x2, y2), .........., (xn, yn). Then, its area is

12

x1 y1

x2 y2

x3 y3

xn yn

x1 y1

........

= + + + + + +1 2 2 3 1 1 2 2 3 11{( ...... ) ( .... )}

2 n nx y x y x y y x y x y x

If the two opposite vertex of a square are 1 1( , )A x y and 2 2( , )C x y , then its area

= = + 2 2 22 1 2 11 1

[( ) ( ) ]2 2AC x x y y

Collinearity of Three Given PointsThree given points A, B, C are collinear, if any one of

the following conditions is satisfied:(i) Area of ABC is zero.

(ii) Slope of AB = Slope of BC = Slope of AC(iii) AC = AB + BC(iv) Find the equation of line passing through 2 given

points, if the third point satis es the given equation of the line, then three points are collinear.

Illustration 21 The area of quadrilateral whose vertices are (1, 1), (3, 4), (5, 2) and (4, 7), is

(a) 41 sq unit (b) 412

sq unit

(c)312

sq unit (d) 7 sq unit

Solution. Area of quadrilateral1

|[(1 4) (3 1) (3 ( 2)) (5 4)2

= +

(5 ( 7)) (4 ( 2)) (4 1) (1 ( 7))]|+ + 41 412 2

= = sq unit


Illustration 22 If the coordinates of two opposite vertices of a square are (6, 3) and (2, 3), then area of square is(a) 50 sq unit (b) 25 sq unit(c) 35 sq unit (d) None of these

Solution. Area of square 212

d=

2 21 {(6 2) ( 3 3) }

2= + + 1 (64 36) 50

2= + = sq unit


Some Standard Points of a TriangleIn this section, we shall discuss problems on finding

centroid, circumcentre, incentre and orthocentre of a triangle.

(i) Centroid The centroid of a triangle is the point of intersection of its medians. It divides the medians in the ratio 2 : 1. If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a ABC, then the coordinates of its centroid G are

EG

bc F

Da

C (x , y )3 3B (x , y )2 2

A (x , y )1 1

1 2 3 1 2 3,3 3

x x x y y y+ + + +

(ii) Incentre The point of intersection of the internal bisectors of the angles of a triangle is called its incentre.

If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a ABC such that BC = a, CA = b and AB = c, then the coordinates of the incentre are

1 2 3 1 2 3,ax bx cx ay by cy

a b c a b c+ + + + + + + +

(iii) Circumcentre The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of its sides.

A (x , y )1 1

B (x , y )2 2 C (x , y )3 3

O

It is the centre of the circle passing through the vertices of a triangle and so it is equidistant from the vertices of the triangle.

Thus, if O is the circumcentre of a ABC, then OA = OB = OCLet A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices

of ABC and let O (x, y) be its circumcentre. Then, the coordinates of O are obtained by solving

OA2 = OB2 = OC2ie, 2 2 2 21 1 2 2( ) ( ) ( ) ( )x x y y x x y y + = +

= + 2 23 3( ) ( )x x y y


The coordinates of the circumcentre are also given by

1 2 3sin2 sin2 sin2 ,sin2 sin2 sin2

x A x B x CA B C+ +

+ +1 2 3sin2 sin2 sin2

sin2 sin2 sin2y A y B y C

A B C+ + + +

(iv) Orthocentre The orthocentre of a triangle is the point of intersection of its altitudes.

In order to find the coordinates of the orthocentre of a triangle, we first find the equations of its altitudes and then we find the coordinates of the point of intersection of any two of them.

If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a ABC, then the coordinates of its orthocentre are

1 2 3 1 2 3tan tan tan tan tan tan,tan tan tan tan tan tan

x A x B x C y A y B y CA B C A B C+ + + + + + + +

Important !The orthocentre of a right angled triangle is at the vertex forming the right angle.The orthocentre O, circumcentre O and centroid G of a triangle are collinear and G, divides O, O in the ratio 2 : 1ie, OG : OG = 2 : 1.The circumcentre of a right angled triangle is the mid point of its hypotenuse.

(v) Excentre Coordinate of excentre opposite to A is

given by 1 2 3 1 2 31

,

ax bx cx ay by cyI

a b c a b c+ + + + + + + +

and similarly for excentres (I2 and I3) opposite to B and C are given by

1 2 3 1 2 32

,

ax bx cx ay by cy

Ia b c a b c

+ + + +

A

C

bc

BL

I1

,BL cLC b

= also 11

AI b cI L a

+=

1 2 3 1 2 33

,

ax bx cx ay by cyI

a b c a b c+ + + +

(vi) Nine point circle Let ABC be a triangle such that AD, BE and CF are its altitudes drawn from A, B and C

on the opposite sides, H, I, J are the mid points of the sides BC, CA and AB respectively and K, L, M are the mid points of the line segments joining the orthocentre O' to the angular points A, B, C. These nine points (D, E, F, H, I, J, K, L, M) are concyclic and the circle passing through these nine points is called the nine point circle. Its centre is know as the nine point centre.

The nine point centre of a triangle is collinear with the circumcentre O and the orthocentre O' and bisects the segment joining them. The radius of nine point circle of a triangle is half the radius of the circumcircle.

Important !The orthocentre, nine point centre, centroid and circumcentre are collinear.

Illustration 23 Centroid of the triangle whose vertices are (0, 0), (2, 5) and (7, 4), is(a) (3, 3) (b) (2, 3)

(c) (3, 2) (d) (2, 2)Solution. The centroid of a triangle are

0 2 7 0 5 4, (3, 3)

3 3+ + + + =


Illustration 24 Incentre of triangle whose vertices are A ( 36, 7), B (20, 7), C (0, 8), is (a) (0, 1) (b) (0, 0)(c) (1, 0) (d) None of these

Solution. 2 2(20 0) (7 8) 25a BC= = + + =

2 2( 36 0) (7 8) 39b CA= = + + =

and 2 2(36 20) (7 7) 56c AB= = + + =

25 ( 36) 39(20) 56(0) 25(7) 39(7) 56( 8),

25 39 56 25 39 56I

+ + + + = + + + +( 1, 0)I =


Illustration 25 If (1, 4) is the centroid of a triangle and two vertices are (4, 3) and (9, 7), then third vertices is(a) (8, 7) (b) (7, 8)(c) (8, 8) (d) (6, 8)Solution. Let the third vertex of triangle be (x, y), then

4 91

3x + = x = 8 and 3 74

3y += y = 8

( , ) (8, 8)x y =Hence, (c) is the correct answer.

Illustration 26 If (0, 1), (1, 0) and (1, 0) are middle points of the sides of a triangle, then its incentre is(a) (2 2, 2 2) + (b) (2 2, 2 2) (c) (2 2, 2 2)+ (d) (2 2, 2 2)+ +


Solution. Let 1 1 2 2( , ), ( , )A x y B x y and 3 3( , )C x y vertices of a triangle, then

1 2 2 3 3 10, 2, 2x x x x x x+ = + = + = 1 2 2 3 3 12, 2, 0y y y y y y+ = + = + =Solving these equations, we get

(0, 0), (0, 2)A B and (2, 0)CNow, 2 2, 2, 2a BC b CA c AB= = = = = =Thus, incentre of ABC is (2 2, 2 2). Hence, (b) is the correct answer.

Illustration 27 The orthocentre of the triangle formed by the line 2 22 6 0x xy y = and the line 1 0x y+ + = is

(a) 4 4

,3 3

(b) (4, 4)

(c) ( 4, 4) (d) None of these

Solution. We have, 2 22 6 0x xy y = (2 3 ) ( 2 ) 0x y x y+ =

2x + 3y =0B (3, 2)

x2y=0x + y + 1 = 0

A FH

IK

23

, 13

X' X

Y

Y'

2x + 3y = 0 and x 2y = 0Thus, the equations of the sides of the triangle are 2x + 3y = 0, x 2y = 0 and x + y + 1 = 0The equation of the altitude through the vertex O (0, 0) is y 0 = 1 (x 0)or y = x ...(i)The equation of the altitude through vertex B (3, 2) is y 2 = 2 (x + 3)or 2x + y + 4 = 0 ...(ii)On solving Eqs. (i) and (ii), we obtain

4 4,

3 3x y= =

Hence, the coordinates of the orthocentre are4 4

, .3 3


Illustration 28 The orthocentre of the triangle formed by the lines ,i

i

ay m x

m= + i = 1, 2, 3, m1, m2, m3 being the

roots of the equation 3 23 2 0x x + = , is

(a) ,2a

a (b) , 2a

a (c) (2a, a) (d) ( a, a)

Solution. Let ABC be the triangle such that the equations of its sides BC, CA and AB are 1 2

1 2,

a ay m x y m x

m m= + = +

and 33

ay m x

m= +

respectively.

A

B C

y m3x am3 y m2x

am2

y m1xam1

Solving these equation in pairs, we obtain the coordinates of

the vertices 2 32 3 2 3

( ), ,

a a m mA

m m m m+

1 3 1 2

1 3 1 3 1 2 1 2

( ) ( ), and ,

a a m m a a m mB C

m m m m m m m m+ +

The equation of the altitude through the vertex A is

2 3

2 3 1 2 3

1m m ay a x

m m m m m+ =

or 2 1 3 11 1 2 3

(1 )x a

y m m m mm m m m

+ = + + ...(i)

Similarly, the equation of the altitude through the vertex Bis 1 2 3 2

2 1 2 3(1 )

x ay m m m m

m m m m+ = + + ...(ii)

Solving Eqs. (i) and (ii), we obtain that the coordinates of the orthocentre which are

+ + + 1 2 2 3 3 11 2 3, (1 )

aa m m m m m m

m m m

It is given that m1, m2, m3 are the roots of the equation + =3 23 2 0x x

+ + =1 2 3 3m m m1 2 2 3 3 1 0m m m m m m+ + =

and 1 2 3 2m m m = So, the coordinates of the orthocentre are

, .2a

aHence, (b) is the correct answer.

LocusIt is the path or curve traced by a moving point

satisfying the given condition.Equation to the locus of a point The equation to the

locus of a point is the algebraic relation which is satisfied by the coordinates of every point on the locus of the point.


Step Taken to Find the Equation of Locus of a Point

(a) Assumes the coordinates of the point say (h, k) whose locus is to be nd.

(b) Write the given condition involving (h, k)(c) Eliminate the variable(s), if any(d) Replace h x and k y The equation so obtained is the locus of the point

which moves under some definite conditions.

Illustration 29 The sum of the squares of the distance of a moving point from two xed point (a, 0) and (a, 0) is equal to a constant. Then, the equation of its locus is (a) 2 2 2x y C a = + (b) 2 2 2x y C a+ = (c) 2 2 2 2x y C a+ = + (d) 2 2 2x y C a = Solution. Let P (h, k) be any position of the moving point and let A (a, 0) and B ( a, 0) be the given points. Then,

2 2PA PB+ = (given) 2 2 2 2( ) ( 0) ( ) ( 0)h a k h a k + + + + = 2 2 2 2 2 22 2h ah a k h ah a k + + + + + + =

Locus of (h, k) is 2 2 2,x y C a+ = where 2

C=


Illustration 30 The locus of a point such that the sum of its distances from the points (0, 2) and (0, 2) is 6, is

(a) 2 29 5 32x y+ =(b) 2 29 5 45x y =(c) 2 29 5 45x y = (d) 2 29 5 45x y+ =Solution. Let P (h, k) be any given point on the locus and let A (0, 2) and B (0, 2) be given points.By given condition, PA + PB = 6

2 2 2 2( 0) ( 2) ( 0) ( 2) 6h k h k + + + + =

2 2 2 2 2 2( 2) 36 12 ( 2) ( 2)h k h k h k+ = + + + + +

2 28 36 12 ( 2)k h k = + + 2 2 2(2 9) 9[ ( 2) ]k h k+ = + + 2 29 5 45h k+ = Locus of (h, k) is 2 29 5 45x y+ =Hence, (d) is the correct answer.

Illustration 31 A rod of length l sides with its ends on two perpendicular lines. Then, the locus of its mid point is

(a) 2

2 24l

x y+ = (b) 2

2 22l

x y+ =

(c) 2

2 24l

x y = (d) None of these

Solution. Let the two perpendicular lines be the coordinate axes. Let AB be rod of length l and the coordinates of A and B be (a, 0) and (0, b) respectively.

OX' X

Y'

Y

B (0, b)

P (h, k)

(a, 0) A

Let P (h, k) be the mid point of the rod AB in one of the infinite position it attains, then

02

ah

+= and 02

bk

+=

and 2 2a b

h k= = ...(i)

From OAB, we have2 2 2AB OA OB= +

2 2 2a b l+ = 2 2 2(2 ) (2 )h k l+ = 2 2 24 4h k l+ =

2

2 24l

h x+ =

The equation of locus is2

2 24l

x y+ =Hence, (a) is the correct answer.

Illustration 32 A (a, 0) and B (a, 0) are two xed points in a ABC. If its vertex C moves in such a way that cot A + cot B = ,where is constant. Then, the locus of the point C is(a) y = 2a (b) ya = 2(c) y = a (d) None of theseSolution. We may suppose that coordinates of two xed points A, B are (a, 0) and ( a, 0) respectively and coordinates of variable point C are (h, k).

C (h, k)

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

Y

X' X

Y'

In CDA, from the adjoining figure

cotDA a h

ACD k

= =


and in DBC, cot BD a hBCD k

+= =

But cot cotA B+ =

So, we have a h a hk k ++ =

2 2a a kk

= =

Locus of C is 22 or ay a y = =


Transformation of Axes(i) To alter the origin of coordinates without altering

the direction of the axes Let origin O (0, 0) be shifted to a point (a, b) by moving the x-axis and y-axis parallel to themselves. If the coordinates of point P with reference to old axis are 1 1( , )x y , then coordinates of this point with respect to new axis will be 1 1( , )x a y b .

O X

YY'

(a, b)O'

P (x , y )11

X'

(ii) To change the direction of the axes of coordinates without changing origin Let OX and OY be the old axes and OX' and OY' be the new axes obtained by rotating the old OX and OY through an angle , then the coordinates of P (x, y) with respect to new coordinate axes will be given by

x = ON NL = x' cos y' sin y = PQ + QL = y' cos + x' sin

O X

Y

a M

NLQ

Q

X'

Y'

Thus, if the axes are rotated through an angle .

Then, the coordinates of a point with respect to new axis will be

cos sinx x y= + and sin cosy x y= + The above relation between (x, y) and (x', y' ) can

also be obtained from the table.

x y x cos sin y sin cos

Important !x , y are old coordinates, x', y' are new coordinates.The axes rotation in anti-clockwise is positive and clockwise rotation of axes is negative.

(iii) To change the direction of the axes of coordinates by changing the origin If P (x, y) and the axes are shifted parallel to the original axis so that new origin is (, ) and then the axes are rotated about the new origin (, ) by angle in the anti-clockwise (x', y' ), then the coordinates of P will be given by

O X

Y

F

P (x

, y)

O

Y' X'

x = a + x' cos y' sin y = b + x' sin + y cos ,where x, y are old coordinates and x', y' are new

coordinates.

Important !If P (x, y) and two mutually perpendicular lines ax + by + c = 0 and bx ay + d = 0 are taken as new axes such that new coordinates of P are (x , y ), then

OX' X

Y'

Y

bx

ay

+ d

=0

ax +

by + c

=0

O'

+ + += = + +2 2 2 2

andbx ay d ax by c

x yb a a b


eg, If the axes are shifted to the point (1, 2) without rotation. What do the following equation becomes?(i) 2 22 4 4 0x y x y+ + = (ii) 2 4 4 8 0y x y + + =Solution. (i) Substituting 1, ( 2) 2x X y Y Y= + = + = in the equation 2 22 4 4 0,x y x y+ + = we get

2 22( 1) ( 2) 4( 1) 4( 2) 0X Y X Y+ + + + = 2 22 6,X Y+ = where ,X Y new coordinates

,x y old coordinates.(ii) Substituting x = X + 1, y = Y 2 in the equation 2 4 4 8 0,y x y + + = we get 2( 2) 4( 1) 4( 2) 8 0Y X Y + + + = 2 4Y X=

Illustration 33 If the axes are transformed from origin to the point ( 2, 1), then new coordinates of (4, 5) are

(a) (2, 6) (b) (6, 4) (c) (6, 6) (d) (2, 4)

Solution. Required coordinates are[4 (2), (51)] = (6, 6)


Illustration 34 Keeping the origin constant axes are rotated at an angle 30 in negative direction, then new coordinates of (2, 1) with respect to old axis are

(a) 2 3 1 2 3

,2 2

+

(b) 2 3 1 2 3

,2 2

+ +

(c) 2 3 3

,2 2

+

(d) None of these

Solution. 2 3 12cos ( 30 ) 1sin ( 30 )2

X= + =

2 3

2sin 30 cos 302

Y += + =

Hence, (d) is the correct answer.

Illustration 35 If the axes be turned through an angle tan12. What does the equation 4xy 3x2 = a2 become?

(a) 2 2 24X Y a = (b) 2 2 24X Y a+ = (c) 2 2 24X Y a+ = (d) None of these

Solution. Here, tan = 2

So, 1 2

cos , sin5 5

= =for x and y, we have

2

cos sin5

X Yx X Y

= =

and 2

sin cos5

X Yy X Y

+= + =

The equation 2 24 3xy x a = reduces to

224( 2 ) (2 ) 2. 3

5 5 5X Y X Y X Y

a + =

2 2 2 2 24(2 2 3 ) 3( 4 4 ) 5X Y XY X XY Y a + = 2 2 25 20 5X Y a = 2 2 24X Y a =Hence, (a) is the correct answer.

Illustration 36 If (x, y) and (X, Y ) be the coordinate of the same point referred to two sets of rectangular axes with the same origin and if ux + vy, where u and v are independent of x and y, becomes VX + UY, show that 2 2 2 2.u v U V+ = +Solution. Let the axes rotate an angle , and if (x, y) be the point with respect to old axes and (X, Y ) be the coordinates with respect to new axes, then

we get cos sin

.sin cos

x X Yy X Y=

= + Then, ( cos sin ) ( sin cos )ux vy u X Y v X Y+ = + + = ( cos sin ) ( sin cos )u v X u v Y + + + But given new curve VX UY+Then, +VX UY = ( cos sin ) ( sin cos )u v X u v Y + + + On comparing the coefficients of X and Y, we get cos sinu v V + = (i)and sin cosu v U + = (ii)On squaring and adding Eqs. (i) and (ii), we get 2 2 2 2u v U V+ = +

Points to Remember 1. Re ection (Image) of a point Let (x, y) be any point, then its image with respect

to (i) x-axis is (x, y) (ii) y-axis is ( x, y) (iii) origin is ( x, y) (iv) line y = x is ( y, x) 2. A triangle is isosceles, if any two of its median are

equal. 3. Triangle having integral coordinates can never be

equilateral. 4. If 0 ( 1, 2, 3)r r ra X b Y c r+ + = = are the sides of a

triangle, then the area of the triangle is given by

21 1 1

2 2 21 2 3

3 3 3

12

a b ca b c

C C Ca b c

where 1 2 3, ,C C C are the cofactor of 1 2 3, ,c c c in the determinant.

5. Area of parallelogram (i) Whose sides are a and b and angle between

them is , is given by area of parallelogram ABCD = ab sin

QA

D C

Ba

b

a

b


(ii) Whose length of perpendicular from one vertices to the opposite sides are P1 and P2 and angle between sides is , is given by area

of parallelogram ABCD 1 2sinPP=

.

QA

D C

B

P2

P1

6. A triangle having vertices 2 21 1 2 2( , 2 ), ( , 2 )at at at at and 23 3( , 2 )at at , then area of triangle

2 1 2 2 3 3 1[( ) ( ) ( )]a t t t t t t=

7. Area of triangle formed by coordinate axes and the line

ax by c+ + is 2

2cab

.

8. Area of rhombus formed by| | | | | | 0ax by c+ + =

is 22

.cab

9. Three points 1 1 2 2 3 3( , ), ( , ), ( , )x y x y x y are collinear,

if 3 22 12 1 3 2

y yy yx x x x

=

.

10. To remove the term of xy in the equation 2 22 0.ax hxy by+ + = The angle through which

the axis must be turned

11 2tan .2

ha b

=

Objective Exercise 1 1. If the line segment joining (2, 3) and ( 1, 2) is divided

internally in the ratio 3 : 4 by the line x + 2y = k, then k is (a) 41

7 (b) 5

7

(c) 367

(d) 317

2. The polar coordinates of the vertices of a triangle are

(0, 0), 3, 6

and 3, 2. Then, the triangle is

(a) right angled (b) isosceles (c) equilateral (d) None of the above 3. The incentre of the triangle formed by the axes and the

line + = 1x ya b

is

(a) ,2 2a b

(b) + + + +,

ab aba b ab a b ab

(c) ,3 3a b

(d) + + + + + +2 2 2 2

,ab ab

a b a b a b a b

4. In the ABC, the coordinates of B are (0, 0), AB = 2, =3

ABC and the middle point of BC has the

coordinates (2, 0). The centroid of the triangle is

(a) 1 3

,2 2

(b) 5 1

,3 3

(c) + 4 3 1

,3 3

(d) None of these

5. The coordinates of three consecutive vertices of a parallelogram are (1, 3), (1, 2) and (2, 5). The coordinates of the fourth vertex are

(a) (6, 4) (b) (4, 6) (c) (2, 0) (d) None of these

6. A point moves in the x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to 3. The area enclosed by the locus of the point is

(a) 18 sq unit (b) 92

sq unit

(c) 7 sq unit (d) None of these

7. Let A = (1, 2), B = (3, 4) and let C = (x, y) be points such that (x 1) (x 3) + (y 2) (y 4) = 0. If ar ( ABC) = 1, then maximum number of positions of C in the x-y plane is

(a) 2 (b) 4 (c) 8 (d) None of these

8. The points (, ), (, ), (, ) and (, ) taken in order, where , , , are different real numbers, are

(a) collinear (b) vertices of a square (c) vertices of a rhombus (d) concyclic

9. The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x 2y = 7. Then, PQRS must be a

(a) rectangle (b) square (c) cyclic quadrilateral (d) rhombus

10. The coordinates of the four vertices of a quadrilateral are ( 2, 4), ( 1, 2), (1, 2) and (2, 4) taken in order. The equation of the line passing through the vertex ( 1, 2) and dividing the quadrilateral in two equal areas, is

(a) x + 1= 0 (b) x + y = 1 (c) x y + 3 = 0 (d) None of these


11. The graph of the function + +2.cos cos ( 2) cos ( 1)x x x is a

(a) straight line passing through the point 2(0, sin 1) with slope 2

(b) straight line passing through the origin (c) parabola with vertex 2(1, sin 1) (d) straight line passing through the point

2, sin 12

12. If the points 1

1,( 2, 0),3

and (cos , sin ) are

collinear, then the number of values of [0, 2 ] is (a) 0 (b) 1 (c) 2 (d) in nite 13. The limiting position of the point of intersection of the

lines + =3 4 1x y and + + =2(1 ) 3 2c x c y as c tends to 1, is

(a) (5, 4) (b) (5, 4) (c) (4, 5) (d) None of these 14. If a vertex of an equilateral triangle is the origin and the

side opposite to it has the equation x + y = 1, then the orthocentre of the triangle is

(a) 1 1

,3 3

(b) 2 2

,3 3

(c) 2 2

,3 3

(d) None of these

15. Three vertices of a quadrilateral in order are (6, 1) (7, 2) and (1, 0). If the area of the quadrilateral is 4 sq unit, then the locus of the fourth vertex has the equation

(a) =7 1x y (b) + =7 15 0x y (c) + =2( 7 ) 14( 7 ) 15 0x y x y (d) None of the above

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