STRAIGHT LINE AND PAIR OF LINES DISTANCE BETWEEN TWO POINTS Let P and Q are two points whose coordinates are (x1, y1) and (x2, y2), then

PQ = 22 QQ'PQ' = 221

221 y-(yx-(x ))

Note: Distance is always positive. Therefore we often write PQ instead of PQ. KEY TOOLS In order to prove that a given figure is a 1. Square: Prove that the four sides are equal and the diagonals are equal. 2. Rhombus (but not a square): Prove that the four sides are equal but the diagonals are not equal. 3. Rectangle: Prove that the opposite sides are equal and the diagonals are also equal. 4. Parallelogram (but not a rectangle): Prove that the opposite sides are equal but diagonals are not equal. Note: That in each of these cases diagonals bisects each other. 5. Equilateral Triangle: Prove all three sides are equal. 6. Isosceles Triangle: Prove two opposite side are equal. (Opposite angle also equal) SECTION FORMULA The coordinates of a point which divides the line segment joining two given points A(x1, y1) and B(x2, y2) internally in the given ratio m : n are

nmnymyy

nmnxmxx 1212

,

The coordinates of the point P(x, y) which divides A (x1, y1) and B (x2, y2) externally in the ratio m:n are

n-mny-myy

n-mnx-mxx 1212 ,

Illustration 1: Find the ratio in which x-axis and y-axis divides the line segment joining

(–2, – 3) and (–1, 2). Key concept: If the ratio, in which a given line segment is divided, is to be determined,

then for convenience instead of taking m:n, we take the ratio k:1. If the value of k turns out to be positive, it is an internal division and if k is negative it is an external division.

Solution: Let x-axis divides the line segment in k : 1. Then the coordinate of point on x-

axis which divides PQ in K : 1 is

1k2k3,

1kk2

. But if any point is lying on the x-axis, then its y-

coordinate will be zero

23k

1k2k3

0

Hence x – axis divides PQ in 3:2 internally. Similarly coordinates of any point on y-axis which divides PQ in k :1 is

1k2k3,

1k2 k

. But if any point is lying on the y-axis, then it’s x-

coordinate will always zero

12k

1kk2

0

Hence y-axis divides PQ in 2:1 externally SOME BASIC DEFINITIONS Centroid (G): The centroid of a triangle divides each median in the ratio 2:1(2 from the vertex

and 1 from the opposite side). The coordinates of centroid are given by

G

3

yyy ,3

xxx 321321 .

Incentre (I): The coordinates of the in-centre of a triangle with vertices (x1, y1), (x2, y2) and

(x3, y3) are given by I 1 2 3 1 2 3ax +bx + cx ay + by + cy, a + b + c a + b + c

where a, b and c

are length of the sides BC, CA and AB respectively. Orthocentre (H): The co-ordinates of the orthocentre of the triangle A(x1, y1), B(x2, y2), C(x3,y3) are

tanC+tanB+tanACtanyBtanyAtany ,

CtanBtanAtanCtanxBtanxAtanx 321321 .

Circumcentre (C):

The coordinates of the circum-centre of the triangle with vertices A(x1, y1), B(x2, y2), C(x3, y3) is given by

O = 1 2 3 1 2 3sin2 sin2 sin2 y sin2 sin2 sin2, sin2 sin2 sin2 sin2A+sin2B+sin2C

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

.

Note: The circum centre of a right angled triangle is the mid point of its hypotenuse. In an equilateral triangle centroid, in-centre, orthocentre and circum centre all

coincides. Centroid divides the line joining orthocentre and circum centre in 2:1 internally.

Illustration 2: If the vertices of a triangle be (2 3 , 6), (3, – 3 ) and (0, 0). Then find its orthocentre and circumcentre.

Key concept: In a right angled triangle orthocentre is the right angled vertex, and circumcentre is the mid point of the hypotenuse.

Solution: Clearly, the given triangle is right angled at vertex(0, 0), so its orthocentre is (0,

0) and circumcentre will be the mid point of hypotenuse which is

2

36,2

332.

Illustration 3: Find the orthocentre of the triangle whose

vertex are (0, 0), (1, 4) and (3, 0). Solution: Let the orthocentre be H. Clearly x co-ordinate of orthocentre is 1 and let

the y co-ordinate be k. Then from OB’H and BB’A

2 1tan k =

1 4 2k

Hence the orthocentre is 11, 2

(1,k)

(0, 0)

()

Y

XO A(3, 0)B'

B

H

AREA OF A TRIANGLE Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle

ABC. Then the area of triangle ABC, is 1 1

2 2

3 3

x y 11 x y 12

x y 1

LOCUS The equation to a locus is the relation which exists between the coordinates of any point on the path,

and which holds for no other point except those lying on the path. Working rule to find the locus of a point: Step 1: Let the coordinates of the moving point be P(h, k). Step 2: Write down the given geometrical condition and express these conditions in terms of

h and k. Step 3: Eliminate the variable to get the relation in h and k i.e. this relation must contain only

h, k and known quantities. Step 4: Express the given relation in h and k in the simplest form and then put x for h and y

for k. The relation thus obtained, will be the required equation of the locus of P(h, k). Illustration 4: Find the locus of a point P which divides the line joining (1, 0) and (2 cos,

2 sin) internally in the ratio 2 : 3 for all . Solution: Let the coordinate of point P which divides the line joining (1, 0) and (2cos,

2sin) in the ratio 2 : 3 be (h, k) then

5

sin4k,5

3cos4h

cos = 4

3h5 , sin =

4k5

22

22

4k5

43h51sincos

= (5h-3)2 + (5k)2 = 16 Hence locus is (5x – 3)2 + (5y)2 = 16 Illustration 5: The ends of a rod of length move on two mutually perpendicular lines.

Find the locus of the point on the rod, which divides it in the ratio 1 : 2. Solution: Suppose the two perpendicular lines are x = 0 and y = 0 and intercepts a and b

are cut respectively on the two lines. Then the two points on these lines are (0, a) and (b, 0).

The point P has coordinates given by

h = 120.2b

, k = 12

0.1a.2

Also 2 = a2 + b2

22

2 h32k3

Thus the required locus is

x2 + 94

y 22 .

2

(0, a)

(b, 0) O

P 1

x

y

STRAIGHT LINE GENERAL EQUATION OF A STRAIGHT LINE Any equation of first degree of the form ax + by + c = 0, where a, b, c, are constants always represents a straight line (at least one out of a and b is non-zero). Slope or gradient of a line: If is the angle at which a straight line is inclined to the positive direction of x-axis, then tan is called the slope or gradient of the line, where 0 < 180 ( 90). Note: (i) If a line is parallel to x-axis, then its slope is equal to tan0 = 0.

(ii) Slope of a line perpendicular to x-axis is not defined. Whenever we say that the slope of a line is not defined, we mean that the line is perpendicular to x-axis.

1.1 INTERCEPT OF A STRAIGHT LINE ON THE AXES Intercept of a line on x-axis: If a line cut the axis at (a, 0), then ‘a’ is called the intercept of a line on x-axis or the x-intercept. If a line intersect the x-axis at (–2, 0), then the length of intercept is –2 = 2. Intercept of a line on y-axis: If a line cuts y-axis at (0, b) then b is called the intercept of the line on y-axis or y intercept. DIFFERENT FORMS OF THE STRAIGHT LINES Slope intercept form: Equation of a line whose slope is m (m = tan) and which cut an intercept c on the y- axis (i.e. which passes through the point (0, c) is given by y = mx + c. Intercept form: The equation of the line which cuts off intercepts a and b on x-axis and y-

axis respectively is given by x 1ya b .

Thus intercept of a straight line on x-axis can be found by putting y = 0 in the equation of the line and then finding the value of x. Similarly intercept on y-axis can be found by putting x = 0 in the equation of the line and then finding the value of y. Normal Form: The equation of a straight line upon which the length of perpendicular from the origin is p and the perpendicular makes an angle with the positive direction of x-axis is given by

x cos + y sin = p Note: In normal form of equation of a straight line p is always taken as positive and is measured from positive direction of x-axis in anticlockwise direction between 0 and 2 . Note: In the normal form x cos +y sin =p, p is always taken as positive. Point-slope form: The equation of a line passing through the point (x1, y1) and having slope m is

given by 1 1y y m x x . Two points form: The equation of a straight line passing through two given points (x1, y1) and (x2, y2)

is given by 2 11 1

2 1

y yy y x xx x

Note: Slope of the line passing through two points (x1, y1) and (x2, y2) is given by 2 1

2 1

.y yx x

p

y

x O

y

y = mx + c

x where m = tan

1b

y

a

x

y

x (a, o)

(o, b)

Parametric form: The equation of line passes through a given point P(x1, y1) and makes a given angle with the

positive direction of the x-axis is given by 1 1

cosθ sinθx x y y r

where r is the distance between the variable point Q (x, y) and the fixed point P(x1, y1).

Any point on the line will be of the form (x1 + r cos, y1 + r sin). For different values of r, we will get different points on the line

Here r will gives the distance of the point Q from the fixed point P(x1, y1).

If P(x1, y1) is any point on the line which makes an angle with the positive direction of x axis, then there will be two points on the line at a distance r from P(x1, y1). One will be relatively upward if r is taken positive and other will be relatively downward if r is taken negative.

Illustration 6: Find the points on the line y = x – 2, which are lying at a unit distance from

the point (5, 3). Key concept 1: In general case if in any question there is something related to distance, we

use the parametric form of line. Solution: Clearly (5, 3) is lying on the given line. Now we will write the parametric equation

of the given line passing through (5, 3), which is

x - 5 y - 3= = r1 1

2 2

Here r will represent the distance between (5, 3) and (x, y)

Any point on the line is

23,

25 rr

. Now if we take r = 1, we get the

coordinate of the point which is at a unit distance from (5, 3) and relatively upward, and if we take r = –1, we will get the point which is also at a unit distance from (5, 3) but relatively downward. Hence the points are

1 1 1 1P' 5 , 3 and '' 5 , 32 2 2 2

P

Key concept 2: Take any point on the given line in terms of a single variable and use the

distance formula. Solution: Any point on the given line can be taken as (, - 2). Given that the distance

between the points (, - 2) and (5, 3) is 1.

2 2( - 5) + ( - 5) 11 = 5 2

Hence the required points are 1 1 1 15 , 3 and 5 , 32 2 2 2

θ

rQ(x,

y)

RP(x1, y1)θ

y -y1

x -x1

Illustration 7: If the straight line through the point P (3, 4) makes an angle /6 with x-axis

and meets the line 12x + 5y + 10 = 0 at Q, find the length of PQ. Solution: The equation of a line passing through P (3, 4) and making an angle /6 with the

x-axis is

r

6sin

4y

6cos

3x

, where r represents the distance of any point on this line

from the given point P (3, 4). The co-ordinates of any point Q on this line are

46

sinr,36

cosr

If Q lies on 12x + 5y + 10 = 0, then

0102r45

23r312

5312

132r

length PQ =

5312132

POSITION OF A POINT WITH RESPECT TO A GIVEN LINE Let the given line be ax + by + c = 0, and we have to check whether the point (, ) lies above the line or below the line. To check first draw a perpendicular line from P(, ) on the x-axis which intersect the given line at Q. x-coordinate of Q will be and corresponding to , y-coordinate of Q will be

aα + cβ' = -b

. Now

if β > β' , the point will lie above the line if β = β' point will lie on the line

and if β < β' point will be lie below the line. Illustration 8: If the point (, 2) lies between the acute angle region formed between the

lines y = 2x and y = 3x, then find the range of . Key concept 1: Since y co-ordinate of (, 2) is always non negative, point (, 2) will

always lie above the x-axis. Now for acute angle region the point (, ) should lie below the line y = 3x and above the line y = 2x.

I I

I ( , 2 )

( , 3 )

y = 3x

y = 2x

2( , )

Solution: (, ) should lie below the lie y = 3x 3 < 2 and (, )should also lie above the line y = 2x 2 > 2 2 < 2 < 3 2 < < 3 Key concept 2: Join P with the origin now slope of OP should lie between the slope of the

lines y = 2x and y = 3x.

X

Y

Here (2,3)

O

2P( , )

y = 2x

Solution: Slope of OP (slope of the line y = 2x, slope of the line y = 3x) (2, 3): Key concept 3: First identity the locus of point )2P(α, α , that means curve on which P

lies for different values of . Then find the portion of the curve lying between the acute angle region between the lines.

From the graph it is clear that (2,3) Solution: First we will find the locus of P x = and y = 2. Now on eliminating we get y = x2. Hence the point P will

always lie on the curve y = x2.

2y x

2 3 X

Y

2( , )

y = 2x

y = 3x

P

POSITION OF TWO POINTS WITH RESPECT TO A LINE Let the given line be ax + by + c = 0 and P(x1, y1), Q (x2, y2) be two given points. Now

if 1 1

2 2

0ax by cax by c

1 1

2 2

0ax by cax by c

both ax1 + by1 + c and ax2 + by2 + c are of opposite sign, then the points P and Q will lie on the opposite side of the line ax + by + c = 0 and,

if 1 1

2 2

0ax by cax by c

1 1

2 2

0ax by cax by c

both ax1 + by1 + c and ax2 + by2 + c are of the same sign, then the points P and Q will lie on the same side of the given line. Illustration 9: Find the range of such that the point (2, 2) and (2, 6) lies on the same

side of the line x + y –1 = 0. Key concept: Sign of x + y –1 should be same for both the points. Solution: For (2, 6) x + y – 1 = 2 + 6 –1 is + ve. Hence 2 + 2 –1 > 0 ( + 1)2 – 2 > 0 ( a + 1 + 2) ( a + 1 - 2)> 0

α ( , 1 2) ( 2 1, ) Common mistake: Generally students solve the inequality ( + 1)2 > 2 like in this way ( + 1)2 > 2 + 1 > 2 > 2 –1 but this is wrong ANGLE BETWEEN TWO STRAIGHT LINES If is the acute angle between two lines, then

tan = 21

21

mm1mm

where m1 and m2 are the slopes of the two lines

and are finite.

y = m1x + c1

y = m2x + c2

180-

Notes:

If the two lines are perpendicular to each other then m1m2 = -1. Any line perpendicular to ax + by + c = 0 is of the form bx – ay + k = 0. If the two lines are parallel or are coincident, then m1 = m2. Any line parallel to ax + by + c = 0 is of the form ax + by + k = 0. If any of the two lines is perpendicular to x-axis, then the slope of that line is not define

(infinite).

Let m1 = , then 21

21

mm1mmtan

= 2

21

1

2

m1

mm1

mm1

or = |90 - |, where tan = m2 i.e. angle is the complimentary to the angle which the oblique line makes with the x-axis.

Illustration 10: Find the equation to the sides of an isosceles right-angled triangle, the

equation of whose hypotenuse is x - 2y = 3 and the opposite vertex is the point (2, 2).

Solution: Clearly other two sides of the triangle will be making an angle of 45o with the given line. That means we have to find the equation of line passing through the point (3, 2) and making an angle of 45o with the given line. Slope of the given line x – 2y = 3 is m1 = 1/2.

3x + 4y = 4 45°

45°

(2, 2)

tan 45° = 1

1

m-1/2, i.e., 1= 11 12

m mm m m

1 m = 3, - .3

Hence the required equations of the two lines are 3x – y – 7 = 0 and x + 3y – 9 = 0

EQUATION OF THE STRAIGHT LINE PASSING THROUGH A POINT AND INCLINED TO A GIVE LINE The equations of the lines through the point (x1, y1) and making equal

angles with the given line are y – y1 = mA(x – x1), y – y1 = mB(x - x1).

(x1, y1)

THE DISTANCE BETWEEN TWO PARALLEL LINES The distance between two parallel lines:

ax + by + c1 = 0 and ax + by + c2 = 0 is 22

21

ba

cc

.

PERPENDICULAR DISTANCE OF A POINT FROM A LINE The length of the perpendicular from

P(x1, y1) on ax + by + c = 0 is 22

11

bacbyax

.

TO FIND THE IMAGE OF A POINT IN A LINE Let the given line be ax + by + c = 0 and we have to find the image of the point P(, ).

ax + by + c1 = 0

ax + by + c2 = 0

ax + by + c = 0

P (x1 y1)

P

P'

R

', '

,

Working rule: Step 1: Find the equation of line PP’ (where P’ is image of P). Slope of PP’ will be b/a, because PP’ is perpendicular to given line. Hence equation of PP’ is y - β = b/a(x - α) Step 2: Find the point of intersection of line PP’ with the given line and let it be R(p, q). Clearly R is the mid point of PP’.

Step 3: Now

pp 2'2

'

Also

qq 2'2

'

Hence the image is (2p – , 2q – ) Illustration 11: Find the image of (5, 7) about line x –y + 6 = 0 Solution: slope of line PP’ = –1 Hence equation of PP’ y – 7 = – 1 (x – 5) x + y – 12 = 0 Point of intersect of x + y – 12 = 0 with x – y

+ 6 = 0 is (3, 9).

Hence α' + 5 β' +7= 3 and = 9

2 2

α', β' (1,11) CO-LINEARITY OF THREE POINTS Following methods can be used to prove that three given points P(x1, y1), Q(x2, y2), R(x3, y3) are collinear. Method 1: If P, Q and R are collinear, then area of triangle formed by these three points

should be equal to zero

P Q R

1 1

2 2

3 3

x y 11 x y 1 02

x y 1

Method 2: Any one point of the given three points should divide other two points either internally or externally.

2 1 2 13 3

mx +nx my +nyx = and y =m+n m+n

for some real m and n.

Method 3: Slope of PQ = Slope of QR

P(5, 7)

P'

R

', '

x - y + 6

3 22 1

2 1 3 2

y -yy -y =x -x x -x

CONCURRENCY OF THREE LINES The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be concurrent is

0cbacbacba

333

222

111

.

BISECTORS OF ANGLE BETWEEN TWO GIVEN LINES Let a1x+b1y+c1 = 0……..(1)

a2x + b2y + c2 = 0….(2) are two intersecting lines. Let any point p(x, y) be any point on either of the two bisectors of angles of (1) and (2).

Then p is equidistance from (1) and (2) 22

22

222

21

21

111

ba

cybxa

ba

cybxa

22

22

222

21

21

111

ba

cybxa

ba

cybxa

which are the required equations of the two bisectors of angles between (1) and (2). If the two given lines are not perpendicular i.e. a1 a2 + b1 b2 0, then one of these equation is the equation of the bisector of acute angle and the other that of the obtuse angle. The equation of acute and the obtuse angle bisectors: Method 1

Step 1: Take one of the given lines and let its slope be m1 and take one of the bisectors and let it’s slope be m2.

Step 2: If be the acute angle between them, then find 21

21

mm1mmtan

Step 3: If tan > 1 then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle.

If tan < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angle.

θ

N

A

M

P (x, y)

q

Q

O

a1 x + b

1 y + c1 = 0

a 2x +

b 2y +

c 2 = 0

1

2

Method 2:

If the constant term c1 and c2 in the two equations a1x+b1y+c1 = 0 and a2x + b2y + c2 = 0 are of the same sign, then

Case 1: if then will give the equation of

obtuse angle bisector and2

22

2

22

21

21

11

ba

cybxa

ba

cybxa

will give the equation of

acute angle bisector.

Case 2: if then will give the equation of

acute angle bisector and will give the equation of

obtuse angle bisector. Note: Whether both the lines are perpendicular or not but the angle bisectors of these lines will

always be mutually perpendicular. The equation of the bisector of the angle which contain a given point: The equation of the bisector of the angle between the two lines containing the point ( , ) is

if 222111 cbaandcba are of the same signs

or if 222111 cbaandcba are of the opposite

signs The equation of the bisector of the angle containing the origin: Write the equations of the two lines so that the constants c1 and c2 are positive. Then the equation

22

22

22

21

21

11

ba

cybxa

ba

cybxa

is the equation of the bisector containing the origin.

Note: if ,02121 bbaa , then the origin will lie in the acute angle and if ,02121 bbaa then origin will lie in the obtuse angle. Illustration 12: For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0 find equation of

the bisector of the (a) acute angle (b) angle which contain (1, 2) (c) angle which contain origin

Solution: (a) The equation of given lines are 4x + 3y – 6 = 0 …(1) and 5x + 12y + 9 = 0 …(2) equation of angle bisectors between these two line are

2 2 2 2

4 3 6 5 12 94 3 5 12

x y x y

….(3)

9x – 7y – 41 = 0 and 7x + 9y – 3 = 0

,02121 bbaa2

22

2

22

21

21

11

ba

cybxa

ba

cybxa

,02121 bbaa2

22

2

22

21

21

11

ba

cybxa

ba

cybxa

22

22

22

21

21

11

ba

cybxa

ba

cybxa

22

22

22

21

21

11

ba

cybxa

ba

cybxa

22

22

22

21

21

11

ba

cybxa

ba

cybxa

Here we will get two equation of bisectors. Two find the acute angle bisector take one equation of the given line and one equation of bisector. Given line be 4x + 3y – 6 = 0 and one bisector be 7x + 9y – 3 =

0. Now

4 73 9tanθ = 14 71 .

3 9

Hence bisector 7x + 9y – 3 = 0 will be acute angle bisector.

(b) Equation of bisector which contain the point (1, 2) since 4.1 + 3.2 – 6 > 0 and 5.0 + 12.2 + 9 > 0 Hence equation of bisector which bisect the angle which contains the point (1, 2) is 4 3 6 5 12 9

5 13x y x y

9x – 7y – 41 = 0

(c) Since 4.0 + 3.0 – 6 < 0 and 5.0 + 12.0 +9 > 0. Hence the equation of the bisector which contain the origin is

4 3 6 5 12 95 13

x y x y

7x + 9y – 3 = 0 FAMILY OF LINES Suppose L1= a1x+b1y+c1 = 0 and L2=a2x + b2y + c2 = 0 are two intersecting lines and let the point of their intersection be ( , ). Now if we write these two equations in this form 1 2L + L 0 (where is a parameter)

1 1 1 2 2 2(a x +b y + c ) (a x + b y + c ) 0 ………(1)

L1 (ax

1 + by1 + c

1 )

L2(ax2 + by2 + c2)

then for different values of , (1) will give different straight lines. Now 1 1 1 2 2 2( ) ( ) 0a b c a b c

00.0 ( α,β ) always lies on (1) whatever be the value of . Hence (1) represent a family of straight lines passing through the point of intersection of a1x+b1y+c1 = 0 and a2x + b2y + c2 = 0.

Note: Whenever we have to show that a line always passes through a fixed point, we use the

concept of family of lines. Family of lines perpendicular to a given line ax+by+c = 0 is given by bx-ay+k=0, where k is a

parameter. Family of lines parallel to a given line ax+by+c = 0 is given by ax+by+k=0, where k is a

parameter. Illustration 13: Find the equation of the straight line which belongs to both of the following

family of lines 5x + 3y – 2 + 1 (3x – y – 4) = 0 and x - y + 1 + 2 (2x – y – 2) = 0

Solution. Lines of first family are concurrent at (1, -1) and that of second at (3, 4) Required line passes through both of these points Equation is 5x – 2y – 7 = 0 Illustration 14: If a, b and c are three consecutive odd integers then prove that the variable

line ax + by + c = 0 always passes through (1, –2). Solution: Since a, b and c are three consecutive odd integers, these must be in form of

2n + 1, 2n +3 and 2n + 5 respectively where n I. The given line can be written as (2n + 1)x + (2n +3)y + (2n + 5) = 0

n/2 (x + y + 1) + (x + 3y + 5) = 0. Now we can say this will represent a family of line passing through the point of

intersection x + y + 1 = 0 and x + 3y + 5 = 0, which is (1, -2) Illustration 15. The base BC of ABC is bisected at (p, q) and equation of sides AB and

AC are px + qy = 1 and qx + py = 1. Then show that the equation of

median through A is ( px + qy –1) – 01pyqx1pq2

1qp 22

.

Solution. Any line through A is given by (px + qy – 1) + (qx + py – 1) = 0. If this line represent the median through A then this will be passing through (p,

q). Hence = )1pq2(

)1qp( 22

.

Thus the required line is (px + qy –1) – 01pyqx1pq2

1qp 22

PAIR OF STRAIGHT LINES The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of

straight lines if 0cfgfbhgha and h2 ab.

abc + 2fgh - af2 - bg2 - ch2 = 0 and h2 ab. The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin if h2 ab.

If the lines through the origin whose joint equation is ax2 + 2hxy + by2 = 0, are y = m1x and

y = m2x, then y2 - (m1 + m2)xy + m1m2x2 = 0 and y2 + bh2

xy + 2xba

= 0 are identical, so that

bamm,

bh2mm 2121 .

If be the angle between two lines, through the origin, then

21

212

21

mm1mm4mm

tan

=

baabh2 2

.

The lines are perpendicular if a + b = 0 and coincident if h2 = ab. Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Curve and a Line: If the line lx + my + n = 0, ((n 0) i.e. the line does not pass through origin) cut the curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e.

ax2 + 2hxy + by2 + (2gx + 2fy) 0nmyxc

nmyx 2

is the equation of the lines OA and OB

B

A

O Illustration 16: If the lines joining origin to the points of intersection of the curve

2x2 + 3y2 + m = 0 and the line y = 2x + 3 are perpendicular, then find the value of m.

Solution. Homogenise the given curve with line, we get the required line

03

x2ymy3x22

22

Since, the lines are perpendicular Coefficient of x2 + coefficient of y2 = 0 m = -9 Illustration 17: Find the value of unknowns for which the given equation represents

two coincident lines x2 + 9y2 + pxy + qx + ry + 1 = 0. Solution: Let the line is ax + by + c = 0 Equation of pair of lines (ax + by + c)2 = 0 a2x2 + 2abxy + b2y2 + 2acx + 2bcy + c2 = 0 Comparing this with given equation a2 = 1 a = 1 b2 = 9 b = 3 c2 = 1 c = 1 p = 2ab = 6 q = 2ac = 2

r = 2bc = 6 ROTATION OF CO-ORDINATE AXES Let OX, OY be the original axes and OX’ and OY’ be the new axes obtained after rotating OX and OY through an angle in the anticlockwise direction. Let P be any point in the plane having coordinates (x, y) with respect to axes OX and OY and (x’, y’) with respect to axes OX’ and OY’. Then

x = x’ cos – y’ sin, y = x’ sin + y’ cos ...(1) and x’ = x cos + y sin, y’ = – x sin + y cos …(2) Note: The above transformation can also be displaced by a table.

x’ y’ x cos –sin y sin cos

Y

XO

y'x'

X'Y'

x

P(x, y) (x', y' )

y

If f(x, y) = 0 is the equation of a curve then it’s transformed equation is

f( x’ cos – y’ sin, x’ sin + y’ cos) = 0