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Maths / Straight Lines

Straight Lines

01. Introduction

02. Straight Lines

03. Pair of Straight Lines

CONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOTESTESTESTESTES



Straight Lines

Co-ordinate geometry is a marriage of pure geometry and algebra, and is indispensable in all branches of sciencetoday. Many of you must be pretty familiar with the general outline of this subject.

We will restrict ourselves to 2-dimensional (plane) co-ordinate geometry in the following pages. Later on, we willalso get the chance of studying 3-dimensional co-ordinate geometry.

The basic idea in co-ordinate geometry, as has been mentioned earlier, is to study the properties of geometricalfigures such as straight lines, circles, parabolas etc through the use of numbers. The core concept is that on a2-dimensional (Euclidean) plane, any point can be represented by a pair of real numbers, using two non-parallelstraight lines. The point where these two non-parallel reference lines meet is termed the origin of the reference axis.By convention, one axis is called the x-axis and one the y-axis. Any point on the plane can now be determined inreference to this reference axes as described in the figure below:

y-axis

x-axis(origin) O

x

y

The point P lies x units along the x-axis and y units along the y-axis, and therefore, P can be represented as (x, y). Note that x, y x is called the abscissa of P while y is the ordinate of P. x and y together are called the co-ordinates of P.

ε R

P

Fig - 1

Conversely, given the co-ordinates x and y of a point P, we can easily determine its location by moving x unitsalong the x-axis and then y-units parallel to the y-axis.

Notice that as long as the two axes are non-parallel, the entire plane is representable using these two axes asreference. These two axes in general can be at any non-zero angle to each other.

Section - 1 INTRODUCTION



However, it is almost always the case (out of convenience) that the two axis are taken at right angles to each other.Such axes are called Rectangular Axes. We will always be using Rectangular Axes in our discussion from nowonwards.

With this introduction, we start with the most elementary of geometric figures: line segments and lines (and othergeometric figures obtainable from these elementary ones, like polygons).

For a good command over co-ordinate geometry, a lot of results will be required to be memorised since they areencountered so often. For this purpose, each new theorem or result or property that we will encounter in thefollowing pages is discussed in a separate article for ease of reference later.

Art 1 : Distance formula

One of the most basic expressions in co-ordinate geometry is that of the distance between two arbitrary

points ( ) ( )1 1 1 2 2 2, and ,P x y P x y . To obtain the required distance in terms of the co-ordinates of thesepoints, the Pythagoras theorem is employed as described in the figure below:

P (x ,y )2 2 2

A

d

P1

(x ,y )1 1

Note that:(i) OC = x ; OB = x BC = AP = x – x

1 2

1 2 1⇒

(ii) CP = y ; BP = y AP = y – y

1 1 2 2

2 2 1⇒

x

y

O C B

Fig - 2

As explained in the figure, the distances AP1 and AP2 have been obtained. Thus, by the Pythagoras

theorem, the distance d is 2 21 2AP AP+ or

( ) ( )2 22 1 2 1d x x y y= − + −

As a direct consequence of this formula, the distance of an arbitrary point P(x, y) from the origin is2 2x y+ . As an elementary exercise, assume four points anywhere on the co-ordinate plane randomly,

and use the distance formula to calculate the distance between each pair of points.

Art 2 : Section formula

Suppose that we are given two fixed points in the co-ordinate plane, say ( ) ( )1 1 2 2, and ,A x y B x y . Weneed to find the co-ordinates of the point C which divides the line segment AB in the ratio m : n. Observe

Section - 2 STRAIGHT LINES



that two such points will exist. Name them C1 and C2. One of them will divide the line segment ABinternally in the ratio m : n while the other will divide AB in the same ratio externally, as shown in thefigure below:

B(x , y )2 2

C (x , y )1 i i

A(x , y )1 1

Internal divisionC divides AB internally in the ratio m : n. Thus,

1

C (x , y )2 e e

B(x , y )2 2

A(x , y )1 1

External divisionC divides AB externally in the ratio m : n. Thus,

2

AC1

C1Bmn

=AC2

C2 Bmn

=

y

x

y

x

Fig - 3

Let us find the co-ordinates of C1 using the help of the more detailed figure of internal division below:

B(x , y )2 2

C (x , y )1 i i

A(x , y )1 1 DE

(0, y )2

(0, y )i

(0, y)

(x ,0)1 (x ,0)i (x ,0)2

Note that AECand ADB are similar. Thus,

1

AEAD

EC1

DB=

AC1

AB=

It is given that AC : C B = m : n.Thus,

1 1

AC1

ABm

m + n=

y

xO

Fig - 4As described in the figure above,

1AE EC mAD DB m n

= =+

1 1

2 1 2 1

i ix x y y mx x y y m n

− −⇒ = =− − +

2 1 2 1,i imx nx my nyx y

m n m n+ +⇒ = =+ +

Thus, the co-ordinates of the point C1 which divides AB internally in the ratio m : n are

2 1 2 1: ,Internal

Division m nmx nx my ny

m n m n+ ++ +



Using an analogous approach, we can obtain the co-ordinates of the point C2 which divides AB externallyin the ratio m : n

2 1 2 1: ,External

Division m nmx nx my ny

m n m n− −− −

A particular case of internal division is finding the co-ordinates of the mid-point of AB. Since the mid-pointof AB divides the segment AB in the ratio 1 : 1, the co-ordinates of the mid-point will be

( ) ( )1 1 2 2

1 2 1 2

, and ,,

2 2Mid point of AB where

A x y B x y

x x y y−

≡ ≡

+ +

Let us put these rudimentary results to use.

Find the co-ordinates of the centroid of a triangle with the vertices ( ) ( ) ( )1 1 2 2 3 3, , , and , .A x y B x y C x y

Solution: To determine the centroid, we will borrow a result from plane geometry that you might remember fromhigh school: the centroid divides any median in the ratio 2 : 1.

A(x , y )1 1

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

G

AD is a median of ABCG divides AD in the ratio 2 : 1, i.e, AG : GD = 2 : 1

Fig - 5

D

The co-ordinates of D, the mid-point of BC, are 2 3 2 3,2 2

x x y y+ +

. Since : 2 :1,AG GD =

we have the co-ordinates of G by the section formula as

2 3 2 31 12 1. 2 1.

2 2,2 1 2 1

x x y yx yG

+ + + + ≡+ +

1 2 3 1 2 3,3 3

x x x y y y+ + + +≡

vThe expression for the centroid confirms the obvious fact that the co-ordinates of the centroid are‘symmetric’ with respect to the co-ordinates of the three vertices of the triangle.

Example – 1



G is the centroid of triangle ABC. If O is any other point in the plane, prove that2 2 2 2 2 2 23 .OA OB OC GA GB GC GO+ + = + + +

Solution: There’s no loss of generality in taking O as the origin of our reference axis since even if we are given Oto be a non-origin point, we can always translate the axes (keeping the triangle ABC unchanged) sothat its origin coincides with O. Note that this operation will have no effect on the lengths

, , , , , ,OA OB OC OG GA GB GC etc. However, the expressions for distances will become muchmore simplified (In co-ordinate geometry, you will be required to follow such steps often, so that theexpressions you are to deal with can be kept as simple as possible).

Now, we assume some co-ordinates for A, B and C as shown in the figure below:

Fig - 6

y

xO

G (x, y)

B (x , y )2 2

A(x , y )1 1

1 2 3 1 2 3,3 3

x x x y y y+ + + +

As discussed earlier, the co-ordinates of G (x, y) are

C (x , y )3 3

We have,2 2 2 2 2 2 2 2 2

1 1 2 2 3 3OA OB OC x y x y x y+ + = + + + + + ... (1)

while

( ) ( ) ( ) ( )2 2 2 22 2 2 21 1 2 23GA GB GC GO x x y y x x y y+ + + = − + − + − + −

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

2 2 2 2 2 21 2 3 1 2 3x x x y y y= + + + + + ... (2)

Comparing (1) and (2), we see that the two expressions are indeed equal v

____________________________________________________________________________________

Art - 3 Area of a triangle

Suppose we are given three points in the co-ordinate plane : ( ) ( ) ( )1 1 2 2 3 3, , , and , .A x y B x y C x y We

intend to find the area of ABC∆ in terms of the given co-ordinates. How to evaluate this area is describedin the figure below:

Example – 2



y

xO

B (x , y )2 2

P

Fig - 7

Note that

A(x, y

)1

1 C (x , y )3 3

Q R

area ( ABC) = area (trapezium APQB)

area (trapezium BQRC)

area (trapezium APRC)

+

–

Observe how the area of ABC∆ has been written in terms of the area of three trapeziums.

From plane geometry, the area of a trapezium is ( )1 sum of bases height.2

× × Thus,

( ) ( )1area trap.2

APQB AP BQ PQ= × + ×

( )( )1 2 2 112

y y x x= + − ... (1)

Similarly,

( ) ( )1area trap.2

BQRC BQ CR QR= × + ×

( )( )2 3 3 212

y y x x= + − ... (2)

( ) ( )1area trap.2

APRC AP CR PR= × + ×

( )( )1 3 3 112

y y x x= + − ... (3)

From (1), (2) and (3), we have, upon simplification,

( ) ( )2 1 1 2 3 2 2 3 1 3 3 11area2

ABC x y x y x y x y x y x y∆ = − + − + −

( ) ( ) ( ){ }1 2 3 2 3 1 3 1 212

x y y x y y x y y= − + − + −

We used the mod sign in the last expression because area is by definition positive.We can express the area obtained in determinant form very concisely:

( )1 1

2 2

3 3

11area 12

1

x yABC x y

x y∆ = ∆ =



As a consequence of this result, we can now easily find the area of an arbitrary quadrilateral ABCD asdescribe in the figure below:

y

xO

D(x , y )4 4

Fig - 8

area (quad ABCD)= area ( ABC) + area ( ACD)

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

A(x , y )1 1

1 1 1 1

2 2 3 3

3 3 4 4

1 11 1 12

1 1

x y x yx y x yx y x y

= +

In other words, we evaluate the areas of the two trianglesseparately and add them to get the area of the quadrilateral

We can generalise this method easily to find the area of any polygon as a sum of the areas of the constituenttriangles. v

Find the area of the triangle, the co-ordinates of whose vertices are , , , and ,a a aap aq arp q r

.

Solution: Using the result obtained is Art - 3, we have,

1

1 12

1

aappaaqqaarr

∆ =

12

a a a a a aap aq arq r r p p q

= − + − + −

( ) ( ) ( )2

2p r q q p r r q pa

qr pr pq− − −

= + +

( ) ( ) ( )2 2 2 2

2a p q r q r p r p q

pqr− + − + −

=

v

Example – 3



Assume two fixed points in the co-ordinate plane: ( ) ( ),0 and ,0A a B a− . A variable point C(x, y) moves in theplane in such a way that CA + CB is a constant k. Use the distance formula to evaluate the condition that theco-ordinates of C must satisfy.

Solution: We have,

( )2 2CA x a y= − +

and ( )2 2CB x a y= + +

From the constraint specified in the question, we have

CA CB k+ =

2 2 22CA CB CA CB k⇒ + + ⋅ =

( ) ( ) ( )( ) ( )( )2 2 2 22 2 2 2 22x a y x a y x a y x a y k⇒ − + + + + + − + + + =

( ) ( ) ( )22 2 4 2 2 2 2 2 2 22 2 2x a y y x a k x y a⇒ − + + + = − + +

( )4 4 4 2 2 2 2 2 2 2 2 2 22 2 2 2 2x y a x y a x a y k x y a⇒ + + + − + = − + +

Squaring both sides and cancelling out the common terms on both sides, we obtain

( )2 2 4 2 2 2 2 2 28 8 4a x k a x k x y a− = + − + +

2 2 2 2 2 2 4 2 24 16 4 4k x a x k y k k a⇒ − + = −

( ) ( )2 2 2 2 2 2 2 24 4 4 4x k a k y k k a⇒ − + = −

This is the relation that the co-ordinates of the variable point C (x, y) must satisfy. All the pairs (x, y)which satisfy this equation, when plotted on the co-ordinate plane, will trace out the path on which thevariable point C is constrained to move. In other words, this equation specifies the locus of thepoint C.

____________________________________________________________________________________

Art - 4 ( )Equation representing a straight lines

The last three articles dealt with the preliminaries of co-ordinate geometry and certain elementary formulaewhich find widespread use. With this article, we start the discussion of the geometry of straight lines indetail.

On the co-ordinate plane, the simplest case for a straight line would be one in which the line is parallel toone of the co-ordinate axes.

Example – 4



y

y0

O x

A line parallel to the x-axis.Any point on this line has a constant y-co-ordinate equal to y .Thus, we can say that the equationof this line is y = y(There's no constraint on the x-co-ordinate of this line)

0

0

y

x0O x

A line parallel to the y-axis.Any point on this line has a constant x-co-ordinate equal to x .Thus, we can say that the equationof this line is x = x(There's no constraint on the y-co-ordinate of this line)

0

0

Fig - 9

As described in the figure above, the equation of such a line is 0 0ory y x x= = accordingly as the line isparallel to the x-axis or the y-axis respectively.

These are special cases of lines; we want to find the equation of any arbitrary line in general. Visualise anysuch line in your mind. To completely specify such a line, you would need two quantities: the inclination ofthe line (or its slope or the angle it makes with say, the x-axis) and the placement of the line (i.e. where theline passes through with reference to the axes: we can specify the placement of the line by specifying thepoint on the y-axis through which the line passes, or in other words, by specifying the y-intercept.)

θ

We need this angle which tells us the inclination of the line

We need this lengthwhich tells us the placement of the line

c

y

x

Fig - 10

It should be obvious to you that any line can be determined uniquely using these two parameters.

We now find out the equation of this straight line, assuming that we know and .cθ In other words, weintend to find out the relation that the co-ordinates (x, y) of any arbitrary point on the line must satisfy. Thedetermination of this equation is straightforward:



We assume an arbitrary point P(x, y) on the line and try to relate x and yto the known quantities

and c. The relation we require is obtainable from the fact that in APB,

θ

x

Fig - 11

PBAB

tan =θθ

y

θ

(x, y)

B

CO

P

A(0, c)

As described in the figure above, we have in ,APB∆

tan PBAB

θ =

tan θ is a measure of the inclination of the line (its steepness). tan θis therefore termed the slope of the line

and is denoted by m. Thus, m = tan θ . Also, notice that ( ) and .PB y c AB x= − = Therefore,

y cm

x−=

Slope intercept

y mx c

y

⇒ = +

−!!

: Slope - intercept form

This is the general equation of a straight line involving its slope and its y-intercept. This form of the equationof the line is therefore termed the slope-intercept form.

Notice that if the line passes through the origin, its equation would reduce to y = mx.

As you might have guessed by now, this is not the only form to represent a straight line. This form uses theslope and the intercept of the line.

Lets discuss another form. Notice that to uniquely determine any straight line, we either need the slope ofthe line and a point through which this line passes, or we need at least two points through which that linepasses. Thus for example, a line can also be uniquely determined if we are given the two points where thisline intersects the x-axis and the y-axis.

θ

The straight line L can be uniquely determined if we know a and b, i.eif we know the x-intercept and the y - intercept

x

Fig - 12

O

b

a

y

φ

L



Notice that tan ba

φ= so that the slope of the line is ( )tan tan tan .bma

= θ = π −φ = − φ= − Also, the

y-intercept is b. Thus, using the slope intercept form obtained earlier, the equation of the line L isby x ba

= − +

bx ay ab⇒ + =

intercept intercept

1

x y

x ya b

− −

⇒ + =

"!

: Intercept form

Thus, if we know the x and y intercepts, we can directly use this form to write the equation of the line.

Lets consider a third form to represent a line. From the figure below, observe carefully that to uniquelydetermine a line, we could also specify the length of the perpendicular dropped from the origin to that lineand the orientation (inclination) of that perpendicular:

The straight line Lcan be uniquelydetermined if we knowp and α

y

y

p

α

L

Fig - 13

To determine the equation of this line, assume any point P on the line with the co-ordinate (x, y). Observethe geometry described in the figure below carefully:

Observe that:OR = OQ cos = x cos andRA = SP = PQ sin = y sin

α α

αα

x

Fig - 14

O

y

BQα α

R

S

A

P(x, y)p



From the figure, note thatOR RA OA p+ = =

cos sin

length ofinclination ofperpendicular perpendicular

x y p⇒ α + α =! "

: Normal form

Thus, we now know of three forms in which the equation of an arbitrary straight line can be written.

From those three forms, you might be able to deduce that the most general form for the equation of anarbitrary straight line is 0Ax By C+ + = . Let us try to prove this assertion, that is, let us try to show thatAx + By + C = 0 represents the equation of a straight line.

For this purpose, it will suffice to show that if we take any three arbitrary points ( ) ( )1 1 2 2, , ,x y x y and

( )3 3,x y on the curve 0,Ax By C+ + = these three points will turn out to be collinear. Equivalently, thearea of the triangle with the vertices as these three points will turn out to be zero.

Since all the three points satisfy the equation 0,Ax By C+ + = we have

1 1 0Ax By C+ + =

2 2 0Ax By C+ + =

3 3 0Ax By C+ + =

We can eliminate A, B and C from these three equations simultaneously to obtain a relation involving onlythe co-ordinates of the three points. A basic knowledge of elimination in determinant form will tell you thatthe relation we’ll get after elimination is

1 1

2 2

3 3

11 01

x yx yx y

=

which means that the area of the triangle formed by these three points as vertices is zero! Hence, theassertion is true.

With this discussion in mind, you should be able to write the equation for any arbitrary straight line. We willencounter the use of all these forms in the coming examples.

Before concluding this article, do this as a simple exercise based on the discussion we've already done:

(a) Show that the equation of the straight line with slope m and passing through the fixed point (x1, y1)

is ( )1 1y y m x x− = −

(b) Show that the equation of the straight line passing through the two fixed points ( ) ( )1 1 2 2, and ,x y x yis

1 2 1

1 2 1

y y y yx x x x

− −=− −



The following table summarizes the various forms of the straight line that we've encountered.

Known parameters about the line Equation Name of this form

1. Slope m y-intercept c y = mx + c Slope-intercept form

2. x -intercept a y -intercept b 1x y

a b+ = Intercept form

3. Length of perpendicular from origin to the line : p Inclination of perpendicular : α

cos sinx y pα + α = Normal form

4. Slope : m Any point through which the line passes : ( )1 1,x y

( )1 1y y m x x− = − Point-slope form

5. Any two points through : ( )1 1,x y

which the line passes : ( )2 2,x y 1 2 1

1 2 1

y y y yx x x x

− −=

− − Two point form

Most general form : Ax + By + C = 0 where , ,A B C ∈# and at least one of A, B is non-zero

Note that each of the five specific forms mentioned in the table above can be converted easily to the mostgeneral form of the equation of a line. You are urged to do this as an exercise.

Also, the five forms are inter convertible among themselves in most cases too. For example, y = mx + c

can be written in intercept form as ( )

1/

x yc m c

+ =−

so that the x-intercept of this line is cam

= − and the

y-intercept is b = c. You are urged to try out all the (possible) conversions from one form to another.

You should now be able to understand that to determine a straight line uniquely, we must have twoquantities given. Thus, two points could uniquely fix a line, or a point on the line and its slope could do sotoo, and so on. Notice that the general equation of the line also in fact contains only two arbitraryconstants:

0Ax By C+ + =

1 0A Bx yC C

⇒ + + =

1 0Px Qy⇒ + + =contains only twoarbitrary constants

Art 5 Point of intersection ; Angle of intersection

We are given two lines L1 and L2, and we are required to find the point at which they intersect (if they arenon-parallel) and the angle at which they are inclined to one another, i.e., the angle of intersection.Evaluating the point of intersection is a simple matter of solving two simultaneous linear equations. Let the



equations of the two lines be 1 1 1 1 2 2 2 2L 0 and 0a x b y c L a x b y c≡ + + = ≡ + + = (written in the mostgeneral form). Now, let the point of intersection be (x1, y1). Thus,

1 1 1 1 1 0a x b y c+ + =

2 1 2 1 2 0a x b y c+ + =This system can be solved to get

1 1

1 2 2 1 1 2 2 1 1 2 2 1

1x yb c b c c a c a a b a b

= =− − −

From this relation we obtain the point of intersection ( )1 1, asx y

1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

,b c b c c a c ca b a b a b a b

− − − −

: Point of intersection

To obtain the angle of intersection between these two lines, consider the figure below:

y

θ

θ2θ1 x

L = a x + b y + c = 02 2 2 2

L = a x + b y + c = 01 1 1 1

Note thatθ θ − θ= 2 1

Fig - 15

The equation of the two lines in slope-intercept form are

1 1 1 11 1 1

1 1 1 1

wherea c c ay x m x mb b b b

= − + = + = −

and2 2 2 2

2 2 22 2 2 2

wherea c c ay x m x mb b b b

= − + = + = −

Note in Fig - 15 that 2 1θ = θ −θ and thus

( )2 1tan tanθ = θ − θ

2 1

1 2

tan tan1 tan tan

θ − θ=+ θ θ

2 1

1 21m m

m m−=

+ ... (1)

Conventionally, we would be interested only in the acute angle between the two lines and thus we have to

have tan θ as a positive quantity. So in (1) above, if the expression 2 1

1 21m m

m m−

+ turns out to be negative, this

would be the tangent of the obtuse angle between the two lines; thus, to get the acute angle between thetwo lines, we use the magnitude of this expression.



Therefore, the acute angle θ between the two lines is

1 2 1

1 2

tan1m m

m m− −θ =

+ : Acute angle between the two lines

From this relation, we can easily deduce the conditions on m1 and m2 such that the two lines L1 and L2 areparallel or perpendicular.If the lines are parallel, 0θ = so that 1 2m m= which is intuitively obvious since parallel lines must have the

same slope. For the two lines to be perpendicular, 2πθ = so that cot 0;θ = this can happen if 1 21 0m m+ =

or 1 2 1.m m = − Thus,

1 2 : for parallel linesm m=

and

1 2 1 : for perpendicular linesm m = −

If the lines L1 and L2 are given in the general form given in the general form 0,ax by c+ + = the slope of

this line is amb

= − so that the condition for L1 and L2 to be parallel becomes 1 21 2 2 1

1 2

ora a a b a bb b

− = − =

and the condition for L1 and L2 to be perpendicular becomes 1 2

1 2

1a ab b

= − or 1 2 1 2 0a a b b+ = .

For example, the line 1 1L x y≡ + = is perpendicular to the line 2 1L x y≡ − = because the slope of 1Lis –1 while the slope of L2 is 1.

L = x - y = 12

y

x

L = x + y = 11

Fig - 16

As another example, the line 1 2 1 0L x y≡ − + = is parallel to the line 2 2 3 0L x y≡ − − = because the

slope of both the lines is 1 .2

m =



Find the equation to the straight line which passes through (3, –2) and is inclined at 60º to the line 3 1.x y+ =

Solution: Observe carefully that there will be two such lines. Denote the two lines by 1 2andL L

L1

Fig - 17

60º(3, -2)

L2

60º

3x+y = 1

Let the slope of the line(s) we require be m.

The slope of 3 1x y+ = is 1 3m = −Since we want the acute angle between the two lines to be 60º, we must have by Art - 5,

1

1

tan 60º1m m

mm−=

+

331 3

mm

− −⇒ =−

3 31 3m

m+⇒ = ±

−

3 3 3 or 3 3 3m m m m⇒ + = − + = −

0 or 3m m⇒ = =Since we get two values of m, this confirms our earlier assertion that two such lines will exist. We nowhave the slope. We also know that the lines pass through (3, –2). We can therefore use the point-slopeform to write down the required equations:

( ) ( ) ( ) ( )1 22 0 3 ; 2 3 3L y x L y x≡ − − = − ≡ − − = −

1 22 0 and 3 2 3 3 0L y L y x⇒ ≡ + = ≡ − + + =

Find the equation of the straight line which passes through the point ( )3 3cos , sina aθ θ and is perpendicular to

the straight line sec cosecx y aθ + θ = .

Example – 5

Example – 6



Solution: The slope of the given line is 1sec tan

cosecm θ= − = − θ

θTherefore, the slope of the line we require will be given by m2 where

21

1mm

= −

2 cotm⇒ = θWe now know the slope of the line and we are also given a fixed point through which the line passes.We can therefore use the point-slope form to determine its equation:

( )3 3sin cot cosy a x a− θ = θ − θ

( )4 4cos sin sin cosx y a⇒ θ − θ = − θ + θ

( )( )2 2 2 2cos sin cos sina= θ + θ θ − θ

cos 2a= θThus, the required equation is

cos sin cos 2x y aθ − θ = θ ____________________________________________________________________________________

Art 6 Half-planes

Any straight line divides the Euclidean plane into two half planes. In this article, we wish to determine thehalf-plane in which an arbitrary point lies with respect to a given line.

Let the equation of the given line be 0.ax by c+ + = Consider two points ( ) ( )1 1 2 2, and ,x y x y that lie indifferent half-planes with respect to this line:

Half-plane B(x , y )2 2

(x , y' )2 2

(x , y' )1 1

(x , y )1 1

Half-plane A

ax + by + c = 0

Q

P

Fig - 18

y

x

The point ( )1 1,x y lies in the lower half-plane while ( )2 2,x y lies in the upper half plane. We require acondition on these co-ordinates which must be satisfied if the points lie in opposite half-planes. In Fig - 18,

we have dropped verticle line segments from ( ) ( )1 1 2 2, andx y x y to the given line, intersecting it in P andQ respectively.

The co-ordinates of P and Q are 1 1'( , )x y and 2 2

'( , )x y respectively where 1 1 2 2' 'andy y y y≠ ≠ .

Since P, Q lie on the given line, their co-ordinates must satisfy the equation of the line. Thus,



( )11 1 10

ax cax by c y

b+

+ + = ⇒ = −' '

and( )2

2 2 20ax c

ax by c yb+

+ + = ⇒ = −' '

Now, from Fig - 18 we have

1 1y y< ' and 2 2y y> '

( )11

ax cy

b+

⇒ < − and( )2

2

ax cy

b+

> −

1 1 0ax by cb

+ +⇒ < and 2 2 0ax by cb

+ + >

1 1ax by c⇒ + + and 2 2ax by c+ + are of opposite signs.

This is the required condition. Translated into words, it says that for two points lying in oppositehalf-planes, their co-ordinates when substituted respectively into the equation of the line must giveexpressions of opposite signs. (For two points in the same half-plane, the signs would be the same).

As a corollary, observe that a point (x1, y1) lies in the same half–plane or opposite half–plane in which the

origin lies accordingly as ( )1 1ax by c+ + and c are of the same sign or opposite signs respectively.

Art 7 Length of perpendicular

Suppose that we are given the equation of a line L and we are required to find the length of the

perpendicular dropped from an arbitrary point ( )1 1, on .P x y L

Suppose that the equation of L is in normal form, i.e, cos sin .L x y p≡ α + α =

Fig - 19

y

x

R

P (x , y )1 1

L = x cos + y sin - p = 0αα

αp

S

Q

O L'

We are required to findPQ. Note that PQ = OR - OS = OR - pTo determine OR, wedraw a line L' parallel toL through P (x , y )Let OR = p

1 1

1

Based on the discussion in the figure above, the equation of the line 1is cos sin 0.L x y pα + α − ='

Since L1 passes through P, the co-ordinates of P must satisfy the equation of L1. Thus,

1 1 1cos sin 0x y pα + α − =

Thus, we get 1p as ( )1 1cos sin .x yα + α The length of perpendicular PQ is now simply

1 1 1cos sinp p x y p− = α + α − . (Modulus sign is used since PQ is a length so it must be positive).

1 1cos sinPQ x y p= α + α − : Length of perpendicular



Let us now assume the case where L is given in the general form, i.e. 0.L ax by c≡ + + =

We can easily adjust the equation of L so that c is negative. We do this so that we can convert L into thenormal form:

0ax by c+ + = c < 0ax by c⇒ + = −

2 2 2 2 2 2

a b cx ya b a b a b

−⇒ + = + + +

cos sinx y p⇒ α + α = ... (1)

where 2 2 2 2 2cos , sin anda b cp

a b a b a b2

−α = α = =+ + +

The equation in (1) is in the normal form; we can now use the result obtained in the preceding discussionto obtain the length of the perpendicular PQ:

1 1cos sinPQ x y p= α + α −Modulus sign isused since must be +ve

PQ

1 1

2 2 2 2 2 2

ax by ca b a b a b

= + ++ + +

1 1

2 2

ax by cPQ

a b

+ +=

+ : Length of perpendicular

Find the distance between two parallel lines 1 2L and L given by 1 1 0L ax by c≡ + + = and 2 2 0L ax by c≡ + + =

Solution:

L1

L2

P(x , y )1 1

d

Assume any point P on the line L ; we are required to find d

1

Fig - 20Since P lies on L1, we have

1 1 1 0ax by c+ + =

1 1 1ax by c⇒ + = −

1 1 2 2 1ax by c c c⇒ + + = − ... (1)

Example – 7



By the previous article, the length of the perpendicular dropped from P upon the line L2 is

1 1 2

2 2

ax by cd

a b

+ +=

+

1 2

2 2

c c

a b

−=

+(By (1) above)

This is the required distance between the two lines

If p and p' be the perpendiculars from the origin upon the lines 1 sec cosec 0L x y a≡ θ+ θ − = and

2 cos sin cos 2 0,L x y a≡ θ − θ − θ = show that 2 2 24 'p p a+ =

Solution: The length of the perpendicular dropped from (0, 0) to L1 is by Art - 7

2 2sec cosec

ap =

θ + θ

sin cosa= θ θ ... (1)Similarly, p', the length of the perpendicular from (0, 0) to L2 is

2 2

cos 2'

cos sin

ap

θ=

θ + θ

cos 2a= θ ... (2)We now have from (1) and (2)

2 '2 2 2 2 24 4 sin cos 2p p a a+ = θ + θ 2 2 2 2sin 2 cos 2a a= θ + θ 2a=

____________________________________________________________________________________

Art 8 Concurrency

Consider three different straight lines L1, L2 and L3:

1 1 1 1 0L a x b y c≡ + + = ... (1)

2 2 2 2 0L a x b y c≡ + + = ... (2)

3 3 3 3 0L a x b y c≡ + + = ... (3)

We need to evaluate the constraint on the coefficients , andi i ia s b s c s' ' ' such that the three lines areconcurrent.

Let us first determine the point P of intersection of L1 and L2. By Art - 5, it will be

1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

,b c b c c a c aPa b a b a b a b

− −≡− −

Example – 8



Thus three lines will be concurrent if L3 passes through P too, that is P satisfies the equation of L3. Thus,

1 2 2 1 1 2 2 13 3 3

1 2 2 1 1 2 2 1

0b c b c c a c aa b ca b a b a b a b

− −+ + = − −

( ) ( ) ( )3 1 2 2 1 3 1 2 2 1 3 1 2 2 1 0a b c b c b c a c a c a b a b⇒ − + − + − =

( ) ( ) ( )1 2 3 3 2 1 2 3 3 2 1 2 3 3 2 0a b c b c b c a c a c a b a b⇒ − + − + − =

This relation can be written compactly in determinant form as

1 1 1

2 2 2

3 3 3

0a b ca b ca b c

=

This is the condition that must be satisfied for the three lines to be concurrent.

For example, consider the three lines 2 3 5 0,x y− + = 3 4 7 0x y+ − = and 9 5 8 0x y− + = . These threelines are concurrent because the determinant of the coefficients is 0, i.e,

2 3 53 4 7 09 5 8

−− =

−

Prove that the three lines L1, L2 and L3 whose equations have been mentioned in the preceeding discussion, areconcurrent if we can find three constants 1 2 3, andλ λ λ such that

1 1 2 2 3 3 0L L Lλ + λ + λ =

Solution: Assume that L1 and L2 intersect at the point P whose co-ordinates are (x0, y0) P should satisfy theequations of both L1 and L2.

( )1 1 0 1 0 1at 0L P a x b y c≡ + + = ... (1)

( )2 2 0 2 0 2at 0L P a x b y c≡ + + = ... (2)

Now assume that we can find three non-zero constants 1 2 3, andλ λ λ such that

1 1 2 2 3 3 0L L Lλ + λ + λ = . We will prove that due to this condition, L3 will definitely have to passthrough P:

1 1 2 2 3 3 0L L Lλ + λ + λ =

1 23 1 2

3 3

L L L λ λ⇒ = − + − λ λ

If we evaluate the value of L3 at P, we get

( ) ( ) ( )1 23 1 2

3 3

at at atL P L P L P λ λ= − × + − × λ λ

1 2

3 3

0 0 λ λ= − × + − × λ λ

( )( )

By 1

and 2

= 0Since the value of L3 is 0 at P, the line L3 must pass through P. Thus, L1, L2 and L3 are concurrent.

Example – 9



Show that the medians of a triangle are concurrent.

Solution: Let the triangle have the vertices ( )1 1,A x y , ( ) ( )2 2 3 3, and ,B x y C x y as in the figure below:

A(x , y )1 1

B(x , y )2 2

C(x , y )3 3

Let D, E, F be the mid-pointsof BC, CA and AB respectively.The co-ordinates of any mid point can easily be evaluated by the section formula. For eg, the co-ordinates of D are

2 3 2 3,2 2

x x y y+ +

Fig - 21

E

D

F

From the two-point form of the equation of a line, we can write down the equations of AD, BE andCF.

The equation L1 of the median AD is:

2 31

11

2 311

2

2

y yyy yL x xx x x

+−−≡ = +− −

( )( ) ( )( ) ( )( )1 1 2 3 1 2 3 1 1 2 32 2 2L y y y x x x x y x y y y⇒ ≡ − + − − + = − + ( )( )1 1 2 32y x x x− − +

By symmetry, we can write down the corresponding equations L2 and L3 of the medians BE and CF.

Observe carefully that when we subsequently add the three equations L1, L2 and L3, their left handsides sum to 0. Thus, we have found three constants 1, 1 and 1 such that

1 2 31 1 1 0L L L⋅ + ⋅ + ⋅ =

1 2 3, andL L L⇒ are concurrent

⇒ The medians of any triangle are concurrent.

Show that the equation of any line passing through the intersection point P of two given lines whose equationsare L1 and L2, can be expressed as 1 2 0, whereL L+ λ = λ is a real parameter.

Solution: Let 1 1 1 1 0L a x b y c≡ + + = and 2 2 2 2 0L a x b y c≡ + + =

Consider the equation 1 2 0L L+ λ = ... (1)

( )1 1 1 2 2 2 0a x b y c a x b y c⇒ + + + λ + + =

( ) ( ) ( )1 2 1 2 1 2 0a a x b b y c c⇒ + λ + + λ + + λ =

Example – 10

Example – 11



This is definitely the equation of a straight line because it is of the form 0.ax by c+ + = Also, notice inaddition that the intersection point P will satisfy this equation, because if we substitute the co-ordinatesof the intersection point P in (1), both L1 and L2 vanish.

Thus, 1 2 0L L+ λ = is the equation of an arbitrary straight line that passes through the intersectionpoint P of L1 and L2. (As we vary λ , the slope of this line will vary but it will always pass through P).

L + L = 01 2λ

P

L = 02

L = 01

The equation of linepassing through P can bewritten as L + L = 0where is a real parameter

any

1 2λ λ

Fig - 22

This result is very beneficial in certain cases. We’ll see such cases in some subsequent examples

Find the equations to the straight lines passing through

(a) (3, 2) and the point of intersection of 2 3 1x y+ = and 3 4 6x y− =

(b) Origin and the point of intersection of 1 and 1x y x ya b b a

+ = + = .

Solution: (a) The equations of the two given lines in standard form are :

1 2 3 1 0L x y≡ + − =

2 3 4 6 0L x y≡ − − =

Any line passing through the intersection point of L1 and L2 is

1 2 0L Lλ+ =

⇒ (2 3 1) (3 4 6) 0x y x yλ+ − + − − =

⇒ (2 3 ) (3 4 ) (1 6 ) 0x yλ λ λ+ + − − + = ... (1)

We want this line to pass through (3, 2). Therefore (3, 2) must satisfy the equation of this line, i.e.

(2 3 )3 (3 4 )2 (1 6 ) 0λ λ λ+ + − − + =

⇒ 5 11 0λ− + =

⇒115

λ =

Example – 12



We substitute 115

λ = in (1) to get the required equation:

11 11 11(2 2 ) (3 4 ) (1 6 ) 05 5 5

x y+ ⋅ + − ⋅ − + ⋅ =

⇒ 43 29 71 0x y− − =

(b) We follow the same procedure as in part (a)

1 : 0L bx ay ab+ − =

2 : 0L ax by ab+ − =

The equation of any line passing through the intersection point of L1 and L2 is

1 2 0L Lλ+ =

⇒ ( ) ( ) (1 ) 0b a x a b y abλ λ λ+ + + − + = ... (2)

Since this line must pass through (0, 0), we substitute (0, 0) into (2) to get

(1 ) 0ab λ+ =

⇒ 1λ = −

We substitute 1λ = − into (2) to get the required equation :

( ) ( ) 0b a x a b y− + − =

⇒ 0x y− =

____________________________________________________________________________________

Art 9 Angle Bisectors

Consider two straight lines L1 and L2 with the equations

1 1 1 1: 0L a x b y c+ + =

2 2 2 2: 0L a x b y c+ + =

We intend to find the angle bisector formed at the intersection point P of L1 and L2. Note that there will betwo such angle bisectors

A2

y

xL2

A2L1

We denote the two angle bisectors by A and A1 2

Fig - 23

P



To write down the equations of the two angle bisectors, we first modify the equations of L1 and L2 so thatc1 and c2 are say, both negative in sign. This can always be done. Why this is done will soon become clear.

We first write down the equation of A1, the angle bisector of the angle in which the origin lies.

By virtue of being an angle bisector, if any point ( , )P x y′ ′ lies on A1, the distance of P from L1 and L2 mustbe equal. Using the perpendicular distance formula of Art -7, we have

⇒1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y ca b a b

′ ′ ′ ′+ + + +=+ +

⇒1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y ca b a b

′ ′ ′ ′+ + + += ±+ + ...(1)

Which sign should we select, “+” or “–”, for the bisector of the angle containing the origin?

Since P and origin lie on the same side of L1, 1 1 1a x b y c′ ′+ + and c1 must be of the same sign by Art - 6.

Similarly, 2 2 2a x b y c′ ′+ + and c2 must be of the same sign. But since we have already arranged c1 and c2

to be of the same sign (both negative), we must have ( )1 1 1a x b y c′ ′+ + and ( )2 2 2a x b y c′ ′+ + also of thesame sign.

Thus, it follows from (1) that to write the equation of the angle bisector of the angle containing the origin,

we must select the “+” sign since ( )1 1 1a x b y c′ ′+ + and ( )2 2 2a x b y c′ ′+ + are of the same sign. The “–”sign gives the angle bisector of the angle not containing the origin, i.e., the equation of A2.

To summarize, we first arrange the equations of L1 and L2 so that c1 and c2 are both of the same sign.Subsequently, using the property of any angle bisector, we obtain

1 1 1 2 2 22 2 2 2

1 1 2 2

: a x b y c a x b y c

a b a b+ + + += +

+ +Angle bisector of anglecontaning the origin

and

1 1 1 2 2 22 2 2 2

1 1 2 2

: a x b y c a x b y c

a b a b+ + + += −

+ +Angle bisector of angle notcontaning the origin

Find the angle bisector of the angle between the straight lines 1 2: 3 4 7 0 and :12 5 8 0L x y L x y− + = − − = whichcontains the origin.

Example – 13



Solution: Following the discussion of the preceeding article, we first modify the equations L1 and L2 so that theconstant terms in both the equations are of the same sign (say both positive):

1 : 3 4 7 0L x y− + =

2 : 12 5 8 0L x y− + + =

The angle bisector of the angle containing the origin is

2 2 2 2

(3 4 7) ( 12 5 8)3 4 12 5

x y x y− + − + += ++ +

⇒ 99 77 51x y− + = 0

Evaluating the other angle bisector is left to the reader as an exercise.

Find the bisector of the angle between the lines 2 11 0x y+ − = and 3 6 5 0x y− − = which contains thepoint (1, –3).

Solution: Again we first arrange the equations of the two lines such that constant terms are positive

1 : 2 11 0L x y− − + =

2 : 3 6 5 0L x y− + + =

Note that

1 1(at origin) 0 and (at(1, 3)) 0L L> − >

⇒ Origin and (1, –3) are on the same side of L1.

2 2(at origin) 0 and (at(1, 3)) 0L L> − <

⇒ Origin and (1, –3) are on the opposite sides of L2.

This means that the point (1, –3) does not lie in the same region as the origin, since (1, –3) must be onthe opposite side of the origin with respect to L2.The example figure below will make this clear:

L1

L2

(1, -3 )

Origin

Fig - 24

We see that (1, -3) lies in the angle not containing the origin

Example – 14



Thus, it is clear that (1, –3) lies in the angle not containing the origin.

L1

L2

Origin

Fig - 25

P (x, y)

There is one point worth mentioning here. If suppose a point P (x, y) lies on the opposite side of the origin with respect to both L and L , it would lie in the vertically opposite angle to the angle in which the origin lies ; in such a case, the angle bisector of the angle containing P is the same as that of the one containing the origin.

1 2

To determine the angle bisector of the angle containing (1, –3), we simply determine the angle bisectorof the angle not containing the origin, i.e.

2 11 3 6 55 3 5

x y x y− − + − + += −

⇒ 6 5 38 5x =

⇒193

x =

Note that to determine the angle bisector of the angle containing the point P as in Fig.-25, we wouldhave chosen the angle bisector of the angle containing the origin.



Art 10 Polar / Distance form of a line

Sometimes, it is very convenient to write the equation of a straight line in polar / distance form.Suppose we know that the line passes through the fixed point ( , )P h k and is at an inclination of :θ

Let PQ = r

y

x

P h, k( )

θ

Fig - 26

r

Q x, y( )

For any point ( , )Q x y at a distance r from P along this line, we can write the simple relation

cos sinx h y k r− −= =

θ θ

This is the required equation of the line. The point ( , ),Q x y at a distance r from P, has the coordinates

( , ) ( cos , sin ).Q x y h r k r≡ + θ + θ

Obviously, there will be another point, say ( , ), at a distance from along this line but on the opposite side of ; thus ( , ) will have the coordinates ( , ) ( cos , sin )

Q x y r PQ Q x y

Q x y h r k r

′ ′ ′ ≡ − θ − θ

A line through ( 5, 4)A − − meets the lines 3 2, 2 4 0x y x y+ = + + = and 5 0x y− − = at the points B, C and Drespectively. If

2 2 215 10 6AB AC AD

+ = ,

find the equation of the line.

Example – 15



Solution:

We want to find the equationof the line .Assume

L

AB = r1

AC = r AD = r

2

3

x - y - 5 = 0

Fig - 27

x + y 3 = 2

2 4 = 0x + y + D

A(-5,-4)

C

B

L = 0

The figure above roughly sketches the situation described in the equation. Let B, C and D be atdistances 1 2,r r and 3r from A along the line 0,L = whose equation we wish to determine. Assume theinclination of L to be .θ Thus, B, C and D have the coordinates (respectively):

1 1( 5 cos , 4 sin )B r r≡ − + θ − + θ

2 2( 5 cos , 4 sin )C r r≡ − + θ − + θ

3 3( 5 cos , 4 sin )D r r≡ − + θ − + θ

Since these three points(respectively) satisfy the three given equations, we have :

Point B : 1 1 115( 5 cos ) 3( 4 sin ) 2 0

cos 3sinr r r− + θ + − + θ + = ⇒ =

θ + θ

Point C : 2 2 2102( 5 cos ) ( 4 sin ) 4 0

2cos sinr r r− + θ + − + θ + = ⇒ =

θ + θ

Point D : 3 3 36( 5 cos ) ( 4 sin ) 5 0

cos sinr r r− + θ − − + θ − = ⇒ =

θ − θIt is given that

2 2 215 10 6AB AC AD

+ =

i.e.22 2

1 2 3

15 10 6r r r

+ =

2 2 2(cos 3sin ) (2cos sin ) (cos sin )⇒ θ + θ + θ + θ = θ − θ

2 24cos 9sin 12sin cos 0⇒ θ + θ + θ θ =

2(2cos 3sin ) 0⇒ θ + θ =

2tan3

−⇒ θ =

23

m −⇒ =



Thus, we obtain the slope of L as 2 .

3−

The equation of L can now be easily written :

2: ( 4) ( ( 5))3

L y x−− − = − −

: 2 3 22 0L x y⇒ + + =



1. A variable straight line drawn through the intersection of the lines 1x ya b

+ = and 1x yb a

+ = meets the

axes in A and B. Show that the locus of the mid-point of AB is 2 ( ) ( )xy a b ab x y+ = +

2. The line bx ay ab+ = cuts the axes in A and B. Another variable line cuts the axes in C and D suchthat ,OA OB OC OD+ = + where O is the origin. Prove that the locus of the point of intersection ofthe lines AD and BC is the line .x y a b+ = +

3. A point P moves so that the square of its distance from (3, –2) is equal to its distance from the line5 12 13.x y− = Find the locus of P.

4. A line intersects the x-axis in A(7, 0) and the y-axis in B(0, –5). A variable line perpendicular to ABintersects the x-axis in P and the y-axis in Q. If AQ and BP intersect in R, find the locus of R.

5. If the sum of the distances of a point from two perpendicular lines in a plane is 1, prove that its locusis a square.

6. A vertex of an equilateral triangle is (2, 3) and the opposite side is 2.x y+ = Find the equations of theother sides.

7. A ray of light along the line 2 3 0x y− − = is incident upon the mirror-line 3 2 5 0.x y− − = Find theequation of the reflected ray.

8. If the vertices of a triangle have integral coordinates, show that it cannot be equilateral.9. Show using coordinate geometry that the angle bisectors of the sides of a triangle are concurrent.

10. The sides of a triangle are 4 3 7 0, 5 12 27 0x y x y+ + = + − = and 3 4 8 0.x y+ + = By explicitlyevaluating the medians in this triangle, show that they are concurrent.

11. A rod APB of constant length meets the axes in A and B. If AP = b and PB = a and the rod slidesbetween the axes, show that the locus of P is 2 2 2 2 2 2b x a y a b+ =

12. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and

b, show that 2 2 2

1 1 1 .p a b

= +

13. The lines 3 4 8 0x y+ − = and 5 12 3 0x y+ + = intersect in A. Find the equations of the lines passingthrough (3, 4),P which intersect the given lines at B and C, such that .AB AC=

14. The equal sides AB and AC of an isosceles triangle ABC are produced to the points P and Q such that2.BP CQ AB⋅ = Prove that the line PQ always passes through a fixed point.

15. One side of a square is inclined to the x-axis at an angle α and one of its extremities is at the origin;prove that the equations to its diagonals are

(cos sin ) (sin cos )y xα − α = α + α

and (sin cos ) (cos sin )y x aα + α + α − α =

where a is the length of the side of the sqaure.

TRY YOURSELF - I



Consider two lines

1 1 1 0L y m x c≡ − − =

2 2 2 0L y m x c≡ − − =

What do you think will 1 2 0L L = represent ? It is obvious that any point lying on L1 and L2 will satisfy 1 2 0,L L =

and thus 1 2 0L L = represents the set of points constituting both the lines, i.e.,

1 2 1 20 represents the pair of straight lines given by 0 and 0L L L L= = =

For example, consider the equation 2 2 0.y x− = What does this represent ? We have

2 2 0y x− = ...(1)

( )( ) 0y x y x⇒ + − =

⇒ (1) represents the pair of straight lines x y= and 0.x y+ =

y

x

x = y

The pair of lines given by 0 y - x = 2 2

x + y = 0

Fig - 28

There is nothing special about considering a pair. We can similarly define the joint equation of n straight lines0 ( 1,2..., )i i iL y m x c i n≡ − − = = as

1 2... 0nL L L =

1 1 2 2( )( )...( ) 0n ny m x c y m x c y m x c⇒ − − − − − − = ...(2)

Any point lying on any of these n straight lines will satisfy (2), and thus (2) represents the set of all points constitutingthe n lines, i.e. (2) represents the joint equation of the n straight lines.

What is relevant to us at this stage is only a pair of straight lines and it is on a pair of lines that we now focus ourattention.

Section - 3 PAIR OF STRAIGHT LINES



PAIR OF LINES PASSING THROUGH THE ORIGIN

We first consider a special (and simple)case. Both the lines in our pair pass through the origin. Thus, their equationscan be written as

1 1: 0L y m x− =

2 2: 0L y m x− =

y

L1Pair of lines passing through the origin with slopes

tan m = 1 1

θm = 2 2tan θ

Fig - 29

θ1

θ2

θ

L2

O x

The joint equation of this pair is :

1 2 0L L =

1 2( )( ) 0y m x y m x⇒ − − =

2 21 2 1 2( ) 0y m m x m m xy⇒ + − + = ...(3)

(3) suggests that the general equation of a pair of straight lines passing through the origin is

2 22 0ax hxy by+ + = ...(4)

(4) is a homogenous equation of degree 2, implying that the degree of each term is 2.

It should now be apparent that any homogenous equation of degree 2 will represent two straight lines passingthrough the origin (we’ll soon see that the two straight lines might be imaginary; the meaning of this will becomeclear in a subsequent example).

Generalising, any nth degree homogenous equation of the form

1 2 20 1 2 ... 0n n n n

na x a x y a x y a y− −+ + + + = ...(5)

represents n straight lines (real or imaginary) passing through the origin. To obtain the slopes of these n lines, we

divide by nx in (5) and substitute :y mx

=

11 0... 0n n

n na m a m a−−+ + + =

The n values of m gives us the slopes of the n lines.



Find the straight lines represented by

(a) 2 25 6 0y xy x− + =

(b) 2 23 10 3 0y xy x− + =

(c) 2 2 0y xy x+ + =

Solution: Note that the homogenous nature of these equations tells us that the lines will pass through the origin.

(a) 2 25 6 0y xy x− + =

We either factorize this equation straightaway :

( 2 )( 3 ) 0y x y x− − =

so that the lines are

2y x= and 3y x=

OR,

we divide it by 2x and substituteymx

= to obtain :

2 5 6 0m m− + =

2, 3m⇒ =

2, 3yx

⇒ =

2y x⇒ = or 3y x=

Both alternatives are entirely equivalent.

(b) 2 23 10 3 0y xy x− + =

23 10 3 0m m⇒ − + =ymx

=

13,3

m⇒ =

3 ,3xy x y⇒ = =

Example – 16



(c) 2 2 0y xy x+ + =

2 1 0m m⇒ + + = Again, ymx

=

This has no real roots and thus physically, no lines will exist with the joint equation 2 2 0.y xy x+ + =We sometimes say that this equation represents imaginary lines.

Note that in the entire plane, only (0, 0) satisfies this equation.

____________________________________________________________________________________

Consider now that we’ve been given the equation of a pair of straight lines passing through the origin as :2 22 0ax hxy by+ + = ...(1)

We wish to determine the angle between these two lines. Let m1 and m2 be the slopes of these two lines. By

dividing(1) by x2 and substituting ,y mx

= we have

2 2 0bm hm a+ + =

This quadratic in m will have its roots as m1 and m2. Thus,

1 2 1 22 ;h am m m mb b

−+ = = ...(2)

The angle between the two lines, say ,θ is given by

1 2

1 2

tan1m m

m m−θ =

+

2

1 2 1 2

1 2

( ) 41

m m m mm m

+ −=

+

22 h ab

a b−=

+ (Using (2))

As a consequence of this formula, we see that the lines represented by (1), are :

2 (in fact coincident since both pass through the origin) if if 0

h aba b

=+ =

ParallelPerpendicular

The importance of this condition must be mentioned; it is very widely used and should be committed to memory.

As an example, the locus given by2 23 8 3 0y xy x− − = ...(3)

represent two perpendicular straight lines since

(3) ( 3) 0a b+ = + − =

Verify this by the explicit factorization of (3).



Find the equation of the pair of lines through the origin and perpendicular to the pair of lines 2 22 0.ax hxy by+ + =

Solution: Let the slopes of the two lines represented by the given equation be m1 and m2. As explained earlier,m1 and m2 are the roots of the quadratic

2 2 0bm hm a+ + =

so that

1 2 1 22 ,h am m m mb b

−+ = = ...(1)

The slopes of the lines whose joint equation we require will simply be 1

1m−

and 2

1m−

so that their

equations will be :

1 2

1 1,y x y xm m− −= =

1 20, 0x m y x m y⇒ + = + =

The required joint equation is

1 2( )( ) 0x m y x m y+ + =

2 21 2 1 2( ) 0x m m y m m xy⇒ + + + =

2 2 2 0a hx y xyb b

⇒ + − = (Using (1))

2 22 0bx hxy ay⇒ − + =

he equation 3 2 2 3 0ax bx y cx dy+ + + = is a third degree homogenous equation and hence represents three straightlines passing through the origin. Find the condition so that two of these three lines may be perpendicular.

Solution: We divide the given equation by 3x and substitute y mx

= to obtain:

3 2 0dm cm bm a+ + + = ...(1)

This has three roots, say 1 2 3, , ,m m m corresponding to the three straight lines. Since we want two ofthese lines to be perpendicular, we can assume

1 2 1m m = −

Example – 17

Example – 18



From (1), we have

1 2 3am m m

d−=

3amd

⇒ =

Substituting this value of m3 back in (1), (since m3 is a root of (1)), we obtain

3 2

3 2 0da ca ba ad d d

+ + + =

2 2 0a ac bd d⇒ + + + =

Find the area of the triangle formed by the lines 2 29 18 0y xy x− + = and 9.y =

Solution: The joint equation can be factorized to obtained

( 3 )( 6 ) 0y x y x− − =

Thus, the three lines forming the sides of the triangle are

3 , 6 , 9y x y x y= = =

The three intersection points can easily be seen to be

3(0, 0), (3, 9), ,92

Thus, the area of the triangle is

30 321 0 9 9

21 1 1

∆ =

27 sq. units4

=

The slope of one of the two lines represented by 2 22 0ax hxy by+ + = is the square of the other. Prove that

28 6a b hh ab+ + =

Example – 19

Example – 20



Solution: Let the two slopes be m and m2; these are the roots of the quadratic

2 2 0bM hM a+ = =

so that

2 2 ,hm mb

−+ = 3 amb

=

Cubing the first relation, we have

33 6 3 2

3

83 ( ) hm m m m mb

−+ + + =

2 3

2 3

3 2 8a a a h hb b b b b

− − ⇒ + + =

2 2 38 6ba ab h hab⇒ + + =

28 6a b hh ab+⇒ + =

_____________________________________________________________________________________

GENERAL EQUATION OF A PAIR OF LINES

Consider the equations of two arbitrary lines L1 and L2:

1 1 1 1: 0L l x m y n+ + =

2 2 2 2: 0L l x m y n+ + =

The joint equation of the two lines is

1 2 0L L =

1 1 1 2 2 2( )( ) 0l x m y n l x m y n⇒ + + + + =

2 21 2 1 2 1 2 1 2 1 2 2 1 1 2 2 1 1 2( ) ( ) ( ) ( ) ( ) 0l l x l m m l xy m m y n l n l x n m n m y n n⇒ + + + + + + + + = ...(1)

(1) suggests that the most general equation to a pair of straight lines has the form

2 22 2 2 0ax hxy by gx fy c+ + + + + = ...(2)

It might be apparent to you that (2) will not always represent a pair of straight lines. For (2) to indeed represent apair of straight lines, it must be able to be factorised into two linear factors; as an exercise for the reader, show that(2) can be expressed as a product of linear factors if the following condition is satisfied:

2 2 22 0abc fgh af bg ch+ − − − =

0a h gh b fg f c

⇒ =



We now re-evaluate the conditions for parallel and perpendicular lines, in the general case :

Let L1 and L2 be two lines with slopes s1 and s2; their equations have already been mentioned above. L1 and L2 areparallel if

1 2s s=

1 2

1 2

l lm m− −⇒ =

1 2 2 1l m l m⇒ =

( )21 2 2 1 0l m l m⇒ − =

( )21 2 2 1 1 2 1 24l m l m l l m m⇒ + = ...(3)

From (2), observe that the constraint in (3) can be specified as

24 4h ab=

2h ab⇒ = : Parallel lines

This is the same condition as the one for the homogenous case.

For L1 and L2 to be perpendicular,

1 2 1s s⋅ = −

1 2

1 2

1l lm m− −⇒ ⋅ = −

1 2 1 2 0l l m m⇒ + =

0a b⇒ + = : Perpendicular lines

Again, this condition is the same as the one in the homogenous case.

If fact, you can verify that the angle θ subtended between the two lines is also given by the same formula as in thehomogenous case, i.e.,

22tan h aba b

−θ =+

That these formulae in the homogenous and the general case are the same should be obvious since the slope of anyline is independent of the constant term appearing in its equation.

Prove that the equation 2 26 13 6 8 7 2 0x xy y x y+ + + + + = represents a pair of straight lines. Find the point ofintersection and the angle between these two lines.

Example – 21



Solution: To show that this equation represents a pair of straight lines, we use the determinant condition mentionedearlier:

136 42

13 762 2

74 22

a h gh b fg f c

=

( )49 13 916 12 14 13 4 244 2 4

= − + − + −

3 13 52 2

= − + −

= 0

which confirms the stated assertion.

The angle between these two lines is given by

22tan h aba b

−θ =+

5

12=

1 5tan

12− ⇒ θ =

To find the point of intersection, we must factorise the joint equation to obtain the separate equationsof the lines. This task can be made easy be observing that since the homogenous part of the givenequation is

2 26 13 6 0x xy y+ + =

which can be factorised as

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

the actual factors of the (original) equation will be of the form

(2 3 ) (3 2 ) 0x y x y+ + α + +β =

Convince yourself about this argument. α and β can easily be evaluated using comparison of coefficientsto be 2 and 1 respectively. Thus, the two lines are

1 : 2 3 2 0L x y+ + =

2 : 3 2 1 0L x y+ + =

so that their point of intersection is, by solving this system of equations, 1 4, .5 5

−



Show that the four lines given by the equations

2 23 8 3 0x xy y+ − =

2 23 8 3 2 4 1 0x xy y x y+ − + − − =

form a square. What is the length of the sides of the square ?

Solution: The first joint equation can be easily factorised to yield

(3 )( 3 ) 0x y x y− + =

3 0, 3 0x y x y⇒ − = + = ...(1)

These are perpendicular lines intersecting at the origin. As described in the previous example, thesecond joint equation can be factorised as

(3 ) ( 3 ) 0x y x y− + α + +β =

where α and β can be determined by the comparison of coefficients :

1, 1α = − β =

The other two sides are thus

3 1 0, 3 1 0x y x y− − = + + = ...(2)

From (1) and (2), it should be evident that the four lines form a square. The length l of the sides of thissquare can be evaluated by determining the perpendicular distance between any pair of oppositesides, say 3 0x y− = and 3 1 0 :x y− − =

2 2

0 ( 1) 1101 3

l− −

= =+

Find the joint equation of the angle bisectors of the lines represented by 2 22 0.ax hxy by+ + =

Solution: Let the slopes of the two lines represented by 2 22 0ax hxy by+ + = be m1 and m2, so that m1 and m2

are the roots of the quadratic

2 2 0bm hm a+ + =

Thus,

1 2 1 22 ,h am m m mb b

−+ = =

Example – 22

Example – 23



It should be obvious that the angle bisectors will also pass through the origin, as shown below:

A2

y

x

Fig - 30

A1

L1

L2

φ1θ1

θ2φ2

L1 and are the original pair of lines while and are their angle bisectors = tan

= tan

LA A

mm

2

1 2

1 1

2 2

θθ

From the figure, observe that

1 2 1 21 2,

2 2θ + θ π θ + θφ = φ = +

2

1 1 2 2 1 22 , 2⇒ φ = θ + θ φ = π+ θ + θ

1 2 1 2tan 2 tan 2 tan( )⇒ φ = φ = θ + θ

1 2

1 2

tan tan1 tan tan

θ + θ=− θ θ

1 2

1 21m m

m m+=

−

2h

a b=

− ... (1)

Now, for any point ( , )x y on the angle bisector 1 2(or ),A A

1tan yx

φ = 2or tan yx

φ =

2 2

2tan 2 xyx y

⇒ φ =− 1 2

We have used to represent both and

φ φ φ

From this relation and (1), we have

2 2x y xya b h

− =−



This is the joint equation of the angle bisectors; as expected, it is a second degree homogenousequation.

As a corollary, suppose we are required to find the joint equation of the angle bisectors of the linesL1 and L2 represented by

2 22 2 2 0ax hxy by gx fy c+ + + + + =

We first find the point of intersection of L1 and L2, say ( , ).P α β If we now shift our coordinate system(translation) so that P is the origin (Refer to Appendix - 1), and denote the coordinates in the newsystem by ( , ),X Y we will have the joint equation of the angle bisectors of L1 and L2 as

2 2X Y XYa b h

− =−

But since , ,X x Y y→ − α → −β the joint equation in the original frame is

2 2( ) ( ) ( )( )x y x ya b h

− α − −β − α −β=−

____________________________________________________________________________________

We now discuss a very useful application of the concept of pair of straight lines.

Consider a second degree curve ( , )S x y with the equation

2 2( , ) 2 2 2 0S x y ax hxy by gx fy c≡ + + + + + =and a straight line

0L px qy r≡ + + =intersecting 0S = in A and B. Let O be the origin.

What is the joint equation of OA and OB ?y

x

Fig - 31

B

A

What is the joint equation of the pair of straight lines

and ?

OA OB

OL = 0

S x, y =( ) 0



The insight that we use here is that since both OA and OB pass through the origin, their joint equation will behomogenous. We now construct a homogenous equation and show that both A and B satisfy it; that equation isthen guaranteed to jointly represent OA and OB.

First of all, observe that since A and B satisfy the equation of L, i.e. 0,px qy r+ + = they will also satisfy therelation

1px qyr

+ =− : Both A and B will satisfy this relation.

Now, we homogenize the equation of the second degree curve ( , )S x y using the relation above; consider thisequation :

22 22 2 2 0px qy px qy px qyax hxy by gx fy c

r r r+ + + + + + + + = − − −

...(1)

Can you understand why we’ve done this? The equation we obtain above is a second degree homogenous equation,and so it must represent two straight lines passing through the origin. Which two straight lines? Since A and B

satisfy the equation of the original curve as well as the relation 1,px qyr

+ =− A and B both satisfy the homogenized

equation in (1).

What does this imply ? That (1) is the joint equation of OA and OB!

Go over this discussion again if you find this confusing. You must fully understand the described technique whichwill find very wide usage in subsequent chapters.

Find the joint equation of the straight lines passing through the origin O and the points of intersection of the line3 4 5 0x y+ − = and the curve 2 22 3 5.x y+ =

Solution: One approach is of course to explicitly determine the two points of intersection, say A and B, writingthe equations of OA and OB, thereby obtaining the required joint equation. You are urged to do this asan exercise.

However, we’ll use the homogenizing technique just described :

Fig - 32

y

x

2 3 = 5x + y2 2

B

A We wish to determine the joint equation of

andOA OBO

Example – 24



The required equation can be obtained by first writing the line as

3 4 15

x y+ =

and then using this relation to homogenize the equation of the curve :

22 2 3 42 3 5

5x yx y + + =

2 2 2 210 15 9 16 24x y x y xy⇒ + = + +

2 224 0x xy y⇒ − − =

This is the required joint equation of OA and OB.

Find the value of m, if the lines joining the origin O to the points of intersection A, B of 1y mx= + and 2 2 1x y+ =are perpendicular.

Solution: The joint equation of OA and OB is

2 2 2( )x y y mx+ = −

2 2(1 ) 2 0m x mxy⇒ − + = ... (1)

The condition for perpendicularity is

0a b+ =

which when applied to (1) yields

21 0m− =

1m = ±

This example was more or less trivial and a little knowledge of circles would have enabled you to solve thisquestion without resorting to the homogenizing approach; however, the fact that this approach is very powerful inmany cases will become apparent in later examples.

Example – 25



Q. 1 Find the values(s) of m for which the following equation(s) represents a pair of straight lines:

(a) 2 22 3 1 0x xy y y+ λ − + − =

(b) 2 24 10 5 10 0x xy y x y+ + λ + + =

Q. 2 Find the angle of intersection of the straight lines given by the equation

2 23 7 2 9 2 12 0x xy y x y− + + + − =

Q. 3 Show that the lines joining the origin to the points common to 2 2 0x hxy y gx fy+ − + + = and

fx gy− = λ are at right angles for all values of .λ

Q. 4 Find the angle between the lines given by 2 22 0.x pxy y− + =

Q. 5 Prove that the lines joining the origin to the points of intersection of the line 2x yh k

+ = with the curve

2 2 2( ) ( ) ,x h y k c− + − = are perpendicular if 2 2 2.h k c+ =

Q. 6 Find the joint equations of the straight lines passing through (1, 1) and parallel to the lines given by2 25 4 2 2 0.x xy y x y− + + + − =

Q. 7 Evaluate the point of intersection for the lines represented by the general equation2 22 2 2 0.ax hxy by gx fy c+ + + + + =

Q. 8 Find the joint equation of the images of the pair of lines 2 22 0ax hxy by+ + = in the mirror 0.y =

Q. 9 Find the joint equation of the angle bisectors of the lines given by 2 22 sec 0.x xy y+ θ + =

Q. 10 If the pairs of straight lines

2 22 0ax hxy by+ + =

2 22 0a x h xy b y′ ′ ′+ + =

have a line in common, show that

2

( )( )2

ab a b ha h a h b hb′ ′− ′ ′ ′ ′= − −

TRY YOURSELF - II



Consider a fixed point O and n fixed straight lines. Through O, a (variable) line is drawn intersecting the fixed linesin 1 2, ....., .nP P P On this variable line, a point P is taken such that

1 2

1 1 1..... .n

nOP OP OP OP

= + + +

Find the locus of P.

Solution:

PnP3P2P1

L1 L2 L3 Ln

Fig - 33

A figure illustratingthe situation describedO ...

Assume the equations of the fixed lines to be

0, 1, 2,.....i i i iL a x b y c i n≡ + + = =

and the coordinates of the fixed point O to be ( , ).h k

Let the slope of the variable line be represented by tan .θ Thus, the points iP have the coordinates

( cos , sin ) 1,2.....i i iP h OP k OP i n≡ + θ + θ =

Since each iP satisfies ,iL we have

( cos ) ( sin ) 0i i i i ia h OP b k OP c+ θ + + θ + =

( )cos sin

i i ii

i i

ha kb cOPa b− + +⇒ =

θ + θ

Assume the coordinates of P (whose locus we wish to determine) to be ( , ).x y Thus, we have

cos , sinx h OP y k OP= + θ = + θ ...(1)

SOLVED EXAMPLES

Example – 1



From the relation given in the equation, we have

1

i

nNP OP

= ∑

cos sini i

i i i

a bnNP ha kb c

θ + θ⇒ = −+ +∑

cos sini i

i i i i i i

a bha kb c ha kb c

= − θ + − θ + + + +

∑ ∑

cos sin= λ θ +µ θThese substitutions havebeen doen for convenience

From (1), we have

( ) ( )n x h y kOP OP OP

λ − µ −= +

( ) 0x y h k n⇒ λ + µ − λ + µ + =

This is the locus of the point P; it is evidently a straight line.

Lines are drawn to intersect n concurrent lines at the points 1 2, ....., nA A A such that

1

1 constantn

i iOA=

=∑

where O is the point of concurrency. Show that the variable lines all pass through a fixed point.

Solution: There’s no loss of generality in assuming O to be the origin since we are dealing only with lengths whichare invariant with respect to the choice of the coordinate axes.

A2

L1

Fig - 34

LL

i's are all fixed lines. is the variable line

L2Ln

LA1

O

Example – 2



The inclinations of the fixed lines can be assumed to be iθ so that the points iA have the coordinates

( cos , sin )i i i i iA OA OA≡ θ θ

Let the variable line have the equation

0ax by c+ + =

Since all the 'iA s lie on this line, we have

cos sin 0i i i iaOA bOA cθ + θ + =

cos sinii i

cOAa b

−⇒ =θ + θ ...(1)

According to the condition specified in the question,

1

1 costant = (say)n

i i

KOA=

=∑ ...(2)

Thus, using (1) in (2), we have

1

cos sin =n

i i

i

a b Kc=

θ + θ−∑

1 1cos sin

0

n n

i ii ia b c

K K= =

θ θ + + =

∑ ∑...(3)

(3) shows that the variable line L always passes through the fixed point 1 1cos sin

, .

n n

i ii i

K K= =

θ θ

∑ ∑

Prove that the centroid G of a triangle divides the line joining its circumcentre C and its orthocentre H in theratio 1 : 2.

Example – 3



Solution: To make our task simpler, we choose a coordinate frame in which the triangle’s vertices have coordinatesthat are “easy” to work with. One such choice is shown below.

Fig - 35

y

x

R b,c( )

P a,(- 0) Q a,( 0)

Now, we find G, C and H :

The centroid G GG(x , y ) : 3 3Ga a b bx − + += =

0 03 3G

c cy + += =

,3 3b cG ⇒ ≡

The circumcentre C CC(x , y ) : To find C, we need two perpendicular bisectors ( ).B⊥

The B⊥ of PQ is

0x =

Since the slope of PR is ,cb a+ the equation of the B⊥

of PR is

2 2c b a b ay x

c+ − − = − −

The two Bs⊥ intersect at C:

2 2 2

0,2

b a cCc

− +≡



The orthocentre H HH(x , y ) : The altitude from R onto PQ is simply

x b=

Let find the altitude from Q onto PR:

0 ( )b ay x ac+ − = − −

The two altitudes intersect at

2 2

, b aH bc

−≡ −

The point which divides CH in the ratio 1 : 2 is

2 2 2 2 2

1 221 2 0 , ,

3 3 3 3

b a b a cc cb b c

− − +× − + × × + × ≡

which is the same as G. Thus, G divides CH in the ratio 1 : 2.

Find the area of the parallelogram formed by the lines

1 1 1 1 1 10; 0a x b y c a x b y d+ + = + + =

2 2 2 2 2 20; 0a x b y c a x b y d+ + = + + =

Find also the condition for this parallelogram to be a rhombus.

Solution:

a x + b y + d1 1 1 = 0

ax

+ b

y +

c2

22 =

0

a x + b y + c1 1 1 = 0

ax

+ b

y +

d2

22 =

0

We need to find the area of this parallelogram.

α

Fig - 36

Example – 4



We first consider a little geometry for this parallelogram. Let the parallelogram have sides a and b andlet the perpendicular distances between its opposite sides be p1 and p2:

a

p2

p1

α

b

Fig - 37

Then, the area A of the parallelogram is

1sinA ab ap= α = 1( sin )p b= α∵

21sin

p p= ⋅α 2( sin )p a= α∵

1 2

sinp p=

α ...(1)

Thus, the area of the parallelogram can be expressed using the perpendicular distances between itsopposite sides rather than using the length of the sides. This is good for us since we already know howto evaluate the perpendicular distance between two parallel lines.

1 11 2 2

1 1

c dp

a b

−⇒ =

+ ...(2)

2 22 2 2

2 2

c dp

a b

−⇒ =

+ ...(3)

Also,

1 2

1 2

1 2

1 2

tan1

a ab b

a ab b

− − −

α =+

1 2 2 1

1 2 1 2

a b a ba a b b

−=+



so that

1 2 2 1

2 2 2 21 1 2 2

sin( )( )

a b a b

a b a b

−α =

+ + ...(4)

Using (2), (3) and (4) in (1), we have

1 1 2 2

1 2 2 1

( )( )( )

c d c dAa b a b− −=

−

Now, the parallelogram is a rhombus if its diagonals are perpendicular, which also means that thedistances between its opposite sides are equal, i.e.

1 2p p=

1 1 2 2

2 2 2 21 1 2 2

c d c d

a b a b

− −⇒ =

+ +

A rod AB of length l slides with its end on the coordinate axes. Let O be the origin. The rectangle OAPB iscompleted. Find the locus of the foot of the perpendicular drawn from P onto AB.

Solution:

P

BO

A

y

xθ

Fl

We need to find the locusof as slides between the axes

F AB

Fig - 38

In terms of the variable ,θ A and B, and hence P, have coordinates

(0, sin ), ( cos , 0), ( cos , sin )A l B l P l l≡ θ ≡ θ ≡ θ θ

The slope of PF can be observed to be cot θ so that its equation is

sin cot ( cos )y l x l− θ = θ − θ ...(1)

Example – 5



The equation of AB is

sec cosec x y lθ + θ = ...(2)

The intersection of (1) and (2) gives us the point ( , ) :F h k

3 3cos , sinh l k l= θ = θ

1/ 3 1/ 3

cos , sinh kl l

⇒ θ = θ =

Eliminating ,θ we have

2 / 3 2 / 3 2 / 3h k l+ =

Thus, the locus of F is2/3 2/3 2/3x y l+ =

(a) Consider a line segment AB where 1 1( , )A x y≡ and 2 2( , ).B x y≡ In what ratio does a line

0L ax by c≡ + + = divide AB?

Solution: Let the required ratio be :1λ

λ

1

A x y( , )1 1

B x y( , )2 2

L ax by c= + + = 0

Fig - 40

C

The coordinates of C are (from the internal division formula),

2 1 2 1,1 1

x x y yC λ + λ + ≡ λ + λ +

Since this lies on L, we have

2 1 2 1 01 1

x x y ya b cλ + λ + + + = λ + λ +

2 2 1 1( ) ( ) 0ax by c ax by c⇒ λ + + + + + =

1 1

2 2

ax by cax by c− + +⇒ λ =

+ +

Example – 6



1 1

2 2

( , )( , )

L x yL x y

⇒ λ = −

This is a useful result (as we’ll see from part(b), the next example) and it would be worth memorizingit.

(b) A line intersects BC, CA and AB in ABC∆ at P, Q and R respectively. Show that

1BP CQ ARPC QA RB

⋅ ⋅ = −

Solution:

Fig - 39

PB

x , y( )2 2

QL = 0

C x , y ( )3 3

A x , y ( )1 1

R

Using the result of the last example, we have

1 1

2 2

( , )( , )

AR L x yRB L x y

= − ...(1)

2 2

3 3

( , )( , )

BP L x yPC L x y

= − ...(2)

3 3

1 1

( , )( , )

L x yCQQA L x y

= − ...(3)

From (1), (2) and (3), it should be evident that the assertion stated in the question is valid.

The curves

2 21 1 1 1 1: 2 2 0C a x h xy b y g x+ + + =

2 22 2 2 2 2: 2 2 0C a x h xy b y g x+ + + =

intersect at two points A and B other than the origin. Find the condition for OA and OB to be perpendicular.

Example – 6

Example – 7



Solution: Assume the equation of AB to be .y mx c= + Thus, using the homogenizing technique, we can writethe joint equation of OA and OB:

Homogenizing C1:2 2

1 1 1 12 2 0y mxa x h xy b y g xc

− + + + =

2 21 11 1 1

2 22 0mg g xya x h xy b yc c

⇒ − + + + =

This is the joint equation of OA and OB. OA and OB areperpendicular if

11 1

2 0mga bc

− + =

1 1

12m a bc g

+⇒ = ...(1)

Homogenizing C2: Similarly, we can again evaluate the joint equation of OAand OB by homogenizing the equation of 2 :C

2 22 22 2 2

2 22 0mg g xya x h xy b yc c

− + + + =

The perpendicularity condition gives

22 2

2 0mga bc

− + =

2 2

22m a bc g

+⇒ = ...(2)

From (1) and (2), the necessary required condition is

1 1 2 2

1 2

a b a bg g+ +=

Find the orthocentre of the triangle formed by the lines 2 22 0ax hxy by+ + = and 1.px qy+ =

Example – 8



Solution: The two lines given by the joint equation pass through the origin. Assume their slopes to be m1 and m2so that m1 and m2 are the roots of

2 2 0bm hm a+ + =

1 2 1 22 ,h am m m mb b

⇒ + = − = ...(1)

O

N

M

y = m x1

y = m x2

px +qy = 1

y

x

Fig - 41

To evaluate the orthocentre, we need two altitudes. We take one of them to be the one dropped fromO onto MN.

0qx py− = ...(2)

Let us now find the altitude from M onto ON. The coordinates of M are, by solving 1y m x= and

1px qy+ = simultaneously,

1

1 1

1 , mMp qm p qm

≡ + +

The slope of ON is m2 so that the slope of the altitude through M is 2

1 ;m

− thus, its equation is

1

1 2 1

1 1my xp qm m p qm

−− = − + +

1 2

2 2 1

1( )

x m mym m p qm

+⇒ + =+

1 22

1

1 m mx m yp qm+⇒ + =+ ...(3)



The intersection of (2) and (3) yields the orthocentre ( , ) :h k

1 2

1 2

(1 )( )( )

p m mhp qm p qm

+=+ +

1 2

2 21 2 1 2

(1 )( )p m m

p pq m m q m m+=

+ + +

2 2

( )2

p a bbp hpq aq

+=− +

.qk hp

= (from (1))

2 2

( )2

q a bbp hpq aq

+=− +

Thus, the orthocentre has the coordinates

2 2 2 2

( ) ( ),2 2

p a b q a bbp hpq aq bp hpq aq

+ + − + − +

Show that the equation

3 2 3 2( 3 ) 3 0x xy y x yλ − + − =

represents three straight lines equally inclined to one another.

Solution: Observe that since the equation is homogenous, it will represent three straight lines passing through theorigin. Let the slopes of the three lines be 1 2,m m and 3.m

Thus 1 2,m m and 3m are the roots of the equation

2 3(1 3 ) 3 0m m mλ − + − = where ymx

=

3

2

31 3m m

m−⇒ = λ

−

Example – 9



Since tan ,ymx

= = θ where θ is the inclination of the line, we have

3

2

3 tan tan tan 31 3tan

θ − θλ = = θ− θ

tan 3⇒ θ = λ

13 tann −⇒ θ = π + λ

1tan3

n −π + λ⇒ θ =

Since there are three lines corresponding to the joint equation, we’ll have three corresponding anglesof inclination

11 1

1 2 3tan 2, tan , tan

3 3 3

−− −λ π πθ = θ = + λ θ = + λ

The angles of inclination show that the three lines are equally inclined to one another.

Fig - 42

y

x

π 3

π3

π3

L2

L3

L1

θ1

θ2θ3

Show that all the chords of the curve 2 23 2 4 0x y x y− − + = which subtend a right angle at the origin pass througha fixed point. Find that point.

Example – 10

v



Solution: Let y mx c= + be a chord of the curve which subtends a right angle at the origin. The joint equationof the lines joining the origin to points of intersection of y mx c= + and the curve is

2 23 (4 2 ) 0y mxx y y xc

− − + − =

This represents two perpendicular lines if

2 2Coeff. of Coeff. of 0x y+ =

2 43 1 0mc c

+ + − =

2 0c m⇒ + + =

( 2) (1)m c⇒ − = + ...(1)

(1) shows that y mx c= + always passes through the fixed point (–2, 1).



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT[ LEVEL - I ]

1. Through the origin O, a (variable) line is drawn to cut the lines 1 1y m x c= + and 2 2y m x c= + at Qand R. Let there be a point P on this variable line such that OP is the geometric mean of OQ and OR.Find the locus of P.

2. Find the condition so that the pair of straight lines joining the origin to the points of intersection ofy mx c= + and the circle 2 2 2x y a+ = may be perpendicular.

3. A line drawn through the origin intersects the lines 2 2 0x y+ − = and 2 2 0x y− + = in A and B. Let

M be the mid-point of AB. Show that the locus of M is 2 22 3 2 3 0.x xy y x y− − + + =

4. Show that the reflection of the line 0,ax by c a b+ + = ≠ in the line 1 0x y+ + = is the line( ) 0.bx ay a b c+ + + − =

5. Find the angle between the straight lines given by 2 2 2 2( )sin ( cos sin ) .x y x y+ α = θ − θ

6. Prove that the lines 2 22 6 0x xy y+ + = are equally inclined to the lines 2 24 18 0.x xy y+ + =

7. Let 2 22 2 2 0ax hxy by gx fy c+ + + + + = represent a pair of parallel straight lines. Prove that the

distance between these lines is 2 2

2 2 .( ) ( )

g ac f bcda a b b a b

− −= =+ +

8. If the equation 2 2 2 0hxy gx fy c+ + + = represents two straight lines, show that they form a rectangle

of area 2

fgh

with the coordinate axes.

[ LEVEL - II ]

9. A straight line is such that the algebraic sum of the perpendiculars drawn upon it from any number offixed points is zero. Show that it always passes through a fixed point.

10. Let 1 1 1 0a x b y c+ + = and 2 2 2 0a x b y c+ + = be two lines intersecting at P. Show that

1 2 1 2 0a a b b+ < ⇒ The angle of intersection at P containing the origin is acute.

1 2 1 2 0a a b b+ > ⇒ The angle of intersection at P containing the origin is obtuse.



11. Show that there exists a point equidistant from the four points 1 2 31 2 3

, , , , ,a a aam am amm m m

and 1 2 31 2 3

, .a am m mm m m

12. The vertices of triangle are ( , tan ), 1,2,3.i i ix x iθ = The circumcentre of this triangle is the origin and

its orthocentre is ( , ).a b Show that

1 2 3

1 2 3

cos cos cos .sin sin sin

ab

θ + θ + θ=θ + θ + θ

13. A rectangle PQRS has its side PQ parallel to the line y mx= and the vertices P, Q and S lie on thelines ,y a x b= = and x b= − respectively. Find the locus of R.

14. For points 1 1( , )P x y≡ and 2 2( , ),Q x y≡ the M-distance ( , )d P Q is defined as

1 2 1 2( , )d P Q x x y y= − + −

Let (0, 0)O ≡ and (3, 2).A ≡ Prove that the set of points in the first quadrant which areequi-M-distant from O and A, consists of the union of a line segment of finite length and an infinite ray.Sketch this set.

15. The sides of a triangle are cos sin , 1,2,3.i i i iL x y p i≡ α + α = = Prove that the orthocentre of thistriangle satisfies

1 2 3 2 3 1 3 1 2cos( ) cos( ) cos( ).L L Lα − α = α − α = α − α

16. If 2 22 2 2 0ax hxy by gx fy c+ + + + + = and 2 22 2 2 0,ax hxy by gx fy c+ + − − + = each represent

a pair of straight lines, prove that the area of the parallelogram they enclose is 2

2.

c

h ab−17. Find the area of the triangle formed by the lines given by 2 22 2 2 0ax hxy by gx fy c+ + + + + = and

the x-axis.

18. If 2 22 2 2 0ax hxy by gx fy c+ + + + + = represents two straight lines, prove that the product of the

perpendiculars drawn from the origin to these lines is 2 2

.( ) 4

c

a b h− +

19. A line L through the origin meets 1x y+ = and 3x y+ = at P and Q respectively. Through P and Q,two straight lines L1 and L2 are drawn parallel to 2 5x y− = and 3 5x y+ = respectively. The lines L1

and L2 intersect in R. As L varies, show that the locus of R is a straight line.

20. Let the sides of a parallelogram be , ,U a U b V a′= = = and ,V b′= where U lx my n= + + and.V l x m y n′ ′ ′≡ + + Show that the equation of the diagonal through the intersection points of

,U a V a′= = and ,U b V b′= = is given by

11 01

U Va ab b

′ =′



APPENDIXAPPENDIXAPPENDIXAPPENDIXAPPENDIXTRANSFORMATRANSFORMATRANSFORMATRANSFORMATRANSFORMATION OF COORDINTION OF COORDINTION OF COORDINTION OF COORDINTION OF COORDINAAAAATESTESTESTESTES

Suppose that a person A is flying a kite from the ground while another person B is observing this kite from the topof a building, as shown below:

y

x

Fig - 44

B (H, K)X

Y

A(0,0)

In A's frame of reference, the coordinates of B are ( , ).H K Now suppose that A and B both specify the positionof the kite relative to themselves. It should be evident that the coordinates of the kite in the two reference frameswill be different.

Let the coordinates of the kite be ( , )x y in A's reference frame and ( , )X Y in B's reference frame. Then, we have

,X x H Y y K= − = − ...(1)

Thus, a translation of the axes implies a corresponding change in the coordinates in the manner specified by (1).In fact, if the kite traces a path ( , ) 0f x y = in A's reference frame, it will trace the path ( , ) 0f X H Y K+ + = inB's frame of reference.



Translation of axes implies a simple shift in the origin without a change in the relative orientation of the axes. Wenow consider the case when the axes is rotated but the origin is the same

y

x

Fig - 45

θ

X

AY

Let a point A have the coordinates ( , )x y in the original frame of reference and ( , )X Y in the rotated frame ofreference. Verify that the following relations hold true :

cos sinx X Y= θ− θ

sin cosy X Y= θ + θ

We can now combine the case of translation and rotation of axes to determine the most general transformedcoordinates. Let the origin of the axes be shifted to ( , )h k and the axes be rotated by an anticlockwise angle .θThe original coordinates ( , )x y and the coordinates in the new frame of reference ( , )X Y will satisfy the relations

- cos sinx h X Y= + θ− θ

sin cosy k X Y= + θ + θ

straight lines ok.pdf

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