15. Cartesian Coordinate Geometry and Straight Lines-3

7/31/2019 15. Cartesian Coordinate Geometry and Straight Lines-3

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Cartesian Coordinate Geometry

And

Straight Lines

Session


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Session Objectives


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1. Equations of bisectors of angles betweentwo lines

2. Acute/obtuse angle bisectors3. Position of origin w.r.t bisecors4. Equation of family of lines through

intersection of two lines5. Pair of lines - locus definition6. Pair of lines represented by second

degree equation7. Angle between two lines, represented as a

second degree equation

Session Objectives


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Equation of Bisector

XX

Y

O

Y

A

M

N

P(h,k)

Consider two lines a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0

We are required to find the equations of the bisectors of theangle between them.

The required equations are the equations to the locus of apoint P(h, k)

equidistant from the given lines.


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Equation of Bisector

XX

Y

O

Y

A

M

N

P(h,k)

PM = PN1 1 1 2 2 2

2 2 2 21 1 2 2

a h b k c a h b k c

a b a b

1 1 1 2 2 2

2 2 2 21 1 2 2

a h b k c a h b k c

a b a b

The required equations are 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


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Acute/obtuse Angle Bisectors

Algorithm to determine equations ofbisectors of acute angle and obtuseangle between a pair of lines.

Step I : Rewrite the equations of the lines in general form

a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 such that c1 andc2 are positive.

Step II : Determine sign of expression a1a2+b1b2

Step III : Write the equations of the bisectors

;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

1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y c

a b a b


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Step IV : Case (i) a1a2+b1b2 > 0

1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y c

a 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 c

a b a b

Obtuse anglebisector

Acute anglebisector


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Step IV : Case (i) a1a2+b1b2 < 0

1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y c

a 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 c

a b a b

Acute anglebisector

Obtuse anglebisector


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Illustrative Example

Find the equation of the obtuse

angle bisector of lines 12x-5y+7 = 0and 3y-4x-1 = 0.

Rewrite the equations to make

the constant terms positive,

12x-5y+7 = 0 and

4x-3y+1 = 0

Calculate a1a2+b1b2

12*4+(-5)*(-3) = 63

Solution :


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Solution Cont.

Simplifying,

12x 5y 7 4x 3y 1

13 5

60x 25y 35 52x 39y 13

8x 14y 22 0

4x 7y 11 0

Which is the required equation ofthe obtuse angle bisector.

2 22 2

12x 5y 7 4x 3y 1

12 5 4 3

a1a2+b1b2 > 0, therefore the

obtuse angle bisector is


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Origin w.r.t. Angle Bisectors

Algorithm to determine whetherorigin lies in the obtuse angle orthe acute angle between a pair oflines

Step I : Rewrite the equations of the lines in general forma1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 such that c1 and c2 arepositive.

Step II : Determine sign of expression a1a2+b1b2


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Step III : Case (i) a1a2+b1b2 > 0

Origin lies in obtuse anglebetween the lines


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Step III : Case (i) a1a2+b1b2 < 0

Origin lies in acute anglebetween the lines


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For the straight lines 4x+3y-6 = 0 and

5x+12y+9 = 0 find the equation of thebisector of the angle which contains theorigin.

Rewrite the equations to make

the constant terms positive,

-4x-3y+6 = 0 and

5x+12y+9 = 0

Calculate a1a2+b1b2

(-4)*5+(-3)*(12) = -56

a1a2+b1b2 < 0, therefore origin lies in the acute angle.

Solution :


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Acute angle bisector is given by :

2 2 2 2

4x 3y 6 5x 12y 9

5 124 3

The origin lies in the acute angle and the equation ofthe acute angle bisector is 7x+9y-3 = 0.

4x 3y 6 5x 12y 9

5 13

52x 39y 78 25x 60y 45

77x 99y 33 0

7x 9y 3 0


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Family of Lines ThroughIntersection of a Pair of Lines

Equation of the family of linespassing through the intersection ofthe lines a1x+b1y+c1 = 0 anda2x+b2y+c2 = 0 is given by

1 1 1 2 2 2a x b y c a x b y c 0

where is a parameter

can be calculatedusing some given

condition


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Find the equation of the straight linewhich passes through the point

(2, -3) and the point of intersectionof x+y+4 = 0 and 3x-y-8 = 0.

Required equation can be written as

x+y+4+(3x-y-8) = 0 where is aparameter.

This passes through (2, -3).

2-3+4+ (3*2+3-8) = 0.

= -3

the required equation is x+y+4-3(3x-y-8) = 0

or 8x+4y+28 = 0 or 2x-y-7 = 0

Solution :


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Pair of Lines - Locus Definition

A pair of straight lines is the locus ofa point whose coordinates satisfy asecond degree equationax2+2hxy+by2+2gx+2fy+c = 0 suchthat it can be factorized into two

linear equations.


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Pair of Lines

ax2+2hxy+by2+2gx+2fy+c = 0 in

general represents all the conics inthe x-y plane.

a h g

h b f is called the discri min ant

g f c


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Pair of Lines

ax2+2hxy+by2+2gx+2fy+c = 0

A pair of lines

= 0, h2-ab 0


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Pair of Lines


A hyperbola

0, h2-ab > 0


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Pair of Lines


A parabola

0, h2-ab = 0


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Pair of Lines


An ellipse

0, h2-ab < 0


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Pair of Lines


A circle

0, h = 0, a = b


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Individual Lines

To find the equation of individual

lines in the pair of lines

ax2+2hxy+by2+2gx+2fy+c = 0,

Method I :

Step I : Rewrite the equation as a quadraticin x (or y).

Step II : Solve for x (or y).

Method II :

Will be discussed later.


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Find the separate equations ofthe lines represented by2x2-xy-y2+9x-3y+10 = 0.


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Solution

Rewriting the given equation,

2x2-(y-9)x-(y2+3y-10) = 0

2 2y 9 y 9 4.2 y 3y 10x

4

24x y 9 9y 6y 1

4x y 9 3y 1

4x 4y 8 or 4x 2y 10

x y 2 0 and 2x y 5 0 Which are the required equations.


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Point of Intersection

Point of intersection of the pair oflines represented by

ax2+2hxy+by2+2gx+2fy+c = 0 is

2 2

bg hf af gh,

h ab h ab

No need to memories thisformula. To find the required

point, find the equations of theindividual lines and solve

simultaneously


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Homogeneous Equation

An equation, with R.H.S. 0, in whichthe sum of the powers of x and y inevery term is the same, say n, iscalled a homogeneous equation ofnth degree in x and y.


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Pair of Lines Through Origin

A pair of straight lines passing through

the origin is represented by ahomogeneous equation of second degree

ax2+2hxy+by2 = 0

ax2

+2hxy+by2

= 0 can be rewritten as b(y-m1x)(y-m2x)= 0, where m1 and m2 are the slopes of the two lines.

bm1m2x2-b(m1+m2)xy+by2 = 0

1 2 1 2

a 2hm m ; m m

b b

The above relations can be used to find theequations and the slopes of the individual lines.


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Find the separate equations of the lines

represented by 4x2

+24xy+11y2

= 0.

1 2 1 2

4 24m m ; m m

11 11

21 2 1 2 1 2m m m m 4m m

1 2

576 176m m

121 121

1 220m m11

1 2 2m 2, m11

the required equations are y 2x 0 and 11y 2x 0

Solution :


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Pair of Lines Through Origin

If a pair of straight lines isrepresented by

ax2+2hxy+by2+2gx+2fy+c = 0,then

ax

2

+2hxy+by

2

= 0represents a pair of lines parallel tothem and passing through theorigin.


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Individual Lines

To find the equation of individual

lines in the pair of lines

ax2+2hxy+by2+2gx+2fy+c = 0,

Method II :

Step I : Find slopes of the individual lines m1 and m2using ax2+2hxy+by2 = 0

Step II : compare coefficients in the identityax2+2hxy+by2+2gx+2fy+c b(y-m1x-c1)(y-m2x-c2) to

find c1 and c2.


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Find the separate equations ofthe lines represented by2x2+5xy+3y2+6x+7y+4 = 0.

1 2 1 2

2 5m m ; m m

3 3

2

1 2 1 2 1 2m m m m 4m m

1 2

25 24m m

9 9

1 21m m3

1 2

2m 1, m

3

Solution :


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Solution Cont.

Consider the identity

2 2

1 2

2x 5xy 3y 6x 7y 4

y x c 3y 2x 3c

2 2

2 2 1 2 1 2 1 2

2x 5xy 3y 6x 7y 4

2x 5xy 3y 2c 3c x 3c 3c y 3c c

1 2 1 2 1 22c 3c 6; 3c 3c 7; 3c c 4

1 2

4c 1; c

3

The required equations are x y 1 0 and 2x 3y 4 0


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Pair of Lines With GivenCondition

XX

Y

O

Y

To find the joint equation of a pair of lines joining the

origin to the points of intersection of the curveax2+2hxy+by2+2gx+2fy+c = 0

and the line lx+my+n = 0.


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Pair of Lines With Given Condition

XX

Y

O

Y

Step I : Rewrite lx+my+n = 0 such that R.H.S. = 1

Step II : Make the equation of the curve homogeneous

lx my1

n

2

2 2 lx my lx myax 2hxy by 2 gx fy c 0n n


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Pair of Lines With Given Condition

XX

Y

O

Y

The required equation is the equation arrived at in Step II.

2

2 2 lx my lx myax 2hxy by 2 gx fy c 0n n


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Find the equation of the lines joiningthe origin to the points of intersectionof the straight line y = 3x+2 and thecurver x2+2xy+3y2+4x+8y-11 = 0.

Equation of straight line can be rewritten as

3x y1

2

Using this to make the equation of the curve

homogeneous,

2

2 2 3x y 3x yx 2xy 3y 4 x 2y 11 02 2

Solution :


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Solution

2

2 2 3x y 3x yx 2xy 3y 4 x 2y 11 02 2

2 2 2 2 2 211x 2xy 3y 2 3x 5xy 2y 9x 6xy y 04

On simplifying,

2 27x 2xy y 0

Which is the required equation.


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Angle Between a Pair of Lines

If ax2+2hxy+by2+2gx+2fy+c = 0

represents a pair of lines,

Acute angle between the lines isgiven by :

22 h abtana b

Independent of g, f, c


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22 h abtana b

Lines are parallel orcoincident if h2 = ab


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22 h abtana b

Lines are perpendicularif a+b = 0


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Find the angle between the pair

of lines represented by2x2+5xy+3y2+6x+7y+4 = 0.

Acute angle between the pair of lines is

given by22 h ab

tana b

25

2 62 1tan

2 3 5

1 11 1Acute angle tan ; Obtuse angle tan5 5

Solution :


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Angle Bisectors of a Pair ofLines

If ax2+2hxy+by2 = 0 represents a

pair of lines,

Equation of angle bisectors is givenby :

2 2x y xy

a b h


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Perpendiculars to a Pair of Lines

If ax2+2hxy+by2 = 0 represents a

pair of lines,

Equation of perpendiculars to thepair of lines is given by :

2 2bx 2hxy ay 0


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Find the equation of the bisectors

of the lines represented by135x2-136xy+33y2 = 0.

Equation of bisectors are given by :

Which is the required equation.

2 2x y xy

a b h

2 2x y xy

135 33 68

2 22x 3xy 2y 0

Solution :


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Class Exercise


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Class Exercise - 1

The bisectors of the angle

between the lines y = 3x+3 and3y = x+33 meet the X-axis atP and Q. Find length PQ.

The equations can be rewritten as

3x y 3 0 and x 3y 3 3

Angle bisectors are given by

3x y 3 x 3y 3 3

2 2

Solution :


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Solution Cont.

These lines meet the X-axis atP(3,0) and Q(-3,0).

Clearly length PQ = 6.

or 3 1 x 3 1 y 3 3 1 0

and 3 1 x 3 1 y 3 3 1 0

or x y 3 0 and x y 3 0


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Class Exercise - 2

Prove that the bisectors of the angles

formed by any two intersecting lines areperpendicular to each other.

A. Consider two intersecting lines

Consider the anglebisectors

Q.E.D.

+++ =

+ = /2

Solution :


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Class Exercise - 3

Show that the reflection of the lines

px+qy+r = 0 in the line x+y+1 = 0 is theline qx+py+(p+q-r) = 0, where p -q

Angle bisectors of the lines px+qy+r = 0 and

qx+py+(p+q-r) = 0 are

2 2 2 2

px qy r qx py (p q r)

p q q p

or p q x q p y 2r p q 0

and p q x p q y p q 0

Second equation can be written as x+y+1 = 0(as p+q 0)

Q.E.D.

Solution :


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Class Exercise - 4

A rhombus has two of its sides parallel tothe lines y = 2x+3 and y = 7x+2. If thediagonals cut at (1,2) and one vertex is onthe Y-axis, find the possible values of theordinate of that vertex.

Concept : Diagonals of a rhombus bisect the angle.

Let the sides intersecting at the Y-axis be 2x-y+1 = 0and 7x-y+2 = 0.

These lines will meet the Y-axis at (0,1) and (0,2).1 = 2 = (say)

Thus the sides are 2x-y+ = 0 and 7x-y+ = 0.

Solution :


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Angle bisectors will be

2x y 7x y

5 5 2

or 2 10x 10y 10 7x y

These will pass through (1,2)

10 5

5 1 109

Solution Cont.


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Class Exercise - 5

Find the bisector of the angle between thelines 2x+y-6 = 0 and 2x-4y+7 = 0 whichcontains the point (1,2).

Consider a1x+b1y+c1 = 0 and

a2x+b2y+c2 = 0 (c1, c2 0),

If a1h+b1k+c1 and a2h+b2k+c2 have the same sign,point (h,k) will lie in the angle bisector

1 1 1 2 2 22 2 2 2

1 1 2 2

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

Solution :


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Angle bisector containing (1,2) is

2x y 6 2x 4y 7

5 2 5

or, 6x 2y 5 0

Solution Cont.

Given lines can be rewritten as

-2x-y+6 = 0 and 2x-4y+7 = 0.

Now -2(1)-(2)+6 > 0 and2(1)-4(2)+7 > 0


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Class Exercise - 6

Find the equation of the straight linedrawn through the point of intersectionof the lines x+y = 4 and 2x-3y = 1 andperpendicular to the line cutting offintercepts 5 and 6 on the positive axes.

Family of lines through the intersection ofthe given lines is (x+y-4)+(2x-3y-1) = 0

Line cutting intercepts of 5 and 6 on the

positive axes isx y

1 or 6x 5y 30 05 6

Solution :


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Thus required equation is

3(x+y-4)+11(2x-3y-1) = 0

Or, 25x-30y-23 = 0

Solution Cont.

Slope of line perpendicular to this

line will be -5/6.

1 2 5 11

1 3 6 3


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Class Exercise - 7

Find the separate equations of thepair of lines represented by12x2+5xy-28y2+19x+61y-21 = 0.

Rewrite the equation as a quadratic in x, we

get 12x2+(5y+19)x-(28y2-61y+21) = 0

2 25y 19 25y 190y 361 1344y 2928y 1008x

24

224x 5y 19 1369y 2738y 1369

24x 5y 19 37 y 1

24x 32y 56 0 or 24x 42y 18 0

3x 4y 7 0 or 4x 7y 3 0

Solution :


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Class Exercise - 8

Show that the lines joining the origin to the

points common to x2

+hxy-y2

+gx+fy = 0and fx-gy = are at right angles whateverbe the value of.

Given line can be rewritten asfx gy

1

Using this to make the equation of the curvehomogeneous, we get

2 2fx gy

x hxy y gx fy 0

Solution :


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Sum of coefficients of x2 and y2

= (+fg)-(+fg)

= 0

Thus the required lines are perpendicular to each other,whatever be the value of.

Q.E.D.

Solution Cont.

2 2 2 2Or fg x h f g xy fg y 0


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Class Exercise - 9

Find the angle between the pair of

lines : (x2

+y2

)sin2

= (xcos-ysin)2

.

Given equation can be rewritten as

(cos2-sin2)x2-2sincosxy+sin2-sin2)y2 = 0

Solution :


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Class Exercise - 10

Prove that the lines

a2

x2

+2h(a+b)xy+b2

y2

= 0 areequally inclined to the linesax2+2hxy+by2 = 0.


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Thank you

15. Cartesian Coordinate Geometry and Straight Lines-3

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