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Mathematics. Session. Cartesian Coordinate Geometry And Straight Lines. Session Objectives. Session Objectives. Equations of bisectors of angles between two lines Acute/obtuse angle bisectors Position of origin w.r.t bisecors Equation of family of lines through intersection of two lines

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session
Session

Cartesian Coordinate Geometry

And

Straight Lines

session objectives
Session Objectives
  • Equations of bisectors of angles between two lines
  • Acute/obtuse angle bisectors
  • Position of origin w.r.t bisecors
  • Equation of family of lines through intersection of two lines
  • Pair of lines - locus definition
  • Pair of lines represented by second degree equation
  • Angle between two lines, represented as a second degree equation
equation of bisector

Y

a2x+b2y+c2=0

N

A

P(h,k)

a1x+b1y+c1=0

M

X’

O

X

Y’

Equation of Bisector

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

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

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

equidistant from the given lines.

equation of bisector6

Y

a2x+b2y+c2=0

N

A

P(h,k)

a1x+b1y+c1=0

M

X’

O

X

Y’

Equation of Bisector

PM = PN

 The required equations are

acute obtuse angle bisectors
Acute/obtuse Angle Bisectors

Algorithm to determine equations of bisectors of acute angle and obtuse angle 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 and c2 are positive.

Step II : Determine sign of expression a1a2+b1b2

Step III : Write the equations of the bisectors

;

acute obtuse angle bisectors8
Acute/obtuse Angle Bisectors

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

Obtuse angle bisector

Acute angle bisector

acute obtuse angle bisectors9
Acute/obtuse Angle Bisectors

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

Acute angle bisector

Obtuse angle bisector

illustrative example
Illustrative Example

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

Solution :

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 cont
Solution Cont.

a1a2+b1b2 > 0, therefore the obtuse angle bisector is

Simplifying,

Which is the required equation of the obtuse angle bisector.

origin w r t angle bisectors
Origin w.r.t. Angle Bisectors

Algorithm to determine whether origin lies in the obtuse angle or the acute angle 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 and c2 are positive.

Step II : Determine sign of expression a1a2+b1b2

origin w r t angle bisectors13
Origin w.r.t. Angle Bisectors

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

Origin lies in obtuse angle between the lines

origin w r t angle bisectors14
Origin w.r.t. Angle Bisectors

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

Origin lies in acute angle between the lines

illustrative example15
Illustrative Example

For the straight lines 4x+3y-6 = 0 and 5x+12y+9 = 0 find the equation of the bisector of the angle which contains the origin.

Solution :

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.

illustrative example16
Illustrative Example

Acute angle bisector is given by :

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

family of lines through intersection of a pair of lines
Family of Lines Through Intersection of a Pair of Lines

Equation of the family of lines passing through the intersection of the lines a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 is given by

 can be calculated using some given condition

illustrative example18
Illustrative Example

Find the equation of the straight line which passes through the point (2, -3) and the point of intersection of x+y+4 = 0 and 3x-y-8 = 0.

Solution :

Required equation can be written as x+y+4+(3x-y-8) = 0 where  is a parameter.

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

pair of lines locus definition
Pair of Lines - Locus Definition

A pair of straight lines is the locus of a point whose coordinates satisfy a second degree equation ax2+2hxy+by2+2gx+2fy+c = 0 such that it can be factorized into two linear equations.

pair of lines
Pair of Lines

ax2+2hxy+by2+2gx+2fy+c = 0 in general represents all the conics in the x-y plane.

pair of lines21
Pair of Lines

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

 = 0, h2-ab  0

A pair of lines

pair of lines22
Pair of Lines

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

  0, h2-ab > 0

A hyperbola

pair of lines23
Pair of Lines

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

  0, h2-ab = 0

A parabola

pair of lines24
Pair of Lines

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

  0, h2-ab < 0

An ellipse

pair of lines25
Pair of Lines

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

  0, h = 0, a = b

A circle

individual lines
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 quadratic in x (or y).

Step II : Solve for x (or y).

Method II :

Will be discussed later.

illustrative example27
Illustrative Example

Find the separate equations of the lines represented by 2x2-xy-y2+9x-3y+10 = 0.

solution
Solution

Rewriting the given equation,

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

Which are the required equations.

point of intersection
Point of Intersection

Point of intersection of the pair of lines represented by

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

No need to memories this formula. To find the required point, find the equations of the individual lines and solve simultaneously

homogeneous equation
Homogeneous Equation

An equation, with R.H.S. 0, in which the sum of the powers of x and y in every term is the same, say n, is called a homogeneous equation of nth degree in x and y.

pair of lines through origin
Pair of Lines Through Origin

A pair of straight lines passing through the origin is represented by a homogeneous 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

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

illustrative example32
Illustrative Example

Find the separate equations of the lines represented by 4x2+24xy+11y2 = 0.

Solution :

pair of lines through origin33
Pair of Lines Through Origin

If a pair of straight lines is represented by

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

ax2+2hxy+by2 = 0

represents a pair of lines parallel to them and passing through the origin.

individual lines34
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 m2 using ax2+2hxy+by2 = 0

Step II : compare coefficients in the identity ax2+2hxy+by2+2gx+2fy+c  b(y-m1x-c1)(y-m2x-c2) to find c1 and c2.

illustrative example35
Illustrative Example

Find the separate equations of the lines represented by 2x2+5xy+3y2+6x+7y+4 = 0.

Solution :

solution cont36
Solution Cont.

Consider the identity

pair of lines with given condition

Y

X’

O

X

Y’

Pair of Lines With Given Condition

To find the joint equation of a pair of lines joining the origin to the points of intersection of the curve ax2+2hxy+by2+2gx+2fy+c = 0

and the line lx+my+n = 0.

pair of lines with given condition38

Y

X’

O

X

Y’

Pair of Lines With Given Condition

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

Step II : Make the equation of the curve homogeneous

pair of lines with given condition39

Y

X’

O

X

Y’

Pair of Lines With Given Condition

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

illustrative example40
Illustrative Example

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

Solution :

Equation of straight line can be rewritten as

Using this to make the equation of the curve homogeneous,

solution41
Solution

On simplifying,

Which is the required equation.

angle between a pair of lines
Angle Between a Pair of Lines

If ax2+2hxy+by2+2gx+2fy+c = 0 represents a pair of lines,

Acute angle between the lines is given by :

Independent of g, f, c

angle between a pair of lines43
Angle Between a Pair of Lines

If ax2+2hxy+by2+2gx+2fy+c = 0 represents a pair of lines,

Acute angle between the lines is given by :

Lines are parallel or coincident if h2 = ab

angle between a pair of lines44
Angle Between a Pair of Lines

If ax2+2hxy+by2+2gx+2fy+c = 0 represents a pair of lines,

Acute angle between the lines is given by :

Lines are perpendicular if a+b = 0

illustrative example45

Acute angle between the pair of lines is given by

Illustrative Example

Find the angle between the pair of lines represented by 2x2+5xy+3y2+6x+7y+4 = 0.

Solution :

angle bisectors of a pair of lines
Angle Bisectors of a Pair of Lines

If ax2+2hxy+by2 = 0 represents a pair of lines,

Equation of angle bisectors is given by :

perpendiculars to a pair of lines
Perpendiculars to a Pair of Lines

If ax2+2hxy+by2 = 0 represents a pair of lines,

Equation of perpendiculars to the pair of lines is given by :

illustrative example48
Illustrative Example

Find the equation of the bisectors of the lines represented by 135x2-136xy+33y2 = 0.

Solution :

Equation of bisectors are given by :

Which is the required equation.

class exercise 1
Class Exercise - 1

The bisectors of the angle between the lines y = 3x+3 and 3y = x+33 meet the X-axis at P and Q. Find length PQ.

Solution :

The equations can be rewritten as

Angle bisectors are given by

solution cont51
Solution Cont.

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

Clearly length PQ = 6.

class exercise 2

Class Exercise - 2

Prove that the bisectors of the angles formed by any two intersecting lines are perpendicular to each other.

Solution :

A. Consider two intersecting lines

Consider the angle bisectors

+++ = 

 + = /2

Q.E.D.

class exercise 3
Class Exercise - 3

Show that the reflection of the lines px+qy+r = 0 in the line x+y+1 = 0 is the line qx+py+(p+q-r) = 0, where p  -q

Solution :

Angle bisectors of the lines px+qy+r = 0 and qx+py+(p+q-r) = 0 are

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

Q.E.D.

class exercise 4
Class Exercise - 4

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

Solution :

Concept : Diagonals of a rhombus bisect the angle.

Let the sides intersecting at the Y-axis be 2x-y+1 = 0 and 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 cont55
Solution Cont.

Angle bisectors will be

These will pass through (1,2)

class exercise 5
Class Exercise - 5

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

Solution :

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

solution cont57
Solution Cont.

Given lines can be rewritten as -2x-y+6 = 0 and 2x-4y+7 = 0.

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

Angle bisector containing (1,2) is

class exercise 6
Class Exercise - 6

Find the equation of the straight line drawn through the point of intersection of the lines x+y = 4 and 2x-3y = 1 and perpendicular to the line cutting off intercepts 5 and 6 on the positive axes.

Solution :

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

Line cutting intercepts of 5 and 6 on the positive axes is

solution cont59
Solution Cont.

Slope of line perpendicular to this line will be -5/6.

Thus required equation is

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

Or, 25x-30y-23 = 0

class exercise 7
Class Exercise - 7

Find the separate equations of the pair of lines represented by 12x2+5xy-28y2+19x+61y-21 = 0.

Solution :

Rewrite the equation as a quadratic in x, we get 12x2+(5y+19)x-(28y2-61y+21) = 0

class exercise 8

Given line can be rewritten as

Class Exercise - 8

Show that the lines joining the origin to the points common to x2+hxy-y2+gx+fy = 0 and fx-gy =  are at right angles whatever be the value of .

Solution :

Using this to make the equation of the curve homogeneous, we get

solution cont62
Solution Cont.

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.

class exercise 9
Class Exercise - 9

Find the angle between the pair of lines : (x2+y2)sin2 = (xcos-ysin)2.

Solution :

Given equation can be rewritten as

(cos2-sin2)x2-2sincosxy+sin2-sin2)y2 = 0

class exercise 10
Class Exercise - 10

Prove that the lines a2x2+2h(a+b)xy+b2y2 = 0 are equally inclined to the lines ax2+2hxy+by2 = 0.