Home » Ask & Discuss » Mathematics. » Analytical Geometry « Back to Discussion

Analytical Geometry

Blazing goIITian

Joined:	27 Jul 2007
Post:	3134

21 Dec 2007 22:44:34 IST

4 People liked this

760

Pair of st. lines-equation of angle bisectors

None

Given a pair of st. lines---->

ax² + 2hxy + by² + 2gx + 2fy + c = 0

Is there any short method to write the equation of angle bisectors of these lines. I know the longer method i.e., to find their slope,then finding slope of angle bisectors, then passing the equation of angle bisectors thru the point of intersection of these pair of st. lines Blah Blah Blah...

Its too lengthy.

Is dere any shorter method to find the equation of angle bisectors?

RATES ASSURED.

Share this article on:

Comments (9)

Tarin Bansal

Blazing goIITian

Joined: 27 Jul 2007

Posts: 3134

22 Dec 2007 20:30:42 IST

Like 0 people liked this

Somebody help please. I really need it.
5 pointers assured.

A K

Cool goIITian

Joined: 21 Dec 2007

Posts: 33

22 Dec 2007 21:16:28 IST

Like 2 people liked this

there is no short method.. but there is ofcourse the formula meth as we kno the pt of intersection of the original pair is x1=hf-bg/ab-h^2, y1= gh-af/ab-h^2 Now the equation of the angle bisectors is h[(x-x1)^2-(y-y1)^2)=(a-b)(x-x1)(y-y1)

ravi teja

Scorching goIITian

Joined: 22 Dec 2007

Posts: 293

22 Dec 2007 23:39:04 IST

Like 0 people liked this

this is like sting theory.it cant have short cuts

Tarin Bansal

Blazing goIITian

Joined: 27 Jul 2007

Posts: 3134

24 Dec 2007 17:36:54 IST

Like 0 people liked this

Let the equation be

3x² + 8xy - 3y² - 20x - 10y + 25= 0

The point of intersection is (2,1)

Now what to do to find the equation of angle bisector?

A K

Cool goIITian

Joined: 21 Dec 2007

Posts: 33

24 Dec 2007 18:43:01 IST

Like 1 people liked this

use the formula i gave you .... 4[(x-2)^2-(y-1)^2]=6(x-2)(y-1)

Manish Ranaday

New kid on the Block

Joined: 25 Dec 2007

Posts: 9

26 Dec 2007 15:16:23 IST

Like 0 people liked this

Given a pair of st. lines---->

ax² + 2hxy + by² + 2gx + 2fy + c = 0

Its too lengthy.

Is dere any shorter method to find the equation of angle bisectors?

RATES ASSURED.

Tarin Bansal

Blazing goIITian

Joined: 27 Jul 2007

Posts: 3134

26 Dec 2007 15:44:56 IST

Like 0 people liked this

Why did U copy my question? U must have got the answer in this forum already.

rohit

Blazing goIITian

Joined: 1 Jul 2007

Posts: 301

26 Dec 2007 19:16:37 IST

Like 1 people liked this

look,ive found a way of doing it. take the first 3 terms of the equation

ax² + 2hxy + by² + 2gx + 2fy + c = 0

ie,ax² + 2hxy + by²=0(these lines have the the same slope as the previous lines,only their point of intersection is the origin)

the org. equation is 3x² + 8xy - 3y² - 20x - 10y + 25 = 0

the new eq.is 3x² + 8xy - 3y²=0

the equation of their angle bisectors is (x²- y²)/xy=(a - b)/h

after putting the values ,you get 2x² - 3xy - 2y²=0

now just shift this equation to the org.point of intersection(2,1)

you get

2(x - 2)² - 3(x - 2)(y - 1) - 2(y - 1)=0

expand this eq. and you'll get the eq. of the angle bisectors.

nudge me if im wrong.

madhu sudan dash

New kid on the Block

Joined: 15 Dec 2011

Posts: 1

15 Dec 2011 15:13:01 IST

Like 0 people liked this

Proof that equation of angle bisectors of a pair of straight lines is h[(x-x1)^2-(y-y1)^2]

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team