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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Dec 2007 22:44:34 IST
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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.
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The quality of a person's life is in direct proportion to their commitment to excellence, regardless of their chosen field of endeavor.
It is during our darkest moments that we must focus to see the light.
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(A must see for every student)
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22 Dec 2007 20:30:42 IST
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Somebody help please. I really need it.
5 pointers assured.
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The quality of a person's life is in direct proportion to their commitment to excellence, regardless of their chosen field of endeavor.
It is during our darkest moments that we must focus to see the light.
Check out my blog at:
http://tarinbansal.blogspot.com/
(A must see for every student)
Back to goiit, this time with Baby Veerappan. :D |
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22 Dec 2007 21:16:28 IST
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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)
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22 Dec 2007 23:39:04 IST
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this is like sting theory.it cant have short cuts
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24 Dec 2007 17:36:54 IST
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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?
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The quality of a person's life is in direct proportion to their commitment to excellence, regardless of their chosen field of endeavor.
It is during our darkest moments that we must focus to see the light.
Check out my blog at:
http://tarinbansal.blogspot.com/
(A must see for every student)
Back to goiit, this time with Baby Veerappan. :D |
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24 Dec 2007 18:43:01 IST
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use the formula i gave you .... 4[(x-2)^2-(y-1)^2]=6(x-2)(y-1)
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26 Dec 2007 15:16:23 IST
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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.
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26 Dec 2007 15:44:56 IST
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Why did U copy my question? U must have got the answer in this forum already.
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The quality of a person's life is in direct proportion to their commitment to excellence, regardless of their chosen field of endeavor.
It is during our darkest moments that we must focus to see the light.
Check out my blog at:
http://tarinbansal.blogspot.com/
(A must see for every student)
Back to goiit, this time with Baby Veerappan. :D |
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26 Dec 2007 19:16:37 IST
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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.
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