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Analytical Geometry

varsha valli g.'s Avatar
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9 Apr 2008 14:34:02 IST
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IIT-2004 Q
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area of triangle formed by the angle bisectors of the pair of lines
x2 -y2+2y-1=0 and the line x+y=3 is:
a)1    b)2   c)3   d)4
my doubt is how to find the eqn of the bisectors???????
plz reply soon
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Ambareesh Rishi's Avatar

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9 Apr 2008 14:44:40 IST
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answer is 2
varsha valli g.'s Avatar

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9 Apr 2008 14:45:57 IST
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thats right but how did u find the eqn of the bisectors????pl xplain
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Ambareesh Rishi's Avatar

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9 Apr 2008 14:50:16 IST
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tlet the linew be

ax + by +c and ax1+by1+c1

then equation of bisectors


(ax+by+c)/root(a*a+b*b) =+or - (ax1+by1+c1)/root(a1*a1+b1*b1)


one will give the equation of the acute angle and other will give the equation of the obtuse angle bisector


varsha valli g.'s Avatar

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9 Apr 2008 15:02:54 IST
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thta is the case when the 2 eqns of the lines r given,but here combined eqn is given ,how did uu split it up.
xplain!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!plz
joy francis's Avatar

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9 Apr 2008 16:12:56 IST
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x2 -y2+2y-1=0  is the joint eqn of
 
x-y+ 1 = 0 &
x+y-1 = 0..
 
 
 
 
 
ankit aggarwal's Avatar

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9 Apr 2008 17:29:34 IST
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let a1x+b1y+c1=0 and a2x+b2y+c2=0 are the lines.
then the procedure to find the angle bisectors is

1) make c2&c2 positive.

2)if a1a2+b1b2>0
=> (a1x+b1y+c1)/root(a12+a22) =+ (a2x+b2y+c2)/root(a22+b22)
is obtuse angle bisector

=>(a1x+b1y+c1)/root(a12+a22) =- (a2x+b2y+c2)/root(a22+b22)
is acute angle bisector
-----------------------------------------
if a1a2+b1b2<0
=> (a1x+b1y+c1)/root(a12+a22) =+ (a2x+b2y+c2)/root(a22+b22)
is acute angle bisector

=>(a1x+b1y+c1)/root(a12+a22) =- (a2x+b2y+c2)/root(a22+b22)
is obtuse angle bisector
Conjurer's Avatar

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9 Apr 2008 17:48:33 IST
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Some additional info:

Equations of the bisectors of the angles between two lines represented by ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 are given by

(x-x1)^2 - (y-y1)^2/a-b = (x-x1)(y-y1)/h

where(x1,y1) is the point of intersection of the lines represented by the given equation.



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