IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - IIT-2004 Q

varshavallig

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

rates assured.

!!!!!!!!!!!!!!!!!!!!

ambareesh13

answer is 2

varshavallig

thats right but how did u find the eqn of the bisectors????pl xplain
!!!!!!!!!!

ambareesh13

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

varshavallig

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

joyfrancis

x² -y²+2y-1=0 is the joint eqn of

x-y+ 1 = 0 &

x+y-1 = 0..

ankitagg

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+a_2²) =+ (a₂x+b₂y+c₂)/root(a₂²+b_2²)
is obtuse angle bisector

=>(a₁x+b₁y+c₁)/root(a₁²+a_2²) =- (a₂x+b₂y+c₂)/root(a₂²+b_2²)
is acute angle bisector
-----------------------------------------
if a₁a₂+b₁b₂<0
=> (a₁x+b₁y+c₁)/root(a₁²+a_2²) =+ (a₂x+b₂y+c₂)/root(a₂²+b_2²)
is acute angle bisector

=>(a₁x+b₁y+c₁)/root(a₁²+a_2²) =- (a₂x+b₂y+c₂)/root(a₂²+b_2²)
is obtuse angle bisector

Conjurer

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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