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if ax^3+by^3+cx^2y+dxy^2=0 represent three distinct straight lines,such that each line bisects the angle between other two then prove that 3b+c=0

3 years ago

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Answers : (1)

                                        

Let the lines be


 


y-mix = 0 where i=1,2,3


 


As one line is equally inclined to the other two, we impose the following condition.


(m1-m3)/(1+m1m3) = -(m2-m3)/(1+m2m3)


Simplify to get,


 


(m1+m2+m3)-3m3-m3(m1m2+m2m3+m3m1)+3m1m2m3 = 0       ...........(i)


 


Similarly applying the condition on the other two pairs, we get,


 


(m1+m2+m3)-3m2-m2(m1m2+m2m3+m3m1)+3m1m2m= 0         ............(ii)


and


(m1+m2+m3)-3m1-m1(m1m2+m2m3+m3m1)+3m1m2m= 0          .......... (iii)


 


Adding equations (i),(ii) and (iii), we get,


 


-(m1m2+m2m3+m3m1)(m1+m2+m3) + 9m1m2m3 = 0                 ............(iv)


 


Putting y=mx in the equation of combined equation,


 










ax3+by3+cx2y+dxy2=0




we get,


 


bm3 + dm2 + cm + a = 0


By theory of equations,


 


m1+m2+m3 = -d/b


m1m2+m2m3+m3m1 = c/b


m1m2m3 = -a/b


 


 


Putting the above relations in equation (iv),


 


-(c/b)(-d/b) + 9(-a/b) = 0


 


cd-9ab=0


 


This is the condition that i get. No idea where i went wrong.

3 years ago

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