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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

## 1 Answers

9 years ago

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.

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