if ax3+by2+cx2y+dxy2=0 represents 3 distinct straight lines,so that each line is angle bisector of of the other two,then prove that c+3b=0
62The question probably should have been this:
ax3+by3+cx2y+dxy2=0
Now we divide the whole by x3
a+b(y/x)3+c(y/x)+d(y/x)2 = 0
this is a cubic in (y/x)
It is also clear that the lines given above pass through origin.
y/x gives the slope for each of these three lines (m1,m2,m3)
one is the bisector of two other
use tan (2. @1) = tan (@2+@3)
I think u will get the soln.. do let me know if u dont!
9simple
one of lines has to be x=0 ( coefficient of y3 is 0)
put other lines as y=mx, y=-mx and multiply lines
you get b=c=0
62no celestine.. i think that is a mistyping..
the 2nd term is y cube! that is my guess.. but i may be wrong!
1Assuming second term is y3
we have
ax3+by3+cx2y+dxy2=0
or bm3+dm2+cm+a=0, where m=y/x
implying Σm1m2=c/b ...(i)
Since, each line is angle bisector of the other two implying angle between any two pair of lines will be same.
Now, let this angle be θ (in fact, θ=60°)
tan θ=(m1-m2)/(1+m1m2)
or
m1-m2=tan θ(1+m1m2) ...(ii)
similarly, taking into cyclic order
m2-m3=tan θ(1+m2m3) ...(iii)
m3-m1=tan θ(1+m3m1) ...(iv)
Adding, we get
0=tan θ(3 + Σm1m2)
or 3+c/b=0 ... using (i)
or c+3b=0
