11FOR equilateral triangle,
z12 + z22 + z32 - z1z2 - z2z3 - z3z1 = 0
and
(z1 + z2w + z3w2) (z1 + z2w2 + z3w) = 0.
use this...
might help.. :)
Let a,bER and 0<a<1 and 0<b<1 if z1=-a+i,z2=-1+bi and z30 and z1,z2,z3 form an equilateral triangle then
(A) (a+√3)(b+√3)=4
(B)a2-2a-2b=0
(C)(a-2)(b-2)=3
(D)a4-4(a+b)2=0.
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3 Answers
1111found a simplier method.. :)
z2 - z3z1 - z3 = ei∂
here ∂=60° (as triangle is equilateral).
thus,ei∂ = cos∂ +isin∂ = 1+√3i
2.
thus,
-1 + bi-a + i = 1+√3i2.
thus,on simplifying.
-2+2bi = -√3ai - √3 - a +i.
equating the real and imaginary co-efficients weget,
a=b= 2 - √3
thus...
i get (A) and (C) as answer..
