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Let $a,b$ be real numbers between $0\; and\; 1$ such that $z_1=1+bi,z_2=a+i$ and $z_3=0$ forms an equilateral triangle then

$\begin{array}{1 1}(A)a=b=\sqrt 3 -1 \\ (B) a=b=2- \sqrt 3 \\ (C) a= \sqrt 3 -1 ,b= 2- \sqrt 3 \\(D) None\;of\;these \end{array}$

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$0 < a,b < 1$
$\mid z_1-z_2\mid=\mid z_2-z_3\mid =\mid z_3-z_1\mid$
$\Rightarrow \mid (1-a)+i(b-1)\mid =\mid a+i\mid =\mid 1+bi\mid$
$\Rightarrow (a-1)^2+(b-1)^2=a^2+1=1+b^2$
$\Rightarrow (a-1)^2+(b-1)^2=a^2+1=1+b^2$
$\Rightarrow a^2-2a+1-2b=0$ and $b^2=a^2$
$0 < a,b < 1$
$\therefore a=b$
$\Rightarrow a^2-2a+1-2a=0$
$\Rightarrow a=2\pm \sqrt 3$
As $0 < a < 1\Rightarrow a=b=2-\sqrt 3$
Hence (B) is the correct answer.
answered Apr 9, 2014 by sreemathi.v
edited May 29, 2014 by rohanmaheshwari0831_1
 

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