Question

If a and b are real numbers between 0 and 1 such that the points $z_1=a+i,z_2=1+bi$ and $z_3=0$ form an equilateral triangle. then $a$ and $b$ are equal to

• A. $a=b=2-\sqrt{3}$
• B. $a=b=2+\sqrt{3}$
• C. $a=2-\sqrt{3},b=2+\sqrt{3}$
• D. $a=2+\sqrt{3},b=2-\sqrt{3}$

Answer 1

The correct option is A $a=b=2-\sqrt{3}$

If $z_1,z_2$ and $z_3=0$ be the three sides, then the sides are given by $|z_1-z_2|,|z_1-0|$ and $|z_2-0|$
Since the triangle is equilateral the sides are of equal length
$\therefore {(a-1)}^2+{(b-1)}^2=a^2+1=b^2+1$
From last two, we have $a=b$ and from first two, we have
$\therefore a^2-4a+1=0\because a=b$
$\therefore a=2+\sqrt{3},2-\sqrt{3}.$
Since $0<a<1$
$\therefore a=2-\sqrt{3}=b.$
Ans: A