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 $z_3=0$ from an equilateral triangle then $a$ and $b$ are equal to:-

• A. $a=b=1/2$
• B. $a=b=2-\sqrt{3}$
• C. $a=b=-2+\sqrt{3}$
• D. $a=b=\sqrt{2}-1$

Answer 1

Answer: (B)

$a=b=2-\sqrt{3}$

Given :-

$z_1=a+i$

$z_2=1+bi$

$z_3=0$

If $z_1,z_2,z_3$ form an equilateral triangle, then-

$|z_1-z_2|=|z_2-z_3|=|z_3-z_1|$

${|z_1-z_2|}^2={|z_2-z_3|}^2={|z_3-z_1|}^2$

${|a+i-(1+bi\left)\right|}^2={|1+bi-0|}^2={|0-(a+i\left)\right|}^2$

${(a-1)}^2+{(1-b)}^2={(0-1)}^2+{(0-b)}^2={(a-0)}^2+{(1-0)}^2$

${(a-1)}^2+{(1-b)}^2=1+b^2=a^2+1$

Now,

$1+b^2=a^2+1$

$\Rightarrow b=a$

And,

${(a-1)}^2+{(1-b)}^2=a^2+1$

$a^2+1-2a+1+b^2-2b=a^2+1$

$b^2+1-2a-2b=0$

$a^2+1-4a=0(\because a=b)$

By quadratic formula, we have

$a=\frac{-(-4)\pm \sqrt{16-4}}{2}$

$\Rightarrow a=\frac{4\pm 2\sqrt{3}}{2}$

$\Rightarrow a=2\pm \sqrt{3}$

As $0<a<1$, we get

$a=2-\sqrt{3}$

$\because a=b$

Hence $a=b=2-\sqrt{3}$