Question

If $|z_1-z_0|=|z_2-z_0|=a$ and $amp\frac{z_2-z_0}{z_0-z_1}=\frac{\pi }{2}$, then $z_0$ is equal to

• A. $\frac{1}{2}\left\{\right(1+i)z_1+(1-i\left)z_2\right\}$
• B. $\frac{1}{2}\left\{\right(1-i)z_1+(1+i\left)z_2\right\}$
• C. $\frac{1}{2}\left\{z_1+z_2\right\}$
• D. None of the above

Answer 1

Answer: (A)

$\frac{1}{2}\left\{\right(1+i)z_1+(1-i\left)z_2\right\}$

We have, $\frac{z_2-z_0}{z_0-z_1}=e^{i\frac{\pi }{2}}$
$\Rightarrow z_2-z_0=i{z}_0-i{z}_1$
$\Rightarrow z_0(1+i)=z_2+i{z}_1$
$\Rightarrow z_0=\frac{1}{2}\left(\right(1-i\left)\right(z_2+i{z}_1\left)\right)$
$\Rightarrow z_0=\frac{1}{2}\left(\right(1+i)z_1+(1-i\left)z_2\right)$
Hence, option A is correct