Question

Let $z_1$ and $z_2$ be two distinct complex numbers and let $z = (1 - t) z_1 + tz_2$ for some real number t with $0 < t < 1$. If $arg (w) $ denotes the principal argument of a non-zero complex number $w$, then

• A. $|z-z_1|+|z-z_2|=|z_1-z_2|$
• B. $arg(z-z_2) = arg(z-z_2)$
• C. $\begin {vmatrix} z-z_1 & \overline{z}-\overline{z_1} \\ z_2-z_1 & \overline{z_2}-\overline{z_1} \end {vmatrix} =0$
• D. $arg(z-z_1)=arg(z_2-z_1)$

Answer 1

Answer: (A), (C), (D)

Given , $z=\frac{(1-t)z_1+tz_2}{(1-t)+t}$

Clearly, $z$ divides $z_1$ and $z_2$ in the ratio of $t : (1 - 1), 0 < t < 1$
$\Rightarrow AP+BP+AB i.e., |z-z_1|+|z-z_2|=|z_1-z_2|$
$\Rightarrow $ Option (a) is true.
and $arg (z - z_1) =arg(z_2 - z ) =arg(z_2 - z_1)$
$\Rightarrow $ Option (b) is false and option (d) is true.
Also, $arg(z-z_1)=arg(z_2-z)=arg(z_2-z_1)$
$\rightarrow arg\bigg( \frac{z-z_1}{z_2-z_1}\bigg)=0$
$\therefore \frac{z-z_1}{z_2-z_1}$ is purely real.
$\Rightarrow \frac{z-z_1}{z_2-z_1}=\frac{\overline{z}-\overline{z_1}}{\overline{z_2}-\overline{z_1}}$ or $\begin {vmatrix} z-z_1 & \overline{z}-\overline{z_1} \\ z_2-z_1 & \overline{z_2}-\overline{z_1} \end {vmatrix}=0 $
$\therefore $ Option (c) is correct. Hence, (a, c, d) is the correct option.