Question

Let $z_1$ and $z_2$ be two distinct complex numbers and let $z=(1-t)z_1+t{z}_2$ for some real number t, where $0<t<1.$ If arg(w) denotes the principal argument of a nonzero complex number w then

• A. $|z-z_1|+|z-z_2|=|z_1-z_2|$
• B. $arg(z-z_1)=arg(z-z_2)$
• C. $\left|\begin{array}{cc}z-z_1& \overline{z}-{\overline{z}}_1\\ z_2-z_1& {\overline{z}}_2-{\overline{z}}_1\end{array}\right|$=0
• D. $arg(z-z_1)=arg(z_2-z_1)$

Answer 1

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

The correct options are
A $|z-z_1|+|z-z_2|=|z_1-z_2|$
C $\left|\begin{array}{cc}z-z_1& \overline{z}-{\overline{z}}_1\\ z_2-z_1& {\overline{z}}_2-{\overline{z}}_1\end{array}\right|$=0
D $arg(z-z_1)=arg(z_2-z_1)$

Given $z=\frac{(1-t)z_1+t{z}_2}{(1-t)+t}$
Clearly, $z$ divides $z_1$ and $z_2$ in the ratio $t:(1-t).0<t<1$
$\Rightarrow AP+BP=AB\Rightarrow |z-z_1|+|z-z_2|=|z_1-z_2|$
$\Rightarrow$ option (A) is true
and $arg(z-z_1)=arg(z_2-z)=arg(z_2-z_1)$
$\Rightarrow$ option (B) is false and (D) is true
Also, $arg(z-z_1)=arg(z_2-z_1)\Rightarrow arg\left(\frac{z-z_1}{z_2-z_1}\right)=0$
$\therefore \frac{z-z_1}{z_2-z_1}$ 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 $\left|\begin{array}{cc}z-z_1& \overline{z}-\overline{z_1}\\ z_2-z_1& \overline{z_2}-\overline{z_1}\end{array}\right|=0$
Therefore option (C) is correct.