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 $t$ with $0<t<1$. If $Arg\left(w\right)$ denotes the principal argument of a non-zero complex number $w$, then:

• A. $|z-z_1|+|z-z_2|=|z_1-z_2|$
• B. $Arg(z-z_1)=Arg(z-z_2)$
• 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 $z=(1-t)z_1+t{z}_2$
$\Rightarrow \frac{z-z_1}{z_2-z_1}=t\Rightarrow \arg \left(\frac{z-z_1}{z_2-z_1}\right)=0\cdots$ (1) [As $t$ is real hence the $\frac{z-z_1}{z_2-z_1}$ has $arg$ is $0$ ]

$\Rightarrow \arg (z-z_1)=\arg (z_2-z_1)$

$\frac{z-z_1}{z_2-z_1}=\frac{\overline{z}-\overline{z_1}}{\overline{z_2}-\overline{z_1}}$ As [$t$ is real then $t$ and its conjugate are equal]

$\left|\begin{array}{cc}z-z_1& \overline{z}-{\overline{z}}_1\\ z_2-z_1& {\overline{z}}_2-{\overline{z}}_1\end{array}\right|=0$

$AP+PB=AB$
$\Rightarrow |z-z_1|+|z-z_2|=|z_1-z_2|.$