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$\left(\omega \right)$ denotes the principal argument of a non-zero complex number $\omega$, 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 of t : (1-t), 0<t<1
$\Rightarrow AP+BP=ABi.e.|z-z_1|+|z-z_2|=|z_1-z_2|\Rightarrow$ 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$(z_2-z_1)$
$\Rightarrow arg\left(\frac{z-z_1}{z_2-x_1}\right)=0$
$\therefore \frac{z-z_1}{z_2-z_1}$ is purely real.
$\therefore \frac{z-z_1}{z_2-z_1}=\frac{\overline{z}-\overline{z_1}}{\overline{z_2}-\overline{z_1}}$
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. Hence, ( a, c, d ) is the correct option.