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 num ber 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

The Correct Option is D

Given , z=$\frac{(1-t)z_1+tz_2}{(1-t)+t}$ Clearly, 2 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 a rg (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.