Mathematics - Complex Number Question with Solution | TestHub

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 $t:(1-t),0<t<1$ $$\begin{array}{rl}& \Rightarrow AP+BP=AB\\ & \text{ ie, }\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|\\ & \Rightarrow \text{ Option (a) is true. }\\ & \text{ and }\begin{array}{rl}\arg \left(z-z_1\right)& =\arg \left(z_2-z\right)\\ & =\arg \left(z_2-z_1\right)\end{array}\end{array}$$ $\Rightarrow$ (b) is false and (d) is true. Also, $\arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)$ $$\begin{array}{rl}& \Rightarrow \arg \left(\frac{z-z_1}{z_2-z_1}\right)=0\\ & \therefore \frac{z-z_1}{z_2-z_1}\text{ is purely real. }\\ & \Rightarrow \frac{z-z_1}{z_2-z_1}=\frac{\bar{z}-{\bar{z}}_1}{{\bar{z}}_2-{\bar{z}}_1}\\ & \text{ or }\left|\begin{array}{cc}z-z_1& \bar{z}-{\bar{z}}_1\\ z_2-z_1& {\bar{z}}_2-{\bar{z}}_1\end{array}\right|=0\\ & \end{array}$$ $\therefore$ Option (c) is correct. Hence, (a, c, d) is the correct option.