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 \(\arg (w)\) denotes the principal argument of a non-zero complex number \(w\), then

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 \(t:(1-t), 0 < t < 1\)
\( \Rightarrow A P+B P=A B \)
\( \text { ie, }\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right| \)
\( \Rightarrow \text {Option (a) is true. } \)
\( \text {and } \arg \left(z-z_1\right) =\arg \left(z_2-z\right) \)
\( =\arg \left(z_2-z_1\right) \)
\(\Rightarrow\) (b) is false and (d) is true.
Also, \(\arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)\)
\( \Rightarrow \arg \left(\frac{z-z_1}{z_2-z_1}\right)=0 \)
\( \therefore \frac{z-z_1}{z_2-z_1} \text { is purely real. } \)
\( \Rightarrow \frac{z-z_1}{z_2-z_1}=\frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1} \)
\( \text {or }|z-z_1 \bar{z}-\bar{z}_1 \)
\( z_2-z_1 \bar{z}_2-\bar{z}_1|=0 \)
\(\therefore\) Option (c) is correct.
Hence, (a, c, d) is the correct option.