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

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

Aicharya · Accepted Answer

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{aligned} & \Rightarrow \quad A P+B P=A B \ & 	ext { ie, }\left|z-z_1ight|+\left|z-z_2ight|=\left|z_1-z_2ight| \ & \Rightarrow 	ext { Option (a) is true. } \ & 	ext { and } \begin{aligned} \arg \left(z-z_1ight) & =\arg \left(z_2-zight) \ & =\arg \left(z_2-z_1ight) \end{aligned} \end{aligned} $ $\Rightarrow$ (b) is false and (d) is true. Also, $\arg \left(z-z_1ight)=\arg \left(z_2-z_1ight)$ $ \begin{aligned} & \Rightarrow \quad \arg \left(\frac{z-z_1}{z_2-z_1}ight)=0 \ & 	herefore \frac{z-z_1}{z_2-z_1} 	ext { is purely real. } \ & \Rightarrow \quad \frac{z-z_1}{z_2-z_1}=\frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1} \ & 	ext { 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}ight|=0 \ & \end{aligned} $ $	herefore$ Option (c) is correct. Hence, (a, c, d) is the correct option

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 $t$…

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