[JEE Advanced 2010] let z 1 t z 1 tz 2 for real t in 0 1
Let $z=(1-t)z_1+tz_2$ for real $t\in(0,1)$ with $z_1\ne z_2$. Which are always true?
(a) $|z-z_1|+|z-z_2|=|z_1-z_2|$
(b) $\arg(z-z_1)=\arg(z-z_2)$
(c) $\begin{vmatrix}z-z_1&\bar z-\bar z_1\\z_2-z_1&\bar z_2-\bar z_1\end{vmatrix}=0$
(d) $\arg(z-z_1)=\arg(z_2-z_1)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (a) $|z-z_1|+|z-z_2|=|z_1-z_2|$
$z$ lies on segment $z_1z_2$: distances add (a), $z-z_1\parallel z_2-z_1$ (d), and the collinearity determinant vanishes (c). (b) is false (the args differ by $\pi$).
JEE Advanced 2010 · Complex Numbers — verified solution by the Vidaara Team.
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