Complex Numbers
Argument – Collinearity and Segment Division
Complex Numbers_PYQ
Grade 11
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 number $w$, then
$|z-z_1|+|z-z_2|=|z_1-z_2|$
$\arg(z-z_1)=\arg(z-z_2)$
$\left|\begin{matrix}z-z_1 & \bar{z}-\bar{z}_1\\z_2-z_1 & \bar{z}_2-\bar{z}_1\end{matrix}\right|=0$
$\arg(z-z_1)=\arg(z_2-z_1)$
Step-by-Step Solution
Key Concept: When $z$ lies on the open segment $z_1z_2$, $z-z_1$ is a positive real multiple of $z_2-z_1$, immediately giving (a), (c), (d). The vector $z-z_2$ points in the opposite direction, making (b) false.
**Step 1: Identify the geometric meaning**
$z=(1-t)z_1+tz_2$ with $0<t<1$ places $z$ strictly between $z_1$ and $z_2$ on the segment. Explicitly: $z-z_1=t(z_2-z_1)$ and $z-z_2=-(1-t)(z_2-z_1)$.
**Step 2: Check option (a)**
$|z-z_1|+|z-z_2|=t|z_2-z_1|+(1-t)|z_2-z_1|=|z_1-z_2|$. TRUE ✓
**Step 3: Check option (b)**
$\arg(z-z_1)=\arg(z_2-z_1)$ but $\arg(z-z_2)=\arg(z_2-z_1)+\pi$ (opposite direction). So $\arg(z-z_1)\neq\arg(z-z_2)$. FALSE ✗
**Step 4: Check option (c)**
The determinant $(z-z_1)(\bar{z}_2-\bar{z}_1)-(\bar{z}-\bar{z}_1)(z_2-z_1)=t(z_2-z_1)\cdot t\overline{(z_2-z_1)}-t\overline{(z_2-z_1)}\cdot t(z_2-z_1)=0$. TRUE ✓
**Step 5: Check option (d)**
$z-z_1=t(z_2-z_1)$ with $t>0$ real, so $\arg(z-z_1)=\arg(z_2-z_1)$. TRUE ✓
Correct Answer: 1, 3, 4