Question

$A\left(z_{1}\right)$ and $B\left(z_{2}\right)$ are two points in the argand plane. Then, the locus of the complex number $z$ satisfying $\arg \left(\frac{z-z_{1}}{z-z_{2}}\right)=0$ or $\pi$, is

• A. the circle with $\overline{A B}$ as a diameter
• B. the ellipse with $A, B$ as extremities of the major axis
• C. the perpendicular bisector of $\overline{A B}$
• D. the straight line passing through the points $A$ and $B$

Answer 1

Answer: (D)

Let the point $z(x, y), z_{1}\left(x_{1}, y_{1}\right)$ and $z_{2}\left(x_{2}, y_{2}\right)$.
$\therefore z-z_{1} =x+i y-x_{1}-i y_{1} $
$=x-x_{1}+i\left(y-y_{1}\right) $
and $z-z_{2}=x+i y-x_{2}-i y_{2}$
$=x-x_{2}+i\left(y-y_{2}\right)$
Now, $\frac{z-z_{1}}{z-z_{2}}$
$=\frac{\left[\left(x-x_{1}\right)+i\left(y-y_{1}\right)\right]}{\left[\left(x-x_{2}\right)+i\left(y-y_{2}\right)\right]} \times \frac{\left[\left(x-x_{2}\right)-i\left(y-y_{2}\right)\right]}{\left[\left(x-x_{2}\right)-i\left(y-y_{2}\right)\right]}$
$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)+$
$=\frac{i\left[\left(x-x_{2}\right)\left(y-y_{1}\right)-\left(x-x_{1}\right)\left(y-y_{2}\right)\right]}{\left(x-x_{2}\right)^{2}+\left(y-y_{2}\right)^{2}}$
=Arg $\left(\frac{z-z_{1}}{z-z_{2}}\right)=0$ or $\pi$
$\Rightarrow \left[\frac{\left[\left(x-x_{2}\right)\left(y-y_{1}\right)\right]-\left[\left(x-x_{1}\right)\left(y-y_{2}\right)\right]}{\left[\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)\right]}\right]=0$
$\Rightarrow \left(x-x_{2}\right)\left(y-y_{1}\right)=\left(x-x_{1}\right)\left(y-y_{2}\right)$
$\Rightarrow x y-x y_{1}-x_{2} y+x_{2} y_{1}=x y-x y_{2}-x_{1} y+x_{1} y_{2}$
$\Rightarrow x\left(y_{2}-y_{1}\right)+y\left(x_{1}-x_{2}\right)+\left(x_{2} y_{1}-x_{1} y_{2}\right)=0$
It represents a straight line passing through the points $A$ of $B$