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 $\overline{AB}$ as a diameter
• B. the ellipse with $A,B$ as extremities of the major axis
• C. the perpendicular bisector of $\overline{AB}$
• 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 $z_2\left(x_2,y_2\right)$.
$\therefore z-z_1=x+iy-x_1-i{y}_1$
$=x-x_1+i\left(y-y_1\right)$
and $z-z_2=x+iy-x_2-i{y}_2$
$=x-x_2+i\left(y-y_2\right)$
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 $\pi$
$\Rightarrow \left[\frac{\left[\left(x-x_2\right)\left(y-y_1\right)\right]-\left[\left(x-x_1\right)\left(y-y_2\right)\right]}{\left[\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)\right]}\right]=0$
$\Rightarrow \left(x-x_2\right)\left(y-y_1\right)=\left(x-x_1\right)\left(y-y_2\right)$
$\Rightarrow xy-x{y}_1-x_{2}y+x_2{y}_1=xy-x{y}_2-x_{1}y+x_1{y}_2$
$\Rightarrow x\left(y_2-y_1\right)+y\left(x_1-x_2\right)+\left(x_2{y}_1-x_1{y}_2\right)=0$
It represents a straight line passing through the points $A$ of $B$