Question

Let $(2+i)z+(2-i)\overline{z}=\lambda ,\lambda ϵR$, be a straight line in the complex plane. If $A\left(z_1\right)$ and $B\left(z_2\right)$ are 2 points in the plane such that AB is perpendicular to the given line and also the midpoint of AB lies on the given line, then $\lambda$ is equal to

• A. $(2+i)z_1+(2-i)\overline{z_2}$
• B. $(2-i)\overline{z_1}+(2+i)z_2$
• C. $(2+i)z_1+(2-i)z_2$
• D. $(2-i)\overline{z_1}+(2+i)\overline{z_2}$

Answer 1

Answer: (A), (B)

The correct options are
A $(2+i)z_1+(2-i)\overline{z_2}$
B $(2-i)\overline{z_1}+(2+i)z_2$

Given line is perpendicular to AB.

$\therefore$ Its equation is $|z-z_1|=|z-z_2|$
$\Rightarrow z(\overline{z_1}-\overline{z_2})+\overline{z}(z_1-z_2)=|z_1{|}^2-|z_2{|}^2\dots \dots \left(i\right)$
$(2+i)z+(2-i)\overline{z}=\lambda \dots \dots \left(ii\right)$
From (i) and (ii)
$\frac{\overline{z_1}-\overline{z_2}}{2+i}=\frac{z_1-z_2}{2-i}=\frac{|z_1{|}^2-|z_2{|}^2}{\lambda }=k$
$\Rightarrow \lambda =\frac{|z_1{|}^2-|z_2{|}^2}{k}$
$2+i=\frac{\overline{z_1}-\overline{z_2}}{k},2-i=\frac{z_1-z_2}{k}$
$\therefore (2+i)z_1+(2-i)\overline{z_2}=\frac{\overline{z_1}-\overline{z_2}}{k}{z}_1+\left(\frac{z_1-z_2}{k}\right)\overline{z_2}=\frac{|z_1{|}^2-|z_2{|}^2}{k}=\lambda$
Similarly conjugate of (iii) gives a solution