Question

$Q\left(Z_q\right)$ is the foot of perpendicular drawn from $P\left(Z_p\right)$ to the line $a\overline{z}+z\overline{a}+b=0,bϵR$ and ‘a’ is complex number, then:

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Clearly$z_q-z_p=\lambda a\lambda ϵR\Rightarrow z_q=z_p+\lambda a$

Putting it in the equation of line ,we get
$a(\overline{z_p}+\lambda a)+\overline{a}(z_p+\lambda a)+b=0$

$\Rightarrow a\overline{z_P}+\overline{a}{z}_p+b=2\lambda a\overline{a}=0....\left(i\right)$

We also have

$\overline{z_p}\overline{a}-z_p\overline{a}-\lambda a\overline{a}=0...........\left(ii\right)$

From(i)+2(ii),we get

$2\overline{z_q}\overline{a}+a\overline{z_p}-z_p\overline{a}+b=0$