Question

Two lines $\overline{a}z+a\overline{z}+c=0,\overline{b}z+b\overline{z}+d=0$ where $a,b\in C-\left\{0\right\}$ and $c,d\in R$, are

• A. Parallel if $a\overline{b}=\overline{a}b$
• B. Parallel if $\frac{a}{b}$ is real
• C. Perpendicular if $a\overline{b}+b\overline{a}=0$
• D. Perpendicular if $\frac{a}{b}$ is imaginary

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Parallel if $a\overline{b}=\overline{a}b$
B Parallel if $\frac{a}{b}$ is real
C Perpendicular if $a\overline{b}+b\overline{a}=0$
D Perpendicular if $\frac{a}{b}$ is imaginary

Let $z=x+iy$

Now $\overline{a}z+a\overline{z}+c=0$

Also $\overline{b}z+b\overline{z}+d=0$

$\Rightarrow \overline{a}(x+iy)z+a(x-iy)+c=0$

$\Rightarrow \overline{b}(x+iy)z+b(x-iy)+d=0$

$\Rightarrow x(\overline{a}+a)+iy(\overline{a}-a)+c=0$

$\Rightarrow x(\overline{b}+b)+iy(\overline{b}-b)+d=0$

For lines to be parallel

$\Rightarrow -\left(\begin{array}{c}\frac{\overline{a}+a}{\overline{a}-a}\end{array}\right)=-\left(\begin{array}{c}\frac{\overline{b}+b}{\overline{b}-a}\end{array}\right)$

$\Rightarrow \overline{a}\overline{b}-\overline{a}b+a\overline{b}-ab=\overline{a}\overline{b}+\overline{a}b-a\overline{b}-ab$

$\Rightarrow 2\overline{a}b=2a\overline{b}$

$\Rightarrow \overline{a}b=a\overline{b}$

$\Rightarrow \frac{a}{b}=\frac{\overline{a}}{\overline{b}}$

$\Rightarrow \frac{a}{b}isreal$

For lines to be perpendicular

$\Rightarrow -\left(\begin{array}{c}\frac{\overline{a}+a}{\overline{a}-a}\end{array}\right)=\left(\begin{array}{c}\frac{\overline{b}+b}{\overline{b}-a}\end{array}\right)$

$\Rightarrow \overline{a}\overline{b}+\overline{a}b+a\overline{b}+ab=-\overline{a}\overline{b}+\overline{a}b+a\overline{b}-ab$

$\Rightarrow \overline{a}\overline{b}+ab=0$

$\Rightarrow \frac{a}{b}=-\frac{\overline{a}}{\overline{b}}$

$\Rightarrow \frac{a}{b}$ is imaginary