Question

Let $A\left(z_1\right)$ and $B\left(z_2\right)$represent two complex numbers on the complex plane. Suppose the complex slope of the line $l_1$ joining $A$ and $B$ is defined as $\frac{(z_1-z_2)}{(\overline{z_1}-\overline{z_2})}$. If the line $l_1$ with complex slope ${\omega }_1$ and $l_2$ with complex slope ${\omega }_2$ on the complex plane are perpendicular then

• A. ${\omega }_1+{\omega }_2=-1$
• B. ${\omega }_1+{\omega }_2=0$
• C. ${\omega }_1+{\omega }_2=1$
• D. ${\omega }_1\times {\omega }_2=-1$

Answer 1

The correct option is B ${\omega }_1+{\omega }_2=0$



Let $C\left(z_3\right)$ and $D\left(z_4\right)$ be $2$ points on $l_2$.
$l_1$ is perpendicular to $l_2$
Using the concept of rotation we can say that
$\frac{z_1-z_2}{|z_1-z_2|}=\left(\frac{z_3-z_4}{|z_3-z_4|}\right)e^{\pm i\pi /2}$
$\Rightarrow \frac{z_1-z_2}{z_3-z_4}=\left|\frac{z_1-z_2}{z_3-z_4}\right|e^{\pm i\pi /2}$
$\Rightarrow \frac{z_1-z_2}{z_3-z_4}=\lambda (\pm i)$
Clearly, $\frac{z_1-z_2}{z_3-z_4}$ is purely imaginary. So,
$\frac{z_1-z_2}{z_3-z_4}+\frac{\overline{z_1}-\overline{z_2}}{\overline{z_3}-\overline{z_4}}=0$
$\frac{z_1-z_2}{\overline{z_1}-\overline{z_2}}+\frac{z_3-z_4}{\overline{z_3}-\overline{z_4}}=0$
$\Rightarrow {\omega }_1+{\omega }_2=0$