Show that two lines are perpendicular in the complex plane.

Question

I want to show that

$\overline{AB}\perp\overline{CD}$ if and only if ${\bf Re}\dfrac{a-b}{c-d}=0$, where $a, b, c, d$ are the complex numbers corresponding to the points $A$, $B$, $C$ and $D$.

I am yet to find a proof for this. Could someone give me some hints?

Answer 1

We know the angle of vector $a-b$ with $x$-axis is $\arg(a-b)$ and also the angle of vector $c-d$ with $x$-axis is $\arg(c-d)$, then they are perpendicular if $$\arg(a-b)-\arg(c-d)=\dfrac{\pi}{2}$$ or $$\arg\dfrac{a-b}{c-d}=\dfrac{\pi}{2}$$ this shows that the number $\dfrac{a-b}{c-d}$ is purely imaginary, therfore ${\bf Re}\dfrac{a-b}{c-d}=0$.

Answer 2

The lines have direction vectors $$ AB = b-a \quad CD = d-c $$ We have $$ \DeclareMathOperator{Re}{Re} \Re \frac{b-a}{d-c} = \Re \frac{r_1 e^{i \phi_1}}{r_2 e^{i \phi_2}} = \Re \frac{r_1}{r_2} e^{i(\phi_1 - \phi_2)} = \frac{r_1}{r_2} \cos(\phi_1 - \phi_2) $$ where we switched to polar coordinates.