# Complex numbers in geometry

"Let $$A,B,C,D$$ be pairwise distinct points. Then $$\overline{AB}\perp\overline{CD}$$ if and only if $$\frac{d-c}{b-a}\in\mathbb{iR}$$; i.e. $$\frac{d-c}{b-a} +\overline{\bigg( \frac{d-c}{b-a}}\bigg)\ = 0.$$"

Well, what goes wrong when I say that $$\frac{d-c}{a-b} +\overline{\bigg( \frac{d-c}{a-b}}\bigg)\ = 0.$$?

And could anyone explain the difference between $$arg\Big(\frac{d-c}{b-a}\Big)$$ and $$arg\Big(\frac{d-c}{a-b}\Big)$$?

Note by Dhrubajyoti Ghosh
8 months, 3 weeks ago

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Do you know what is the difference between $$\frac{1}{i}$$ and $$\frac{ 1}{ - i }$$?

What about $$\arg \frac{1}{i}$$ and $$\arg \frac {1}{-i}$$?

Staff - 8 months, 3 weeks ago

Well, $$arg\frac{1}{i} = \frac{3\pi}{2}$$ while $$arg\frac{1}{-i} = \frac{\pi}{2}$$. I hope I am correct. But I still don't see why I should be using $$\frac{d-c}{b-a} +\overline{\bigg( \frac{d-c}{b-a}}\bigg)\ = 0$$ instead of $$\frac{d-c}{a-b} +\overline{\bigg( \frac{d-c}{a-b}}\bigg)\ =0$$

- 8 months, 3 weeks ago

Right. The point is that those numbers are different.

For example, $$(1 + i) + \overline{( -1 + i ) } = 0$$, but $$( 1 + i ) + \overline{ (1-i)} \neq 0$$.

Staff - 8 months, 3 weeks ago