"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 \(arg\Big(\frac{d-c}{a-b}\Big)\)?

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## Comments

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TopNewestDo you know what is the difference between \( \frac{1}{i} \) and \( \frac{ 1}{ - i } \)?

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

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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 \( \frac{d-c}{a-b} +\overline{\bigg( \frac{d-c}{a-b}}\bigg)\ =0\)

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Right. The point is that those numbers are different.

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

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