Prove that three points z_1 , z_2 , z_3 are collinear if determinant ((z_1 ,z_1 ^� ,1),(z_2 ,z_2 ^� ,1),(z_3 ,z


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Question Number 19507 by Tinkutara last updated on 12/Aug/17

$Prove that three points z_{1}, z_{2}, z_{3} are$ $collinear if \| \begin{matrix} z_{1} & {\bar{z}}_{1} & 1 \\ z_{2} & {\bar{z}}_{2} & 1 \\ z_{3} & {\bar{z}}_{3} & 1 \end{matrix} \| = 0$
	Answered by dioph last updated on 12/Aug/17

	$\| \begin{matrix} z_{1} & {\bar{z}}_{1} & 1 \\ z_{2} & {\bar{z}}_{2} & 1 \\ z_{3} & {\bar{z}}_{3} & 1 \end{matrix} \| = 0 \Rightarrow$ $\Rightarrow z_{1} {\bar{z}}_{2} + {\bar{z}}_{1} z_{3} + z_{2} {\bar{z}}_{3} = {\bar{z}}_{1} z_{2} + z_{1} {\bar{z}}_{3} + {\bar{z}}_{2} z_{3}$ $\Rightarrow (z_{1} {\bar{z}}_{2} - {\bar{z}}_{1} z_{2}) + ({\bar{z}}_{1} z_{3} - z_{1} {\bar{z}}_{3}) + (z_{2} {\bar{z}}_{3} - {\bar{z}}_{2} z_{3}) = 0$ $α_{n} + i β_{n} := z_{n}$ $\Rightarrow (α_{2} β_{1} - α_{1} β_{2}) + (α_{1} β_{3} - α_{3} β_{1}) + (α_{3} β_{2} - α_{2} β_{3}) = 0$ $\Rightarrow (α_{2} - α_{3}) β_{1} + (α_{3} - α_{1}) β_{2} + (α_{1} - α_{2}) β_{3} = 0$ $Δ α_{n m} := α_{n} - α_{m}, Δ β_{n m} := β_{n} - β_{m}$ $\Rightarrow Δ α_{31} (β_{1} + Δ β_{21}) = Δ α_{21} (β_{1} + Δ β_{31}) + (Δ α_{31} - Δ α_{21}) β_{1}$ $\Rightarrow Δ α_{31} Δ β_{21} = Δ α_{21} Δ β_{31}$ $\Rightarrow \frac{Δ α_{31}}{Δ β_{31}} = \frac{Δ α_{21}}{Δ β_{21}} ◼$
	Commented by Tinkutara last updated on 12/Aug/17

	$Thank you very much Sir!$
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