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-3-1-0-

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!