Question Number 19507 by Tinkutara last updated on 12/Aug/17
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{three}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{collinear}\:\mathrm{if}\:\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0} \\ $$
Answered by dioph last updated on 12/Aug/17
$$\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \:+\:\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{3}} \:+\:{z}_{\mathrm{2}} \bar {{z}}_{\mathrm{3}} \:=\:\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} +\:{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{3}} +\:\bar {{z}}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow\:\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} −\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} \right)+\left(\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{3}} −{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{3}} \right)+\left({z}_{\mathrm{2}} \bar {{z}}_{\mathrm{3}} −\bar {{z}}_{\mathrm{2}} {z}_{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\alpha_{{n}} \:+\:{i}\beta_{{n}} \::=\:{z}_{{n}} \\ $$$$\Rightarrow\:\left(\alpha_{\mathrm{2}} \beta_{\mathrm{1}} −\alpha_{\mathrm{1}} \beta_{\mathrm{2}} \right)+\left(\alpha_{\mathrm{1}} \beta_{\mathrm{3}} −\alpha_{\mathrm{3}} \beta_{\mathrm{1}} \right)+\left(\alpha_{\mathrm{3}} \beta_{\mathrm{2}} −\alpha_{\mathrm{2}} \beta_{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\left(\alpha_{\mathrm{2}} −\alpha_{\mathrm{3}} \right)\beta_{\mathrm{1}} +\left(\alpha_{\mathrm{3}} −\alpha_{\mathrm{1}} \right)\beta_{\mathrm{2}} +\left(\alpha_{\mathrm{1}} −\alpha_{\mathrm{2}} \right)\beta_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\Delta\alpha_{{nm}} :=\:\alpha_{{n}} −\alpha_{{m}} ,\:\Delta\beta_{{nm}} \::=\:\beta_{{n}} −\beta_{{m}} \\ $$$$\Rightarrow\:\Delta\alpha_{\mathrm{31}} \left(\beta_{\mathrm{1}} +\Delta\beta_{\mathrm{21}} \right)=\Delta\alpha_{\mathrm{21}} \left(\beta_{\mathrm{1}} +\Delta\beta_{\mathrm{31}} \right)+\left(\Delta\alpha_{\mathrm{31}} −\Delta\alpha_{\mathrm{21}} \right)\beta_{\mathrm{1}} \\ $$$$\Rightarrow\:\Delta\alpha_{\mathrm{31}} \Delta\beta_{\mathrm{21}} \:=\:\Delta\alpha_{\mathrm{21}} \Delta\beta_{\mathrm{31}} \\ $$$$\Rightarrow\frac{\Delta\alpha_{\mathrm{31}} }{\Delta\beta_{\mathrm{31}} }\:=\:\frac{\Delta\alpha_{\mathrm{21}} }{\Delta\beta_{\mathrm{21}} }\:\blacksquare \\ $$
Commented by Tinkutara last updated on 12/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$