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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 determinant ((z_1 ,z_1 ^� ,1),(z_2 ,z_2 ^� ,1),(z_3 ,z_3 ^� ,1))= 0
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{three}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{collinear}\:\mathrm{if}\:\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0} \\ $$
Answered by dioph last updated on 12/Aug/17
determinant ((z_1 ,z_1 ^� ,1),(z_2 ,z_2 ^� ,1),(z_3 ,z_3 ^� ,1))= 0 ⇒ ⇒ z_1 z_2 ^� + z_1 ^� z_3 + z_2 z_3 ^� = z_1 ^� z_2 + z_1 z_3 ^� + z_2 ^� z_3 ⇒ (z_1 z_2 ^� −z_1 ^� z_2 )+(z_1 ^� z_3 −z_1 z_3 ^� )+(z_2 z_3 ^� −z_2 ^� z_3 )=0 α_n + iβ_n := z_n ⇒ (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 ⇒ (α_2 −α_3 )β_1 +(α_3 −α_1 )β_2 +(α_1 −α_2 )β_3 = 0 Δα_(nm) := α_n −α_m , Δβ_(nm) := β_n −β_m ⇒ Δα_(31) (β_1 +Δβ_(21) )=Δα_(21) (β_1 +Δβ_(31) )+(Δα_(31) −Δα_(21) )β_1 ⇒ Δα_(31) Δβ_(21) = Δα_(21) Δβ_(31) ⇒((Δα_(31) )/(Δβ_(31) )) = ((Δα_(21) )/(Δβ_(21) )) ■
$$\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \:+\:\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{3}} \:+\:{z}_{\mathrm{2}} \bar {{z}}_{\mathrm{3}} \:=\:\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} +\:{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{3}} +\:\bar {{z}}_{\mathrm{2}} {z}_{\mathrm{3}} \\ $$$$\Rightarrow\:\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} −\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} \right)+\left(\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{3}} −{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{3}} \right)+\left({z}_{\mathrm{2}} \bar {{z}}_{\mathrm{3}} −\bar {{z}}_{\mathrm{2}} {z}_{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\alpha_{{n}} \:+\:{i}\beta_{{n}} \::=\:{z}_{{n}} \\ $$$$\Rightarrow\:\left(\alpha_{\mathrm{2}} \beta_{\mathrm{1}} −\alpha_{\mathrm{1}} \beta_{\mathrm{2}} \right)+\left(\alpha_{\mathrm{1}} \beta_{\mathrm{3}} −\alpha_{\mathrm{3}} \beta_{\mathrm{1}} \right)+\left(\alpha_{\mathrm{3}} \beta_{\mathrm{2}} −\alpha_{\mathrm{2}} \beta_{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\left(\alpha_{\mathrm{2}} −\alpha_{\mathrm{3}} \right)\beta_{\mathrm{1}} +\left(\alpha_{\mathrm{3}} −\alpha_{\mathrm{1}} \right)\beta_{\mathrm{2}} +\left(\alpha_{\mathrm{1}} −\alpha_{\mathrm{2}} \right)\beta_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\Delta\alpha_{{nm}} :=\:\alpha_{{n}} −\alpha_{{m}} ,\:\Delta\beta_{{nm}} \::=\:\beta_{{n}} −\beta_{{m}} \\ $$$$\Rightarrow\:\Delta\alpha_{\mathrm{31}} \left(\beta_{\mathrm{1}} +\Delta\beta_{\mathrm{21}} \right)=\Delta\alpha_{\mathrm{21}} \left(\beta_{\mathrm{1}} +\Delta\beta_{\mathrm{31}} \right)+\left(\Delta\alpha_{\mathrm{31}} −\Delta\alpha_{\mathrm{21}} \right)\beta_{\mathrm{1}} \\ $$$$\Rightarrow\:\Delta\alpha_{\mathrm{31}} \Delta\beta_{\mathrm{21}} \:=\:\Delta\alpha_{\mathrm{21}} \Delta\beta_{\mathrm{31}} \\ $$$$\Rightarrow\frac{\Delta\alpha_{\mathrm{31}} }{\Delta\beta_{\mathrm{31}} }\:=\:\frac{\Delta\alpha_{\mathrm{21}} }{\Delta\beta_{\mathrm{21}} }\:\blacksquare \\ $$
Commented by Tinkutara last updated on 12/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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