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Question Number 139838 by EnterUsername last updated on 01/May/21

If z_1 , z_2 and z_3 are the vertices of a right-angled isos- celes triangle described in counter clock sense and right angled at z_3 , then (z_1 −z_2 )^2 is equal to (A) (z_1 −z_3 )(z_3 −z_2 ) (B) 2(z_1 −z_3 )(z_3 −z_2 ) (C) 3(z_1 −z_3 )(z_3 −z_2 ) (D) 3(z_3 −z_1 )(z_3 −z_2 )

$$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}-\mathrm{angled}\:\mathrm{isos}- \\ $$$$\mathrm{celes}\:\mathrm{triangle}\:\mathrm{described}\:\mathrm{in}\:\mathrm{counter}\:\mathrm{clock}\:\mathrm{sense}\:\mathrm{and} \\ $$$$\mathrm{right}\:\mathrm{angled}\:\mathrm{at}\:{z}_{\mathrm{3}} ,\:\mathrm{then}\:\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\: \\ $$$$\left(\mathrm{A}\right)\:\left({z}_{\mathrm{1}} −{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left({z}_{\mathrm{1}} −{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left({z}_{\mathrm{1}} −{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{3}\left({z}_{\mathrm{3}} −{z}_{\mathrm{1}} \right)\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right) \\ $$

Answered by mr W last updated on 02/May/21

Commented by mr W last updated on 02/May/21

let z_1 −z_3 =re^(θi) ⇒z_3 −z_2 =re^((θ−(π/2))i) ⇒z_1 −z_2 =(√2)re^((θ−(π/4))i) (z_1 −z_2 )^2 =2r^2 e^((2θ−(π/2))i) =2re^(θi) re^((θ−(π/2))i) =2(z_1 −z_3 )(z_3 −z_2 ) ⇒answer (B)

$${let}\:{z}_{\mathrm{1}} −{z}_{\mathrm{3}} ={re}^{\theta{i}} \\ $$$$\Rightarrow{z}_{\mathrm{3}} −{z}_{\mathrm{2}} ={re}^{\left(\theta−\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\Rightarrow{z}_{\mathrm{1}} −{z}_{\mathrm{2}} =\sqrt{\mathrm{2}}{re}^{\left(\theta−\frac{\pi}{\mathrm{4}}\right){i}} \\ $$$$\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} {e}^{\left(\mathrm{2}\theta−\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{re}^{\theta{i}} {re}^{\left(\theta−\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left({z}_{\mathrm{1}} −{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{answer}\:\left({B}\right) \\ $$

Commented by EnterUsername last updated on 02/May/21

Thank You, Sir

$$\mathrm{Thank}\:\mathrm{You},\:\mathrm{Sir} \\ $$

Commented by EnterUsername last updated on 02/May/21

Please Sir, how did you get the angle between z_1 −z_2 and the horizontal axes origin to be θ−(π/4) ? I′m not able to easily notice that.

$$\mathrm{Please}\:\mathrm{Sir},\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{horizontal}\:\mathrm{axes}\:\mathrm{origin}\:\mathrm{to}\:\mathrm{be}\:\theta−\frac{\pi}{\mathrm{4}}\:?\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{able}\:\mathrm{to} \\ $$$$\:\mathrm{easily}\:\mathrm{notice}\:\mathrm{that}. \\ $$

Commented by mr W last updated on 02/May/21

Commented by EnterUsername last updated on 02/May/21

Oh wow ! Thanks a lot.

$$\mathrm{Oh}\:\mathrm{wow}\:!\:\mathrm{Thanks}\:\mathrm{a}\:\mathrm{lot}. \\ $$

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