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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 )

Ifz1,z2andz3aretheverticesofaright−angledisos−celestriangledescribedincounterclocksenseandrightangledatz3,then(z1−z2)2isequalto(A)(z1−z3)(z3−z2)(B)2(z1−z3)(z3−z2)(C)3(z1−z3)(z3−z2)(D)3(z3−z1)(z3−z2)

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)

letz1−z3=reθi⇒z3−z2=re(θ−π2)i⇒z1−z2=2re(θ−π4)i(z1−z2)2=2r2e(2θ−π2)i=2reθire(θ−π2)i=2(z1−z3)(z3−z2)⇒answer(B)

Commented by EnterUsername last updated on 02/May/21

Thank You, Sir

ThankYou,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.

PleaseSir,howdidyougettheanglebetweenz1−z2andthehorizontalaxesorigintobeθ−π4?I′mnotabletoeasilynoticethat.

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

Commented by EnterUsername last updated on 02/May/21

Oh wow ! Thanks a lot.

Ohwow!Thanksalot.

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