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Question
If the complex numbers z1,z2,z3 are the vertices A,B,C respectively of an isosceles right angled triangle with right angle at C, then (z1−z2)2=k(z1−z2)(z3−z2), where k=
If the complex numbers z1,z2,z3 are the vertices A,B,C respectively of an isosceles right angled triangle with right angle at C, then (z1−z2)2=k(z1−z2)(z3−z2), where k=
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Solution
The correct option is C 2
Since |CA|=|CB| and ∠ACB=900
∴(z2−z3)=±i(z1−z3)⇒(z2−z3)2=−(z1−z3)2
⇒z22+z32−2z1z2=−z12−z33−2z1z2
⇒z12+z22−2z1z2=2(z1z2−z1z3−z2z3+z22)
⇒(z1−z2)2=2(z1−z3)(z3−z2)
∴k=2
Since |CA|=|CB| and ∠ACB=900
∴(z2−z3)=±i(z1−z3)⇒(z2−z3)2=−(z1−z3)2
⇒z22+z32−2z1z2=−z12−z33−2z1z2
⇒z12+z22−2z1z2=2(z1z2−z1z3−z2z3+z22)
⇒(z1−z2)2=2(z1−z3)(z3−z2)
∴k=2
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