If $A (z_{1}), B (z_{2}), C (z_{3})$ be the vertices of triangle ABC in which $| A B C - ---- -$ = $\frac{π}{4}$ and $\frac{A B}{B C}$ = $\sqrt{2}$ then $z_{2}$ is equal to

Q. If

A (z_{1}), B (z_{2}), C (z_{3})

are the vertices of the triangle ABC such that

\frac{(z_{1} - z_{2})}{(z_{3} - z_{2})} = (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})

, then triangle

A B C

If A(z₁), B(z₂) and C(z₃) are three points in the argand plane where

| z₁+ z₂ | = | |z₁| - |z₂| | and | (1 - i)z₁+ iz₃ | = | z₁| + | z₃- z₁| ,

where i = √( -1) then

(a) A, B and C lie on a fixed circle with centre (z₂+ z₃)/2

(b) A, B and C are collinear points

(d) ABC form an obtuse angle triangle

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In △ABC,A(z1),B(z2) and C(z3) are inscribed in the circle |z|=5. If H(zH) be the orthocentre of △ABC, then zH=

The correct option is B z1+z2+z3Circumcentre of △ABC is clearly origin. Let G(zG) be its centroid. Then, zG=13(z1+z2+z3) Now we know that, OG:GH=1:2⇒zG=2×0+1×zH3 ⇒zH=3zG=z1+z2+z3

In $△ A B C, A (z_{1}), B (z_{2})$ and $C (z_{3})$ are inscribed in the circle $| z | = 5$ . If $H (z_{H})$ be the orthocentre of $△ A B C$ , then $z_{H} =$