If A z 1- B z 2- C z 3 be the vertices of triangle ABC in which - ABC -4 and AB - BC -2 then z 2 is equal toA- z3-iz1-z3B- z3-iz1-z3C- z3-iz1-z3D- z3-iz2-z1

The correct option is B z_1 z_2/z_3 z_2 = √(2)e^iπ/4 ...

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Byju's Answer

Standard XI

Mathematics

Distance Formula

If A z 1, B z...

Question

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

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Solution

The correct option is B

$\frac{z_{1} - z_{2}}{z_{3} - z_{2}}$ = $\sqrt{2} e^{\frac{i π}{4}}$

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Similar questions

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

are the vertices of the triangle

A B C

(in anticlockwise order). If

∠ A B C = \frac{π}{4}

and

A B = \sqrt{2} (B C)

, then

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

are the vertices of the triangle ABC (in anticlockwise order). If

∠ A B C = π / 4

and

A B = \sqrt{2} (B C)

Then prove that

z_{2} = z_{3} + i (z_{1} - z_{3}) .

Let $A (Z_{1}), B (Z_{2}), C (Z_{3})$ be the vertices of an equilateral triangle ABC, then the value of $a r g (\frac{Z_{2} + Z_{3} - 2 Z_{1}}{Z_{3} - Z_{2}})$ 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

Q. If

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

are vertices of

Δ A B C

in which

∠ A B C = \frac{π}{4}

and

\frac{A B}{B C} = \sqrt{2}

, then find

z_{2}

in terms of

z_{1}

and

z_{3}

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If A(z1),B(z2),C(z3) be the vertices of triangle ABC in which |ABC–––––– = π4 and ABBC = √2 then z2 is equal to

The correct option is B z1−z2z3−z2 = √2eiπ4

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

The correct option is B

$\frac{z_{1} - z_{2}}{z_{3} - z_{2}}$ = $\sqrt{2} e^{\frac{i π}{4}}$