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Angle Bisector and It's Construction

If | z | = ...

Question

If $| z | = 2$ and $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - 2}{z + 2}$ then $z_{1}, z_{2}, z_{3}$ form

A

an equilateral triangle

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B

acute angle triangle

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C

right angled triangle

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D

none of these

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Solution

The correct option is C right angled triangle
$\begin{matrix} | Z | = 2, \frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - 2}{z + 2} x^{2} + y^{2} = 4 a r g f (\frac{x - 2}{x + 2}) = \frac{π}{2} ∴ (\frac{z_{1} - z_{3}}{z_{2} - z_{3}}) = (\frac{z - 2}{z + 2}) = \frac{π}{2} s o, z_{1}, z_{2}, a n d z_{3} f o r m a r i g h t a n g l e t r i a n g l e H e n c e, t h e o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

| z | = 2

\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - 2}{z + 2}

z_{1}, z_{2}, z_{3}

are vertices of a right-angled triangle.

| z | = 1

\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - i}{z + i}

z_{1}, z_{2}, z_{3}

will be vertices of a

z_{1}, z_{2}, z_{3}

be three vertices of an equilateral triangle circumscribing the circle

| z | = \frac{1}{2}

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

z_{1}, z_{2}, z_{3}

are in anticlockwise sense then

z_{2}

z_{1}, z_{2}, z_{3}

are vertices of equilateral triangle inscribed in the circle

| z | = 2

z_{1} = 1 + i \sqrt{3}

z_{2}

z_{3}

z_{1}, z_{2}, z_{3}

are the vertices of an equilateral triangle inscribed in the circle | z | = 2 . If

z_{1} = 1 + i \sqrt{3}

z_{2} = . . . . z_{3} = . . .

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