If $z_{1}, z_{2}, z_{3}, z_{4}$ are four distinct complex numbers representing the vertices of a quadrilateral taken in order such that $z_{1} - z_{4} = z_{2} - z_{3}$ and $a r g (\frac{z_{4} - z_{1}}{z_{2} - z_{1}}) = \frac{π}{2},$ , then quadrilateral MAY BE a

rhombus

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square

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rectangle

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trapezium

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Solution

The correct options are
A rhombus
B square
C rectangle
Since $z_{1} - z_{4} = z_{2} - z_{3}$
$\Rightarrow \frac{z_{1} + z_{3}}{2} = \frac{z_{1} + z_{4}}{2}$ i.e diagonals AC and DB bisect each other.
$∴$ ABCD is a parallelogram.
Again $∠ B A D = A r g (\frac{z_{4} - z_{1}}{z_{2} - z_{1}}) = π / 2$ which means that the parallelogram is a rectangle.

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