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The points $z_{1} , z_{2} , z_{3} , z_{4} $ in the complex plane are the vertices of a parallelogram taken in order if and only if

\[\begin{array}{1 1}(1)z_{1}+z_{4}=z_{2}+z_{3}&(2)z_{1}+z_{3}=z_{2}+z_{4}\\(3)z_{1}+z_{2}=z_{3}+z_{4}&(4)z_{1}-z_{2}=z_{3}-z_{4}\end{array}\]

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Let $z_1,z_2,z_3,z_4$ denotes the vertices of the parallelogram ABCD.
Mid point of $AC=\large\frac{z_1+z_2}{2}$
Mid point of $AC=\large\frac{z_2+z_4}{2}$
In a parallelogram diagonal intersect in the mid point
$\large\frac{z_1+z_3}{2} =\frac{z_2+z_4}{2}$
$z_1+z_3=z_2+z_4$
Hence 2 is the correct answer.
answered May 14, 2014 by meena.p
 
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