The centre of a square ABCD is at orign and the point A is represented by $z_{1}$ The centroid of triangle BCD is represented by

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$\frac{z_{1}}{3}$

$- \frac{z_{1}}{3}$

$\frac{i z_{1}}{3}$

$- \frac{i z_{1}}{3} .$

answer is B.

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Detailed Solution

Given A is represented by $z_{1} .$ Since the diagonals of a square bisect cach other at right angles, therefore, $∠ A O B = ∠ B O C = ∠ C O D = \frac{π}{2}$

And hence $B \equiv i z_{1}, c \equiv i^{2} z_{1}$ and $D \equiv i^{3} z_{1}$

$∴$ Centroid of triangle BCD is represented by

$\frac{i z_{1} - z_{1} - i z_{1}}{3} = - \frac{z_{1}}{3} .$

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