The centre of a square ABCD is at `z_0dot` If `A` is `z_1` , then the centroid of the ABC is `2z_0-(z_1-z_0)` (b) `(z_0+i((z_1-z_0)/3)` `(z_0+i z_1)/3` (d) `2/3(z_1-z_0)`

The centre of a square ABCD is at z_(0). If A is z_(1) ,then the centroid of the ABC is 2z_(0)-(z_(1)-z_(0))(b)(z_(0)+i((z_(1)-z_(0))/(3))(z_(0)+iz_(1))/(3) (d) (2)/(3)(z_(1)-z_(0))

The centre of a square ABCD is at z=0, A is z_(1) . Then, the centroid of /_\ABC is (where, i=sqrt(-1) )

The centre of a square ABCD is at z=0, A is z_(1) . Then, the centroid of /_\ABC is (where, i=sqrt(-1) )

The centre of a square ABCD is at z=0, A is z_(1) . Then, the centroid of /_\ABC is (where, i=sqrt(-1) )

The centre of a square ABCD is at z=0, A is z_(1) . Then, the centroid of /_\ABC is (where, i=sqrt(-1) )

If |z_1-z_0|=|z_2-z_0|=a and amp((z_2-z_0)/(z_0-z_1))=pi/2 , then find z_0

If |z_1-z_0|=|z_2-z_0|=a and amp((z_2-z_0)/(z_0-z_1))=pi/2 , then find z_0

If |z_1-z_0|=|z_2-z_0|=a and amp((z_2-z_0)/(z_0-z_1))=pi/2 , then find z_0

If |z_(1)-z_(0)|=|z_(2)-z_(0)|=a and amp((z_(2)-z_(0))/(z_(0)-z_(1)))=(pi)/(2), then find z_(0)