A) \[2y=x\]
B) \[y=x\]
C) y-axis
D) x-axis
Correct Answer: D
Solution :
\[{{\left| \,z+\frac{i}{2}\, \right|}^{2}}\,=\,{{\left| \,z-\frac{i}{2}\, \right|}^{2}}\]Þ \[{{\left| \,x+iy+\frac{i}{2}\, \right|}^{2}}={{\left| \,x+iy-\frac{i}{2}\, \right|}^{2}}\] Þ \[{{\left| \,x+i\left( y+\frac{1}{2} \right)\, \right|}^{2}}={{\left| \,x+i\left( y-\frac{1}{2} \right)\, \right|}^{2}}\] \[\Rightarrow \,\,\,{{x}^{2}}+{{\left( y+\frac{1}{2} \right)}^{2}}={{x}^{2}}+{{\left( y-\frac{1}{2} \right)}^{2}}\] Þ \[2y=0\] i.e. x-axis.
You need to login to perform this action.
You will be redirected in
3 sec
