Let `z_(1)`, `z_(2)`, `z_(3)` are the vertices of `DeltaABC`, respectively, such that `(z_(3)-z_(2))/(z_(1)-z_(2))` is purely imaginery number. A square on side `AC` is drawn outwardly. `P(z_(4))` is the centre of square, then

Options

A. `|z_(1)-z_(2)|=|z_(2)-z_(4)|`
B. `arg((z_(1)-z_(2))/(z_(4)-z_(2)))+arg((z_(3)-z_(2))/(z_(4)-z_(2)))=+(pi)/(2)`
C. `arg((z_(1)-z_(2))/(z_(4)-z_(2)))+arg((z_(3)-z_(2))/(z_(4)-z_(2)))=0`
D. `z_(1)`, `z_(2)`, `z_(3)` and `z_(4)` lie on a circle

Answer

C, D

Explanation

`(c,d)` `P` is centre of square `implies AP=PC` and `/_APC=(pi)/(2)`
`/_ABC` is also `(pi)/(2)`.
`implies ABCP` is cyclic quadrilateral
`impliesA,B,C,P` lie on a circle.
`implies` In circl chord, `AP=PC`
`implies` arc `AP=` are `PC`
`implies/_ABP=/_PBC`
`implies BP` is angular bisector.
Let `arg((z_(1)-z_(2))/(z_(4)-z_(2)))=-alpha`........`(i)`
`impliesarg((z_(3)-z_(2))/(z_(4)-z_(2)))=alpha`..........`(ii)`
Add `(i)` and `(ii)`, `arg((z_(1)+z_(2))/(z_(4)-z_(2)))+arg((z_(3)-z_(2))/(z_(4)-z_(2)))=0`