In an Agrad plane `z_(1),z_(2) and z_(3)` are, respectively, the vertices of an isosceles trinagle ABC with AC= BC and `/_CAB = theta`. If `z_(4)` is incentre of triangle, then
The value of `(z_(2)-z_(1))^(2)tan theta tan theta //2` is

Options

A. `(z_(1) + z_(2)-2z_(3))`
B. `(z_(1)+z_(2)-z_(3))(z_(1) +z_(2)-z_(4))`
C. `-(z_(1) +z_(2)-2z_(3))(z_(1)+z_(2)-2z_(4))`
D. `z_(4)=sqrt(z_(2)z_(3))`

Answer

C

Explanation

Keeping in mind that `tan theta = CD//AD and tan theta//2 = ID//BD`, we have
`(z_(3) - (z_(1)+z_(2))/(2))/(z_(1) - (z_(1)+z_(2))/(2))=(|z_(3)-(z_(1)+z_(2))/(2)|)/(|z_(1) -(z_(1) +z_(2))/(2)|)e^(-i(pi)/(2)`
`rArr (2z_(3) -z_(1) -z_(2))/(z_(1)-z_(2)) =(CD)/(AD)e^(-i(pi)/(2))" "(1)`
`and (z_(4) -(z_(1)+z_(2))/(4))/(z_(2) -(z_(1)+z_(2))/(2))=(|z_(4) -(z_(1) + z_(2))/(2)|)/(|z_(2) -(z_(1) +z_(2))/(2)|)`
`rArr (2z_(4) -z_(1) -z_(2))/(z_(2)-z_(1)) = (ID)/(BD) e^(-i(pi)/(2))" "(2)`
Multiplying (1) and(2) we have,
`(2z_(3)-z_(1) -z_(2))/(z_(1)z_(2))( 2z_(4) -z_(1) -z_(2))/(z_(2)-z_(1)) = (CD)/(AD)(ID)/(BD) = tan theta tan .(theta)/(2)`
`rArr (z_(2) -z_(1))^(2) tan theta tan.(theta)/(2) = - (z_(1) +z_(2)+2z_(3))(z_(1) + z_(2)+2z_(4))`