Correct Answer - a,c,d
Given `z=((1-t)z_1+jt z_2)/((1-t)+t)`
Clearly z divides `z_1 " and " z_2` in the ratio of `t: (1-t), 0 lt t lt 1 `
`rArr AP + BP =AB ie |z-z_1|+|z+z_2|=|z_1-z_2|`
`rArr` option (a) is true
Also `arg(z-z_1)=arg(z_2-z_1)`
`rArr arg ((z-z_1) (z_2- z_1))=0`
`therefore (z_z_1)/(z_2-z_1)` is purely real .
`rArr (z-z_1)/(z_2-z_1)=(barz-z_1)/(z_2 - bar z_1)`
or `|{:(z-z_1" "barz-barz_1),(z_2-z_1" "barz_2-barz_1):}|=0`
Option (c ) is correct