If `A(z_(1))`, `B(z_(2))`, `C(z_(3))` are vertices of a triangle such that `z_(3)=(z_(2)-iz_(1))/(1-i)` and `|z_(1)|=3`, `|z_(2)|=4` and `|z_(2)+iz_(1)|=|z_(1)|+|z_(2)|`, then area of triangle `ABC` is

Options

A. `(5)/(2)`
B. `0`
C. `(25)/(2)`
D. `(25)/(4)`

Answer

D

Explanation

`(d)` `|z_(2)+iz_(1)|=|z_(1)|+|z_(2)|impliesz_(2)`, `iz_(1)`, `o` are collinear
`:.arg(iz_(1))=argz_(2)`
`impliesargi+argz_(1)=argz_(2)`
`impliesargz_(2)-argz_(1)=(pi)/(2)`
`z_(3)=(z_(2)-iz_(1))/(1-i)`
`implies(1-i)z_(3)=z_(2)-iz_(1)`
`impliesz_(3)-z_(2)=i(z_(3)-z_(1))`
`:.(z_(3)-z_(2))/(z_(3)-z_(1))=i`
`impliesarg((z_(3)-z_(2))/(z_(3)-z_(1)))=(pi)/(2)` and `|z_(3)-z_(2)|=|z_(3)-z_(1)|`
`:.AC=BC` and `AB^(2)=AC^(2)+BC^(2)`
`impliesAC=(5)/(sqrt(2))`
Required area `=(1)/(2)xx(5)/(sqrt(2))xx(5)/(sqrt(2))=(25)/(4)`sq. units