If `|z_2+i z_1|=|z_1|+|z_2|a n d|z_1|=3a n d|z_2|=4,` then the area of ` A B C ,` if affixes of `A ,B ,a n dCa r ez_1, z_2, a n d[(z_2-i z_1)//(1-i)]` respectively, is `5/2` b. `0` c.`(25)/2` d. `(25)/4`

Options

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

Answer

D

Explanation

`|z_(2) +iz_(1)| =|z_(1)|+|z_(2)|=|iz_(1)| + |z_(2)|`
`rArr iz_(1), 0 + i0` and `z_(2)` are collinear

`rArr arg(iz_(2))= agr(z_(2))`
`rArr arg(z_(2))-agr(z_(1))= (pi)/(2)`
Let `z_(3)=(z_(2) - iz_(1))`
or `(1-i)z_(3) = z_(2) - iz_(1)`
or `z_(2) -z_(3) = i(z_(1) - z_(3))`
`therefore/_ACB = (pi)/(2)`
and ` |z_(1) - z_(3)| = |z_(2)-Z_(3)|`
`rArr Ac = Bc`
` therefore AB^(2) = AC^(2) + BC^(2)`
`rArr Ac = (5)/(sqrt(2))`
Therefore, area of `DeltaABC` is `(1//2)AC xxBC= AC^(2)//2= 25//4`sq.units