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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` isA. `(5)/(2)`B. `0`C. `(25)/(2)`D. `(25)/(4)` |
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Answer» Correct Answer - D `(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 |
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