Question

# If the triangle formed by the complex coordinates A(z),B(2+3i),C(4+5i) which satisfy the relations |z−(2+3i)|=|z−(4+5i)| and |z−(3+4i)|≤4, then

A
ar. (ABC)max=42 sq. unit
B
ar. (ABC)max=4 sq. unit
C
if area of ABC is maximum, then unequal angle is <π4
D
if area of ABC is maximum, then unequal angle is π4

Solution

## The correct options are A ar. (△ABC)max=4√2 sq. unit C if area of △ABC is maximum, then unequal angle is <π4 As given |z−(2+3i)|=|z−(4+5i)| ⇒AB=AC Hence △ABC will be isosceles triangle and lenght of perpendicular from A on BC will be meet at mid point of BC D=(z1+z2)/2=3+4i Hence length of perpendicular =|z−(3+4i)|=4 (for maximum value) So required maximum area  =12×4×|z1−z2|=4√2 sq. unit Now let 2+3i−z4+5i−z=eiθ⇒tanθ/2=DCAD=12√2⇒tanθ=4√27∴arg(2+3i−z4+5i−z)=θ<π4

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