If z2+iz1=z1+z2 and z1=3 and z2=4, then the area of △ABC, if affixes of A, B, and C are z1, z2, and [(z2 – iz1)/(1 – i)] respectively, is

If z2+iz1=z1+z2 and z1=3 and z2=4, then the area of ABC, if affixes of A, B, and C are z1, z2, and [(z2 - iz1)/(1 - i)] respectively, is

  1. A

    52

  2. B

    0

  3. C

    252

  4. D

    254

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    Solution:

    z2+iz1=z1+z2=iz1+z2

    iz1,0+i0 and z2 are collinear 

     argiz1=argz2 argz2argz1=π2

    Let z3=z2iz11i

    or   (1i)z3=z2iz1or   z2z3=iz1z3

     ACB=π2

    and z1z3=z2z3

        AC=BC    AB2=AC2+BC2    AC=52                   (AB=5)

    Therefore, area of ABC is (1/2)AC×BC=AC2/2=25/4 sq. units

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