If z1,z2,z3,z4 are complex numbers in an Argand plane satisfying z1+z3=z2+z4. A complex number ‘z’ lies on the line joining z1 and z4 such that Arg(z−z2z1−z2)=Arg(z3−z2z−z2) . It is given that |z−z4|=5, |z−z2|=|z−z3|=6 then area of quadrilaterial z1z2z3z4 is given by :

BC=5+l=12 cosθ 6sinθ=lsin2θ ⇒l=4 Now solve

If z1,z2,z3,z4 are complex numbers in an Argand plane satisfying z1+z3=z2+z4. A complex number ‘z’ lies on the line joining z1 and z4 such that Arg(z−z2z1−z2)=Arg(z3−z2z−z2) . It is given that |z−z4|=5, |z−z2|=|z−z3|=6 then area of quadrilaterial z1z2z3z4 is given by :

BC=5+l=12 cosθ 6sinθ=lsin2θ ⇒l=4 Now solve

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If $z_{1,} z_{2,} z_{3,} z_{4}$ are complex numbers in an Argand plane satisfying $z_{1} + z_{3} = z_{2} + z_{4}$ . A complex number ‘z’ lies on the line joining $z_{1} a n d z_{4}$ such that $A r g (\frac{z - z_{2}}{z_{1} - z_{2}}) = A r g (\frac{z_{3} - z_{2}}{z - z_{2}})$ . It is given that $| z - z_{4} | = 5, | z - z_{2} | = | z - z_{3} | = 6$ then area of quadrilaterial $z_{1} z_{2} z_{3} z_{4}$ is given by :

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$\frac{15 \sqrt{7}}{4}$

$\frac{27 \sqrt{7}}{2}$

$\frac{15 \sqrt{7}}{2}$

$\frac{9 \sqrt{7}}{2}$

answer is C.

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$B C = 5 + l = 12 \cos θ 6 \sin θ = l \sin 2 θ \Rightarrow l = 4$

Now solve

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If z1,z2,z3,z4 are complex numbers in an Argand plane satisfying z1+z3=z2+z4. A complex number ‘z’ lies on the line joining z1 and z4 such that Arg(z−z2z1−z2)=Arg(z3−z2z−z2) . It is given that |z−z4|=5, |z−z2|=|z−z3|=6 then area of quadrilaterial z1z2z3z4 is given by :

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