If the area of a triangle with vertices Z1, Z2 and Z3 is the absolute value of the number λiZ1 Z¯1 1Z2 Z¯2 1Z3 Z¯3 1 then the value of 1/λ is equal to ______.

Let z1=x1+iy1,z2=x2+iy2 and z3=x3+iy3 be the vertices.Consider z1z¯11z2z¯21z3z¯31=x1+iy1 x1−iy1 1x2+iy2 x2−iy2 1x3+iy3 x3−iy3 1C1→C1+C2=2x1 x1−iy1 1x2 x2−iy2 1x3 x3−iy3 1C2→C2−C1=x1y11x2y21x3y31=(−4i)12||x1y11x2y21x3y31∣=−4i× Area of triangle. ∴Area of triangle =i4z1 z¯1 1z2 z¯2 1z3 z¯3 1;∴λ=14⇒1λ=4

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If the area of a triangle with vertices Z₁, Z₂ and Z₃ is the absolute value of the number $λi |\begin{array}{lll} Z_{1} {\bar{Z}}_{1} 1 \\ Z_{2} {\bar{Z}}_{2} 1 \\ Z_{3} {\bar{Z}}_{3} 1 \end{array}|$ then the value of $1 / λ$ is equal to ______.

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answer is 4.

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Detailed Solution

Let $z_{1} = x_{1} + {iy}_{1}, z_{2} = x_{2} + {iy}_{2} and z_{3} = x_{3} + {iy}_{3}$ be the vertices.
Consider $|\begin{array}{ccc} z_{1} & {\bar{z}}_{1} & 1 \\ z_{2} & {\bar{z}}_{2} & 1 \\ z_{3} & {\bar{z}}_{3} & 1 \end{array}| = |\begin{array}{lll} x_{1} + {iy}_{1} x_{1} - {iy}_{1} 1 \\ x_{2} + {iy}_{2} x_{2} - {iy}_{2} 1 \\ x_{3} + {iy}_{3} x_{3} - {iy}_{3} 1 \end{array}| C_{1} \to C_{1} + C_{2}$

$= 2 |\begin{array}{lll} x_{1} x_{1} - {iy}_{1} 1 \\ x_{2} x_{2} - {iy}_{2} 1 \\ x_{3} x_{3} - {iy}_{3} 1 \end{array}| C_{2} \to C_{2} - C_{1}$
$\begin{array}{l} = |\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}| = (- 4 i) \frac{1}{2} | | \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} ∣ \\ = - 4 i \times Area of triangle. \end{array}$
$∴$ Area of triangle $= \frac{i}{4} |\begin{array}{lll} z_{1} {\bar{z}}_{1} 1 \\ z_{2} {\bar{z}}_{2} 1 \\ z_{3} {\bar{z}}_{3} 1 \end{array}|; ∴ λ = \frac{1}{4} \Rightarrow \frac{1}{λ} = 4$