In a Argand plane z1- z2 and z3 are respectively- the vertices of an isosceles triangle ABC with AC - BC and - CAB - - If z4 is the center of triangle I- then the value of z4 - z12 cos- - 1 - is

The correct option is D (z_2 z_1)(z_3 z_1)/(z_4 z_1)^2 AB× AC(IA)^2=ABIA×ACIA ∠ IAB=θ2, ∠ IAC=θ2 z_2 z_1z_4 z_1=|z_2 z_1||z_4 z_1|e^i ...

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Standard XII

Mathematics

Rotation Concept

In a Argand p...

Question

In a Argand plane $z_{1}, z_{2}$ and $z_{3}$ are respectively, the vertices of an isosceles triangle $A B C$ with $A C = B C$ and $∠ C A B = θ$ . If $z_{4}$ is the center of triangle $I$ , then the value of $(z_{4} - z_{1})^{2} (cos θ + 1) sec θ$ is

\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})}

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(z_{2} - z_{1}) (z_{3} - z_{1})

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(z_{2} - z_{1}) (z_{3} - z_{1})^{2}

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\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}}

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Solution

The correct option is D $\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}}$
$\frac{A B \times A C}{(I A)^{2}} = \frac{A B}{I A} \times \frac{A C}{I A}$
$∠ I A B = \frac{θ}{2}, ∠ I A C = \frac{θ}{2}$
$\frac{z_{2} - z_{1}}{z_{4} - z_{1}} = \frac{| z_{2} - z_{1} |}{| z_{4} - z_{1} |} e^{\frac{i θ}{2}}$
and $\frac{z_{3} - z_{1}}{z_{4} - z_{1}} = \frac{| z_{3} - z_{1} |}{| z_{4} - z_{1} |} e^{\frac{i θ}{2}}$
Multiplying,
$\frac{z_{2} - z_{1}}{z_{4} - z_{1}} \frac{z_{3} - z_{1}}{z_{4} - z_{1}} = \frac{| z_{2} - z_{1} |}{| z_{4} - z_{1} |} \frac{| z_{3} - z_{1} |}{| z_{4} - z_{1} |}$
$\Rightarrow \frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}} = \frac{A B \times A C}{I A^{4}}$ (1)
From (1).
$\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}} = 2 {(\frac{A D}{I A})}^{2} (\frac{A C}{A D}) (∵ A B = 2 A D)$
$\Rightarrow (z_{2} - z_{1}) (z_{3} - z_{1}) = (z_{4} - z_{1})^{2} 2 {cos}^{2} \frac{θ}{2} sec θ$
$= (z_{4} - z_{1})^{2} (cos θ + 1) sec θ$

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In a Argand plane z1,z2 and z3 are respectively, the vertices of an isosceles triangle ABC with AC=BC and ∠CAB=θ. If z4 is the center of triangle I, then the value of (z4−z1)2(cosθ+1)secθ is

In a Argand plane $z_{1}, z_{2}$ and $z_{3}$ are respectively, the vertices of an isosceles triangle $A B C$ with $A C = B C$ and $∠ C A B = θ$ . If $z_{4}$ is the center of triangle $I$ , then the value of $(z_{4} - z_{1})^{2} (cos θ + 1) sec θ$ is