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Rotation Concept

On the Argand...

Question

On the 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 equal angles are $θ$ . If $z_{4}$ is the incentre of the triangle, then $\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}} =$

A

1 + cos θ

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B

1 + sec θ

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C

tan θ

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D

1

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Solution

The correct option is B $1 + sec θ$
$\frac{z_{2} - z_{1}}{| z_{2} - z_{1} |} = \frac{(z_{4} - z_{1})}{| z_{4} - z_{1} |} e^{- i θ / 2} (clockwise)$
$\frac{z_{3} - z_{1}}{| z_{3} - z_{1} |} = \frac{(z_{4} - z_{1})}{| z_{4} - z_{1} |} e^{i θ / 2} (anticlockwise)$

Multiplying $(1)$ and $(2)$ , we get
$\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}} = \frac{| z_{2} - z_{1} | | z_{3} - z_{1} |}{| z_{4} - z_{1} |^{2}}$
$= \frac{(A B) (A C)}{(A I)^{2}} = \frac{2 (A D) (A C)}{(A I)^{2}}$
$= \frac{2 (A D)^{2}}{(A I)^{2}} \frac{A C}{A D} = 2 {cos}^{2} \frac{θ}{2} sec θ [∵ ∠ A D C = 90 °] = (1 + cos θ) sec θ = 1 + sec θ$

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