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Rotation Concept

Az1, Bz2, Cz3...

Question

$A (z_{1}), B (z_{2}), C (z_{3})$ are the vertices of the triangle $A B C$ (in anticlockwise order). If $∠ A B C = \frac{π}{4}$ and $A B = \sqrt{2} (B C)$ , then

z_{2} = (z_{1} - z_{3}) + i z_{3}

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z_{2} = z_{3} + i (z_{1} - z_{3})

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z_{2} = z_{1} + z_{3}

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None of these

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Solution

The correct option is B $z_{2} = z_{3} + i (z_{1} - z_{3})$

Rotating about the point $B$ , we get
$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = (\frac{a \sqrt{2}}{a}) e^{i π / 4} = \sqrt{2} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = (1 + i)$
$\Rightarrow z_{1} - z_{2} = (z_{3} - z_{2}) (1 + i)$
$\Rightarrow z_{2} = z_{1} - (z_{3} - z_{2}) (1 + i)$
$\Rightarrow z_{2} (1 - (1 + i)) = z_{1} - z_{3} (1 + i)$
$\Rightarrow z_{2} = \frac{z_{1}}{- i} - \frac{z_{3}}{- i} (1 + i) = i z_{1} - i z_{3} (1 + i)$
$= z_{3} + i (z_{1} - z_{3})$

Alternate Solution:
Given $∠ A B C = \frac{π}{4} \Rightarrow arg (\frac{z_{1} - z_{2}}{z_{3} - z_{2}}) = \frac{π}{4} \dots (1)$
and $\frac{A B}{B C} = \sqrt{2} \Rightarrow \frac{| z_{1} - z_{2} |}{| z_{3} - z_{2} |} = \sqrt{2} \dots (2)$
From $(1)$ and $(2)$ we can say
$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \sqrt{2} e^{i π / 4} = 1 + i \Rightarrow z_{2} = z_{1} - (1 + i) (z_{3} - z_{2}) \Rightarrow z_{2} = \frac{z_{3} (1 + i) - z_{1}}{i} \Rightarrow z_{2} = z_{3} + i (z_{1} - z_{3})$

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Similar questions

A (z_{1}), B (z_{2}), C (z_{3})

are the vertices of the triangle ABC (in anticlockwise order). If

∠ A B C = π / 4

and

A B = \sqrt{2} (B C)

Then prove that

z_{2} = z_{3} + i (z_{1} - z_{3}) .

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∠ A B C = \frac{π}{4}

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Rotation Concept

Standard XIII Mathematics

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A(z1),B(z2),C(z3) are the vertices of the triangle ABC (in anticlockwise order). If ∠ABC=π4 and AB=√2(BC), then

$A (z_{1}), B (z_{2}), C (z_{3})$ are the vertices of the triangle $A B C$ (in anticlockwise order). If $∠ A B C = \frac{π}{4}$ and $A B = \sqrt{2} (B C)$ , then