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Question Number 194197 by Mingma last updated on 30/Jun/23

	Answered by mr W last updated on 01/Jul/23

	Commented by mr W last updated on 01/Jul/23

	$x_{P} = s \cos θ$ $y_{P} = s \sin θ$ $y_{P} = f (x_{P})$ $x_{Q} = s \cos (\frac{π}{3} + θ) = \frac{1}{2} (s \cos θ - \sqrt{3} s \sin θ)$ $x_{P} - \sqrt{3} y_{P} = 2 x_{Q}$ $y_{Q} = s \sin (\frac{π}{3} + θ) = \frac{1}{2} (\sqrt{3} s \cos θ + s \sin θ)$ $\Rightarrow \sqrt{3} x_{P} + y_{P} = 2 y_{Q}$ $\Rightarrow y_{P} = \frac{- \sqrt{3} x_{P} + y_{P}}{2}$ $\Rightarrow x_{P} = \frac{x_{Q} + \sqrt{3} y_{Q}}{2}$ $l o c u s o f p o i n t Q :$ $\frac{- \sqrt{3} x_{P} + y_{P}}{2} = f (\frac{x_{Q} + \sqrt{3} y_{Q}}{2})$ $o r$ $\frac{- \sqrt{3} x + y}{2} = f (\frac{x + \sqrt{3} y}{2})$ $i f l o c u s o f P i s a s e g m e n t y = a x + b,$ $t h e n t h e l o c u s o f Q i s$ $\frac{- \sqrt{3} x + y}{2} = a \times \frac{x + \sqrt{3} y}{2} + b$ $\Rightarrow (a + \sqrt{3}) x + (a \sqrt{3} - 1) y + 2 b = 0$ $t h i s i s a l s o a s e g m e n t .$ $(s a m e s e g m e n t r o t a t e d b y 60 ° a b o u t O)$ $s i m i l a r l y w e c a n s e e : i f t h e l o c u s o f$ $P i s a c i r c l e, t h e n t h e l o c u s o f Q i s$ $a l s o a c i r c l e .$ $(s a m e c i r c l e r o t a t e d b y 60 ° a b o u t O)$
	Commented by mr W last updated on 01/Jul/23

	Commented by mr W last updated on 01/Jul/23

	Commented by mr W last updated on 01/Jul/23

	$g e n e r a l l y :$ $l o c u s o f p o i n t Q i s t h e l o c u s o f p o i n t$ $P r o t a t e d b y 60 ° a b o u t p o i n t O .$
	Commented by Mingma last updated on 02/Jul/23
	Perfect solutions, sir��
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