In-xy-plane-which-of-the-following-is-the-reflection-of-the-graph-of-y-1-x-x-2-1-about-the-line-y-2x-

Question Number 118874 by bemath last updated on 20/Oct/20

I n x y - p l a n e, w h i c h o f t h e f o l l o w i n g

i s t h e r e f l e c t i o n o f t h e g r a p h

o f y = \frac{1 + x}{x^{2} + 1} a b o u t t h e l i n e y = 2 x .

Answered by bramlexs22 last updated on 20/Oct/20

b y t r a n s f o r m a t i o n m a t r i x (\begin{matrix} \cos 2 θ \sin 2 θ \\ \sin 2 θ - \cos 2 θ \end{matrix})

w h e r e \tan θ = m_{s l o p e} = 2 a n d \cos 2 θ = - \frac{3}{5} \land \sin 2 θ = \frac{4}{5}

l e t (x^{^{'}}, y^{^{'}}) i s i m a g e o f p o i n t (x, y)

w e g e t (\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} - \frac{3}{5} \frac{4}{5} \\ \frac{4}{5} \frac{3}{5} \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

t h e n (\begin{matrix} x \\ y \end{matrix}) = - (\begin{matrix} \frac{3}{5} - \frac{4}{5} \\ - \frac{4}{5} - \frac{3}{5} \end{matrix}) (\begin{matrix} x^{'} \\ y^{'} \end{matrix})

\Rightarrow (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} \frac{- 3 x^{^{'}} + 4 y^{'}}{5} \\ \frac{4 x^{'} + 3 y^{'}}{5} \end{matrix})

H e n c e t h e r e f l e c t i o n o f g r a p h o f y = \frac{x + 1}{x^{2} + 1}

a b o u t t h e l i n e y = 2 x i s \frac{4 x + 3 y}{5} = \frac{\frac{- 3 x + 4 y}{5} + 1}{{(\frac{- 3 x + 4 y}{5})}^{2} + 1}

4 x + 3 y = \frac{- 3 x + 4 y + 5}{\frac{9 x^{2} - 24 x y + 16 y^{2} + 25}{25}}

4 x + 3 y = \frac{- 75 x + 100 y + 125}{9 x^{2} - 24 x y + 16 y^{2} + 25}

Commented by benjo_mathlover last updated on 20/Oct/20