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I need to find the matrix of reflection through line $y=- \frac 23 x$ .

I'm trying to visualise a vector satisfying this. The standard algorithm states that we need to find the angle this line makes with $x$ axis and the transformation matrix can be seen as $R_\alpha T_0 R_{-\alpha}.$

I'm not sure how to proceed. I can't visualise the angle it makes with $x$ axis. Is there a procedure to think about such reflections?

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    $\begingroup$ @WesleyGroupshaveFeelingsToo It hasn't been specified. It seems like it's in R^2. $\endgroup$ – S.Rana Nov 2 '18 at 12:03
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    $\begingroup$ @WesleyGroupshaveFeelingsToo Sure! I am not aware of the syntax so I do it in this format. I'll try to learn as I go. $\endgroup$ – S.Rana Nov 2 '18 at 12:06
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    $\begingroup$ If T is your matrix where does it send $\left( \begin{matrix} 1 \\ 0 \\ \end{matrix} \right)$ and $\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right)$? Then $$T\left( \begin{matrix} x \\ y \\ \end{matrix} \right)=xT\left( \begin{matrix} 1 \\ 0 \\ \end{matrix} \right)+yT\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right)$$ $\endgroup$ – Paul Nov 2 '18 at 12:18
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    $\begingroup$ Shouldn't there also be a $T$ on the right side though? $\endgroup$ – Wesley Strik Nov 2 '18 at 12:21
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    $\begingroup$ Fair point! Edited $\endgroup$ – Paul Nov 2 '18 at 12:22
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Notice that vectors on this line have the form $(1, -\frac{2}{3} )$, and an orthogonal vector would be $(\frac{2}{3}, 1) $. A very straightforward procedure could be to reflect the vectors $(0,1)$ and $(1,0)$ orthogonally in this line. Have you done this before? Once you have determined the images of the basis vectors, you can figure out what the matrix columns should look like.

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Alternatively to the comment above (which requires a bit of trig), where does your matrix T send $\left( \begin{matrix} 3 \\ -2 \\ \end{matrix} \right)$ and $\left( \begin{matrix} 2 \\ 3 \\ \end{matrix} \right)$ which are on and perpendicular to your line respectively. Now can you find a and b in terms of x and y so that $$\left( \begin{matrix} x \\ y \\ \end{matrix} \right)=a\left( \begin{matrix} 3 \\ -2 \\ \end{matrix} \right)+b\left( \begin{matrix} 2 \\ 3 \\ \end{matrix} \right)$$ Finally, apply T to find T$\left( \begin{matrix} x \\ y \\ \end{matrix} \right)$

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