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I have a hard time proving this without using any numbers. How do I show that the point $(x,y)$ reflected across $y=\frac12$ is $(x, 1-y)$ ?

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up vote 2 down vote accepted

Points of the form $(x,\frac12\pm\epsilon)$ reflect to $(x,\frac12\mp\epsilon)$. Just rewrite your point as $$(x,y)=(x,\tfrac12-(\tfrac12-y))$$ so that its reflection is $$(x,\tfrac12+(\tfrac12-y))=(x,1-y),$$ right?

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Hint: You can form the matrices representing those operation then multiply them and see how they act on $(x,y)$.

Otherwise: $(x,y)\xrightarrow{\text{reflect across x-axis}} (x,-y)\xrightarrow{\text{reflect across y=1/2}} (x,y+1)$

and the last part is because: $(x,-y)$ has distance $y+1/2$ from the line $y=1/2$ so when you reflect the point will be of distance $y+1/2+1/2=y+1$ from the x-axis, hence you obtain $(x,y+1)$.

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There is a general formula for reflexion about a line but you don't need it here. Let $s_a(x,y)=(f_a(x,y),g_a(x,y))$ be the reflexion about the line $y=a$. Since $y=a$ has to be the bisector of the line segment between $(x,y)$ and $s_a(x,y)$, it is clear that $f_a(x,y)=x$ and $g_a(x,y)=g_a(y)$ depends only on $y$ (draw a picture). Moreover, we have $|y-a|=|g_a(y)-a|$ and $y\leqslant a \leqslant g_a(y)$ or $g_a(y)\leqslant a \leqslant y$ (everything is clear from a drawing).

It follows that $g_a(y)=2a-y$, so $s_a(x,y)=(x,2a-y)$.

In particular, $s_0(x,y)=(x,-y)$ (as one could expect) and $s_{1/2}(x,y)=(x,1-y)$. Therefore the composition $s_{1/2}\circ s_0(x,y)=(x,1+y)$.

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As in my answer above, just note that points $(x,a\pm\epsilon)$ reflect to $(x,a\mp\epsilon)$. Then since $(x,y)=(x,a-(a-y))$, the reflection of $(x,y)$ is $(x,a+(a-y))=(x,2a-y)$. – MPW May 15 '14 at 4:21
Yes, I saw it. Actually, I upvoted your answer. I should have commented to say it is a great answer :D – Taladris May 15 '14 at 4:23
Ah, thanks for the upvote! – MPW May 15 '14 at 4:26

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