analytic geometrygeometryreflection
I know how to reflect a coordinate over the $y$ and $x$ axis, but is there a rule I could use to help me find the reflected point over $x = -1$?
This is what I know already:
Over the $x$-axis: $(x, y) \to (x, –y) $
Over the $y$-axis: $(x, y) \to (–x, y)$
Over the line $y = x$: $(x, y) \to (y, x)$
Through the origin: $(x, y) \to (–x, –y) $
What I don't know is how to solve when reflecting over something like $x = -1$. Is there a rule that would help make solving this easier?
If a point $P(a,b)$ is rotated $180$ degree about the origin, then the resulting image of $P$ is $(−a,−b)$.
1) $(x,y)\rightarrow (-x,-y)\rightarrow (x,-y).$
2) $(x,y)\rightarrow (-x,y)\rightarrow (x,-y).$
Hence, this is the same as 1.
3) $(x,y)\rightarrow (y,x)\rightarrow (-y,x).$
4) $(x,y)\rightarrow (-y,-x)\rightarrow (y,-x).$
Hence, your conjectures are correct.
The correct answer is the one for which you said "I tested it and it doesn't work for me," namely
$$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \\ \end{bmatrix} $$
Here is a diagram for your example.
The line is $y=\frac 43x$, which has the angle of inclination $\theta\approx 53.1301023542°$ with $\cos(2\theta)=-0.28$ and $\sin(2\theta)=0.96$, and the points before and after reflection are
$$P=\begin{bmatrix}2\\1\\\end{bmatrix}, \quad P'=\begin{bmatrix}0.4\\2.2\\\end{bmatrix} $$
(The line and point $P$ were entered in Geogebra: the angle, cosine, sine, and point $P'$ were calculated by Geogebra.) The matrix calculation is then
$$ \begin{bmatrix}\cos(2\theta)&\sin(2\theta)\\\sin(2\theta)&-\cos(2\theta)\\\end{bmatrix}\cdot P =\begin{bmatrix}-0.28&0.96\\0.96&0.28\\\end{bmatrix}\cdot \begin{bmatrix}2\\1\\\end{bmatrix} =\begin{bmatrix}0.4\\2.2\\\end{bmatrix} =P' $$
so it all works out.
Do you want a derivation of that transformation matrix? And what tests did you try that did not work for you?
Best Answer
To reflect over a vertical line, such as $x=a$, first translate so the line is shifted to the y-axis, then reflect over it, then translate back so the line is shifted to its original position.
$$(x,y)\mapsto (x-a,y) \mapsto (a-x,y) \mapsto (2a-x,y)$$
In this case to reflex over $x=-1$ we shift $x\mapsto x+1$, reflect $\mapsto -1-x$ and shift back $\mapsto -2-x$
$$(x,y)\mapsto (-2-x, y)$$
Similarly for reflecting over horizontal lines, such as $y=b$ involves $(x,y)\mapsto (x, 2b-y)$.
Reflecting over a diagonal line is only a bit more complicated, as it involves rotation of the frame of reference.