# Reflection and rotation

I want to describe analytically the reflection $\sigma$ along the line through the points $(4,1)$ and $(5,-3)$ and the rotation $\delta$ with rotation angle $\frac{\pi}{3}$ about the point $(-1,-1)$.



I have done the following:

We connect the two points $(4,1)$ and $(5,-3)$ to get the line that passes through them. The equation of that line is $y-1=\frac{-3-1}{5-4}(x-4) \Rightarrow y-1=\frac{-4}{1}(x-4) \Rightarrow y=-4x+16+1 \Rightarrow y=-4x+17$

To find the reflected point of a point $P$ along the above line, we have to find the point $S$ on the line that has the minimum distance from $P$.

The part from $P$ tp $S$ must be perpendicular to the line. The vector $\vec{PS}$ is therefore perpendicular to the direction vector of the line.

We construct a plane that is orthogonal to the line and contains the point $P$. The direction vector of the line is the normal vector of the plane.

The intersection point of the plane and the line is the point $S$, that we are looking for. It is on the line $\vec{PS}$ and it is orthogonal to the line, since $\vec{PS}$ is o the plane that we have constructed. This point is the foot of P on the line.

Now we want to reflect the point $P$ about the point $S$.

For that we just need the vector $\vec{PS}$. With that vector we get from the point $P$ to the point $S$. Let $P'$ be the reflected point along the line, then the line bisects $PP'$, and so $S$ is the midpoint. So we want to go in the same direction on the other side of $S$ in the same length. So, we go again with the vector $\vec{PS}$, and so we go to the point $P'$.

($\vec{OP'}=\vec{OP}+2\cdot \vec{PS}$)

Is everything correct? Could I improve something?



For the rotation I thought the following:

The rotation is always to left. We draw a line from rotation center to all the points of the object that we want rotate. From each line we draw the rotation angle.

We calculate the distance from the rotation center to each point and draw with the same distance at the line that we draw with the angle the ampped point. .

If we have done this for each point, then we have to connect the points and so we get the mapped object.

Is everything correct? Could I improve something?



EDIT:

I want to check if $\sigma\circ\delta$ and $\delta\circ\sigma$ are reflections or glide reflections.

We hav that a rotation is a composition of two reflections.

Therefore both $\sigma\circ\delta$ and $\delta\circ\sigma$ are a product of three reflections.

But what is the result of that product?

• The ideas you described look fine to me. For a different approach to $\delta$ you can consider the following. You probably know the matrix of the linear transformation the describes the same rotation about the origin. You get $\delta$ by conjugating that linear transformation with a translation that takes $(-1,-1)\mapsto (0,0)$. – Jyrki Lahtonen Jul 16 '17 at 7:42
• Ah ok!! Thank you!! I have also an other question... Are $\sigma \circ \delta$ and $\delta\circ\sigma$ reflections or glide reflections? How could we check that? @JyrkiLahtonen – Mary Star Jul 16 '17 at 8:21
• I don't know off the top of my head. I would write down the formula for the composition, figure out the fixed points, and improvise from there :-) – Jyrki Lahtonen Jul 16 '17 at 8:24
• You appear to be working in $\mathbb R^2$, but talk about perpendicular planes. – amd Jul 16 '17 at 19:10
• Yes. The reasoning is sound. You just need to correct the terminology. – amd Jul 16 '17 at 20:51

In what follows I'm assuming that you are using a cartesian rectangular coordinate system.

If the reflection were along the x-axis (and the coordinate system were cartesian and rectangular) it would be represented by this matrix:

$$R=\left[\begin{array}{r} 1&0\\ 0&-1\\ \end{array}\right]$$

because the x coordinates must be left unchanged while the y coordinates must be turned negative. Indeed the reflection of a point with coordinates $(x_0, y_0)$ would be a point with these coordinates $(x_1,y_1)$:

$$\left[\begin{array}{r} x_1\\ y_1\\ \end{array}\right] =\left[\begin{array}{r} 1&0\\ 0&-1\\ \end{array}\right] \left[\begin{array}{r} x_0\\ y_0\\ \end{array}\right] =\left[\begin{array}{r} x_0\\ -y_0\\ \end{array}\right] \tag{1}$$

So let us make (with a transformation that we write in a while, named coordinate transformation) the line through $(4,1)$ and $(5,-3)$ the $x$-axis directed from $(4,1)$ and $(5,-3)$, and its orthogonal line the $y$-axis oriented such that the coordinate system is right-handed and such that the origin is in $(4,1)$. This coordinate system will be called $x', y'$.

Following this way, in this new coordinate system the said matrix represents the reflection along the line through $(4,1)$ and $(5,-3)$ when points are represented in the $x', y'$ coordinate system. $$\left[\begin{array}{r} x_1'\\ y_1'\\ \end{array}\right] =\left[\begin{array}{r} 1&0\\ 0&-1\\ \end{array}\right] \left[\begin{array}{r} x_0'\\ y_0'\\ \end{array}\right] =\left[\begin{array}{r} x_0'\\ -y_0'\\ \end{array}\right] \tag{2}$$

Let us now write the coordinate transformation following what we have said. Let us proceeds in two steps. The point whose $x, y$ coordinates are $(4,1)$ must have $x', y'$ coordinates $(0,0)$ because it must be interpreted as the origin of the new coordinate system. So this transformation (a coordinate translation) must subtract to the $x,y$ coordinates of any point the constant: $$b=\left[\begin{array}{r} 4\\ 1\\ \end{array}\right]$$

This is the first step. Now we need to iterate this first transformation with a second, that will orient and scale the $x'$- and $y'$-axes. For that, let us consider that the point whose $x,y$ coordinates are $(5,-3)$ can have $x',y'$ coordinates $(1,0)$: this is one among other choices to make the line through $(4,1)$ and $(5,-3)$ the $x'$-axis. But now we need to create the second transformation, therefore we need to apply the first transformation to $(5,-3)$, obtaining that the second transformation we are searching for must change the translated coordinate $(1,4)$ into the $x', y'$ coordinates $(1,0)$. So the first requirement for this second transformation is:

$$\left[\begin{array}{r} 1\\ 0\\ \end{array}\right] \mapsto \left[\begin{array}{r} 1\\ -4\\ \end{array}\right] \tag{3}$$

Additionally, we need to set the $y'$-axis and for this a point whose coordinates after the first transformation are $(0,1)$ must be change into the $x', y'$ coordinates $(4,1)$, that is

$$\left[\begin{array}{r} 0\\ 1\\ \end{array}\right] \mapsto \left[\begin{array}{r} 4\\ 1\\ \end{array}\right] \tag{4}$$

The inverse of this second transformation is then represented by this matrix: $$A^{-1}=\left[\begin{array}{r} 1&4\\ -4&1\\ \end{array}\right]$$

Indeed we can find out the chosen defining mapping $(3)$ and $(4)$: $$\left[\begin{array}{r} 1\\ -4\\ \end{array}\right] =\left[\begin{array}{r} 1&4\\ -4&1\\ \end{array}\right] \left[\begin{array}{r} 1\\ 0\\ \end{array}\right]$$

$$\left[\begin{array}{r} 4\\ 1\\ \end{array}\right] =\left[\begin{array}{r} 1&4\\ -4&1\\ \end{array}\right] \left[\begin{array}{r} 0\\ 1\\ \end{array}\right]$$

Then the iteration of the first and second transformation is given by

$$\left[\begin{array}{r} x'\\ y'\\ \end{array}\right] =A(\left[\begin{array}{r} x\\ y\\ \end{array}\right] -b)\tag{5}$$

where $$A=\frac{1}{17}\left[\begin{array}{r} 1&-4\\ 4&1\\ \end{array}\right]$$

So from $(2)$ and $(5)$:

$$A( \left[\begin{array}{r} x_1\\ y_1\\ \end{array}\right]-b) =RA( \left[\begin{array}{r} x_0\\ y_0\\ \end{array}\right] -b)\tag{6}$$

Solving for $(x_1,y_1)$ we get:

$$\left[\begin{array}{r} x_1\\ y_1\\ \end{array}\right] =A^{-1}RA( \left[\begin{array}{r} x_0\\ y_0\\ \end{array}\right] -b)+b \tag{7}$$

You can check for instance that the point whose $x,y$ coordinates are $(6,-7)$ that is on the $x'$-axis remains unaltered after this transformation, and the point whose $x,y$ coordinates are $(8,2)$ that is on the $y'$-axis is reflected symmetrically around $(4,1)$ in the $x,y$ coordinates to $(0,0)$ in the $x,y$ coordinates.

Now you can follow the same reasoning for the remaining part of the problem (that is simpler).