*x*,

*y*) -> (

*x*+

*h*,

*y*+

*k*) -- which turns out to be

*every*translation in the plane.

But with rotations, we only perform a precious few of them. The only rotations that appear on the PARCC and other Common Core tests are those of magnitude 90, 180, or 270. Yet we've seen a few of these rotations centered at points other than the origin on the PARCC.

We'll begin with rotations that are centered at the origin, though. Just as we used the Two Reflections Theorem for Translations yesterday, today we'll use the Two Reflections Theorem for Rotations. So to perform the rotation of 180 degrees about the origin, we compose two reflections in mirrors that intersect at the origin, at an angle of half of 180, or 90 degrees. The obvious choices for mirrors are the

*x*- and

*y*-axes. We've already proved that the reflection image of (

*x*,

*y*) in the

*x*-axis is (

*x*, -

*y*) and the reflection image of (

*x*,

*y*) in the

*y*-axis is (-

*x*,

*y*). It doesn't matter in which order we compose these as reflections in perpendicular mirrors always commute. So we prove that the rotation image of (

*x*,

*y*) centered at the origin and of magnitude 180 degrees is (-

*x*, -

*y*).

Now our other common rotation magnitude is 90 degrees -- and this time, it will make a difference whether it's clockwise or counterclockwise. The angle between the mirrors will now have to be half of 90, or 45 degrees. There's one mirror to consider that will help us with a 45-degree angle -- the line whose equation is

*y*=

*x*.

We've hinted at several proofs involving reflection over the line

*y*=

*x*. As it turns out, many of these properties can be proved in neutral geometry, so they should be easy in Euclidean geometry:

-- The line

*y*=

*x*forms 45-degree angles with each coordinate axis.

-- The line

*x*=

*a*, reflected over the mirror

*y*=

*x*, is

*y*=

*a*.

-- The point (

*x*,

*y*), reflected over the mirror

*y*=

*x*, is (

*y*,

*x*).

Let's look at the quadrilateral whose vertices are (0, 0), (

*a*, 0), (

*a*,

*a*), and (0,

*a*). Euclidean geometry proves that this quadrilateral is a square -- in neutral geometry, once again, we only know that this figure is a Lambert quadrilateral. But in either case, we can show that this figure is a kite -- and we mentioned this on the day we learned about kites (Day 50). Just as yesterday, we know that two of the sides of this kite (the ones along the axes) have length

*a*, and the other two sides (along the lines with equations

*x*=

*a*and

*y*=

*a*) are perpendicular to the

*x*- and

*y*-axes.

So now we can apply the properties of a kite -- the Kite Symmetry Theorem. The diagonal of our kite running from (0, 0) to (

*a*,

*a*) bisects the angle between the

*x*- and

*y*-axes -- and since we know that the angle between the axes is 90 degrees, the diagonal must form a 45-degree angle with each axis. And reflecting across this symmetry diagonal must map the axes to each other and

*x*=

*a*to

*y*=

*a*.

Recall that at this point, we don't know the equations of lines, so we aren't yet certain that the graph of

*y*=

*x*is even a line (which we'd better figure out before trying to use it as a mirror). But we see that the value of

*a*in the above proof is arbitrary -- it's true for every single real number

*a*(although in case

*a*is negative, we should probably say that the kite has sides of length |

*a*|, not

*a*). Therefore every single point of the graph of

*y*=

*x*lies on the bisector of the angle between the axes -- that is, the graph of

*y*=

*x*is exactly that line. And reflecting in that line maps

*x*=

*a*to

*y*=

*a*and vice versa -- that is, it switches

*x*and

*y*. Therefore the image of (

*a*,

*b*) must be (

*b*,

*a*).

Now that we know how to reflect in the line

*y*=

*x*, let's use it to perform a 90-degree rotation. It's probably easiest just to start with the reflection in

*y*=

*x*first, so (

*x*,

*y*) maps to (

*y*,

*x*). As for the second mirror, it depends on whether we want to go clockwise or counterclockwise. To go clockwise, the second mirror must be 45 degrees clockwise of the first mirror,

*y*=

*x*. That is the

*x*-axis, and to reflect in it, we change the sign of the second coordinate. So (

*y*,

*x*) reflected in the second mirror is the point (

*y*,

*-x*), so mapping (

*x*,

*y*) to (

*y*, -

*x*) rotates points 90 degrees clockwise. To go counterclockwise, the second mirror must be 45 degrees counterclockwise of the first mirror,

*y*=

*x*. That is the

*y*-axis, and to reflect in it, we change the sign of the first coordinate. So (

*y*,

*x*) reflected in the second mirror is the point (-

*y*,

*x*), so mapping (

*x*,

*y*) to (-

*y*,

*x*) rotates points 90 degrees counterclockwise.

Notice that some of the PARCC questions mention 270-degree rotations -- for example, there was a released question that mentions a 270-degree clockwise rotation. Of course, a 270-degree clockwise rotation is equivalent to a 90-degree counterclockwise rotation, so it maps (

*x*,

*y*) to (-

*y*,

*x*). If students forget this, they can still take half of 270 degrees to get 135 degrees clockwise, and they can see that 135 degrees clockwise from the line

*y*=

*x*is still the

*y*-axis, just as it would have been if they'd gone 45 degrees counterclockwise instead.

Last year, I created a quick worksheet to help students perform any of the reflections and rotations mentioned in this post. (This was late in the year when we were covering PARCC questions, but now I'm giving this lesson much earlier.) It takes the coordinate plane labeled with the positive

*x*-, negative

*x-*, positive

*y*-, and negative

*y*-axes. Students can then perform the rotations on the axes to see what happens. For example, let's try our 270-degree clockwise rotation. After we rotate the paper 270 degrees clockwise, we see that where the

*x*ought to be, we see -

*y*instead, and where the

*y*ought to be, it's +

*x*. Thus the image must be (-

*y*,

*x*).

Okay, so we've taken care of all the rotations centered at the origin, But on the PARCC, there are questions with rotations centered at other points. These questions that I've seen direct the students to take a triangle

*ABC*and rotate it around one of its vertices -- let's say

*C*. Well that makes things a little easier, since then the rotation image of

*C*is

*C*itself. So then there are only two points that we need to find,

*A'*and

*B'*.

It's possible, in principle, to find formulas to determine the image of (

*x*,

*y*) under reflections in mirrors parallel to the axes and rotations centered at points other than the origin. We've seen, for example, that the point (

*x*,

*y*) reflected in the line

*x*=

*a*is (2

*a*-

*x*,

*y*). An interesting question is, where exactly does the 2 in 2

*a*-

*x*come from?

To find out, we notice that if we were reflecting in the

*y*-axis (which is parallel to

*x*=

*a*), then the point (

*x*,

*y*) is mapped to (-

*x*,

*y*). Now that extra 2

*a*term looks just like a horizontal translation of exactly 2

*a*units.

So somehow, our reflection in the line

*x*=

*a*appears to be the composite of a reflection in the

*y*-axis and a horizontal translation. (This is not a glide reflection between the mirror is perpendicular to the direction of translation -- we found out last year that such a composite yields a simple reflection, not a glide reflection.)

Using symbols, let's call the composite transformation T. It is the composite of a

*y*-axis reflection, which we'll call r_

*y*, and a horizontal translation of 2

*a*units, H_2

*a*:

T = H_2

*a*o r_

*y*

*But the horizontal translation is itself the composite of two reflections. The two mirrors here must be vertical mirrors spaced exactly half of 2*

*a*, or

*a*units apart. We might as well let the two mirrors be the

*y*-axis itself and the line

*x*=

*a*.

T = H_2

*a*o r_

*y*

= r_(

*x*=

*a*) o r_

*y*o r_

*y*

= r_(

*x*=

*a*) o I

= r_(

*x*=

*a*)

which is exactly what we wanted -- a reflection in the line

*x*=

*a*.

Likewise, we see that the reflection in the line

*y*=

*b*maps (

*x*,

*y*) to (

*x*, 2

*b*-

*y*). The composite of both reflections is a 180-degree rotation about the point (

*a*,

*b*), which maps (

*x*,

*y*) to (2

*a*-

*x*, 2

*b*-

*y*) -- and that's also the composite of a 180-degree rotation about the origin and yet another translation.

Now 90-degree rotations about points other than the origin are even trickier, because now we'd have to reflect about mirrors with equations like

*y*=

*x*+

*b*-- and we don't even know that's a line yet. The algebra involved in this reflection gets very messy.

Of course, if we try to visualize the rotation, another composite transformation jumps at us. To perform a 90-degree rotation (either clockwise or counterclockwise) about the point (

*a*,

*b*), it appears that we can first perform the translation that maps (

*a*,

*b*) to (0, 0), then perform the rotation centered at the origin, and finally translate (0, 0) back to (

*a*,

*b*).

This seems to work, but is there any reason why it should? Let's use symbols again -- in order to remember what the symbols stand for, we let "rot" stand for the rotation and "trans" stand for the translation mapping (0, 0) to (

*a*,

*b*). Then trans^-1 can stand for the inverse translation -- the one mapping (

*a*,

*b*) to (0, 0). This gives us:

T = trans o rot o trans^-1

This composite has a name in classes like linear algebra and above --

*conjugation*. That is, we are conjugating the rotation by the translation.

We now want to rewrite both the translation and origin-rotation with two mirrors each. And as usual, we want to choose the mirrors strategically so that some of the reflections cancel out. For the translation, we'll let

*k*be the line joining the points (0, 0) and (

*a*,

*b*). Then

*l*will be the line perpendicular to

*k*passing through the origin,

*m*will be the line perpendicular to

*k*passing through the midpoint of (0, 0) and (

*a*,

*b*), and

*n*will be the line perpendicular to

*k*passing through (

*a*,

*b*). Then the rotation can be written as r_

*l*o r_

*k*, and the translation can be written as r_

*m*o r_

*l*. Notice that the inverse translation can be written r_

*l*o r_

*m*-- but it can also be written as r_

*m*o r_

*n*(as either

*l*and

*m*, or

*m*and

*n*, are the correct distance apart). So we write it:

T = trans o rot o trans^-1

= r_

*m*o r_

*l*o r_

*l*o r_

*k*o r_

*m*o r_

*n*

= r_

*m*o I o r_

*k*o r_

*m*o r_

*n*

= r_

*m*o r_

*k*o r_

*m*o r_

*n*

= r_

*k*o r_

*m*o r_

*m*o r_

*n*(as reflections in perpendicular mirrors commute)

= r_

*k*o I o r_

*n*

= r_

*k*o r_

*n*

*which is the composite of reflections in perpendicular mirrors intersecting at (*

*a*,

*b*). And so T is in fact the rotation centered at (

*a*,

*b*), which is what we were expecting.

Once again, though, this is not the sort of symbolic manipulation I'd want my students to see. But then, what should we expect students to do when faced with a PARCC question where they have to rotate around a point other than the origin?

Most likely, this is something that can wait until we discuss the Slope Formula -- especially since it's this rotation that leads to the slopes of perpendicular lines. For now, one can consider such rotations only informally -- after all, the PARCC questions usually include graphs, so students might be able to perform the rotations just by counting units on the graph, rather than using an algebraic formula or manipulating mirrors and symbols.

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