Composition/Reflection Problem

[YummyNoodles]

The Problem: Let P be the transformation r_y=b and Q be the transformation r_y=a where a < b. Describe a single transformation that gives the same result as transformation Q followed by transformation P. Assume that the preimage is on neither of the lines y = a or y = b.

I have experimented a little by substituting a and b with different values, as shown below. From this, I found that all of the distances between the preimage and image in all 3 cases of the preimage being below y = a, between y = a & y = b, and above y = b, to be the same. The only thing is that I don't know the significance of this, and how it could be made into just one transformation. Between the three cases, the only similarity seems to be the distance so I am unsure of how to proceed from here. I have included photos, in case they might help.

 

[Dr.Peterson]

Try making it general. What is the image of point (x, y) under the composition? What transformation does that?

Note in particular, though, that there is something other than the distance moved that is the same in every case! How about direction?

 

[YummyNoodles]

Might you be referring to a translation? I looked over my work again based on your response and found that all of the points moved up. I'm not too sure what you mean by generalizing it though, would the variables a and b remain variables, in addition to x and y being variables, or would they continue to have numerical values?

 

[Dr.Peterson]

Might you be referring to a translation? I looked over my work again based on your response and found that all of the points moved up.

Yes, all your evidence pointed toward every point moving up by a fixed distance in each example. (Did you notice how that distance relates to a and b?)

I'm not too sure what you mean by generalizing it though, would the variables a and b remain variables, in addition to x and y being variables, or would they continue to have numerical values?

Not quite. The a and b are parameters, which are considered fixed (for a given instance of the problem). But the question is about what happens to any point, right? So in order to be sure what happens, you can just call the preimage (x, y), or (p, q) if you prefer, and determine what each reflection in turn does to that general point. You'll get a formula for the final image as a function of the preimage, and that will tell you exactly what is happening -- it will be a proof. This is the power of algebra!

If you prefer, you can do this first with specific values for a and b, as in your examples; but you'll see the big picture best when you go to full generality. You'll also be able to consider further generalizations: are the restrictions they gave (the point being on neither line, and a < b) necessary? This is where math becomes fun.

 

[YummyNoodles]

Yes, all your evidence pointed toward every point moving up by a fixed distance in each example. (Did you notice how that distance relates to a and b?)

I found that every point moved up by 2 times the difference between a and b, or 2(b-a).

Not quite. The a and b are parameters, which are considered fixed (for a given instance of the problem). But the question is about what happens to any point, right? So in order to be sure what happens, you can just call the preimage (x, y), or (p, q) if you prefer, and determine what each reflection in turn does to that general point. You'll get a formula for the final image as a function of the preimage, and that will tell you exactly what is happening -- it will be a proof. This is the power of algebra!

After doing some research, I found the formula for reflecting points over y = n lines. In this case, the first reflection would result in (x , 2a-y) and the reflection after that would result in (x , 2b-(2a-y)). Although, I unfortunately fail to see how this is relevant to the problem. Would I not just need to describe the translation above?

 

[Dr.Peterson]

I take it that you have not been taught to do reflections and translations with a formula? I'd certainly expect you to have been taught something specific with which you could prove the conclusion, and I usually see the notation you are using associated with formulas. This is a problem with helping people without being able to see their context.

You almost did it. Just simplify: 2b-(2a-y) = 2b - 2a + y = y + 2(b - a). So the image of (x, y) is (x, y + 2(b - a)).

Do you see the relevance now? The result of the two reflections is to add 2(b-a) to the y-coordinate, which is exactly the translation you decided it was. So the conclusion isn't just a guess based on lots of experiments; it's something you can know for sure, for every preimage you can imagine!

There are other ways to reach the desired conclusion other than algebra; maybe if you told me the definitions you have been given for each transformation, I could show you how to use that to reach the answer.

 

[YummyNoodles]

Thank you so much for your help! Your tips were very helpful in reaching the solution. I am grateful that you helped me with this problem.