## Translations

25 Aug

What do we need for a translation?

An object. And how much to move which way, otherwise known as distance and direction.

My students have some experience with translations before they get to my high school geometry course…but mostly their experience is with translating an object 2 units right and 1 unit down. So the first Quick Poll I posed in our lesson on translations was to see what they would do with a vector translation. Most of them concluded that the vector was giving them the x-component and the y-component of the translation without me having to tell them.

But here is what I really wanted to know. We were about to explore what translations buy us mathematically using TI-Nspire Technology. And we were going to use a vector as the means for giving us the information we needed for our translation. I wanted to know whether the students could use the vector to perform the translation without the technology. So I asked students to predict the image of ΔABC when translated using vector XY.

As you can imagine, most students put ΔA’B’C’ down and to the right of ΔABC. But I was ready for them to attend to precision. I was ready to know if they really knew how far right and how far down the vector was telling us to translate ΔABC. Just for the record, they had no idea how far right and how far down the vector was telling us to translate ΔABC.

But I wasn’t ready to tell them. I asked again. And I waited. How does the vector give us information about distance and direction? It was uncomfortable for the students to have the images in a general plane instead of the x-y coordinate plane. But I began to hear some good words from the group conversations: “right triangle” from one, “hypotenuse” from another. And the students began to make sense of the x- and  y-components of the vector, even though they didn’t initially see them. The students began to look for and make use of structure by drawing in a right triangle with the vector as the hypotenuse. And they began to refine their translation from the initial red figure (right and down) to the actual image of ΔABC using vector XY.

At that point, I let them use their dynamic technology to translate ΔABC.

And I asked them to write down everything they could find that was congruent. This took longer…because they focused first on the triangles. A translation is a rigid motion, and so ΔABC is congruent to ΔA’B’C’. And then that means that segment AB is congruent to segment A’B’. And the perimeter of ΔABC is equal to the perimeter of ΔA’B’C’. And the areas of the triangles are equal. And the corresponding angles are congruent. But what else has to be congruent? Finally, someone suggested that CC’ is congruent to AA’. Really? How do you know? And do others agree? So we took a bit more time, and the students decided that AA’ is also congruent to BB’, which is also congruent to XY. And then several noticed that not only were those segments congruent…but they are also parallel.

And then the bell rang.

And so the journey continues ….

Note: We used the Transformations – Translations lesson from Geometry Nspired as a guide for part of our exploration in this lesson.