So we talked about the slope of the hypotenuse of a 45°-45°-90° triangle with horizontal and vertical legs. One of the students noted to the whole class that the slope triangle was a 45°-45°-90° triangle.
In the midst of talking about what we mean by slope, I happened to move the points so that the slope of the line containing A and B was 1. We started with the Math Nspired Algebra Slope as Rate activity. Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane derive the equation y=mx for a line through the origin and the equation y=mx+b for a line intercepting the vertical axis at b.Īnd so luckily it didn’t take me long to make the connection between the progression of how students should think about slope triangles from grade 8 through high school geometry. I recently came across learning objective 8.EE.6 while working on a project. It just seems like something that I have always known. And then I had to think even longer to realize that I don’t remember ever proving that parallel lines have equal slopes. I read and re-read this learning objective several times before it occurred to me that we weren’t just exploring the slopes of parallel and perpendicular lines to make conclusions about their relationships – we were actually asked to prove the slope criteria for parallel and perpendicular lines.
G-GPE 5, under “Use coordinates to prove simple geometric theorems algebraically”, says that students should “Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).” Note: We used the Transformations – Translations lesson from Geometry Nspired as a guide for part of our exploration in this lesson. And then several noticed that not only were those segments congruent…but they are also parallel. 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. But what else has to be congruent? Finally, someone suggested that CC’ is congruent to AA’. And the corresponding angles are congruent. And the areas of the triangles are equal. And the perimeter of ΔABC is equal to the perimeter of ΔA’B’C’. And then that means that segment AB is congruent to segment A’B’. A translation is a rigid motion, and so ΔABC is congruent to ΔA’B’C’. This took longer…because they focused first on the triangles. And they began to refine their translation from the initial red figure (right and down) to the actual image of ΔABC using vector XY.Īt that point, I let them use their dynamic technology to translate ΔABC.Īnd I asked them to write down everything they could find that was congruent. The students began to look for and make use of structure by drawing in a right triangle with the vector as the hypotenuse. And the students began to make sense of the x- and y-components of the vector, even though they didn’t initially see them. But I began to hear some good words from the group conversations: “right triangle” from one, “hypotenuse” from another.
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. Just for the record, they had no idea how far right and how far down the vector was telling us to translate ΔABC.īut I wasn’t ready to tell them. I was ready to know if they really knew how far right and how far down the vector was telling us to translate ΔABC. But I was ready for them to attend to precision. So I asked students to predict the image of ΔABC when translated using vector XY.Īs you can imagine, most students put ΔA’B’C’ down and to the right of ΔABC. I wanted to know whether the students could use the vector to perform the translation without the technology. And we were going to use a vector as the means for giving us the information we needed for our translation. We were about to explore what translations buy us mathematically using TI-Nspire Technology. 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.īut here is what I really wanted to know. So the first Quick Poll I posed in our lesson on translations was to see what they would do with a vector translation. 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. And how much to move which way, otherwise known as distance and direction.
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