## Tuesday, December 1, 2015

### Lesson 6-2: Translations on the Coordinate Plane (Day 62)

We are revisiting Lesson 6-2, which is on translations. But there are two main differences between today's lesson and the one we had a few weeks ago (Day 52):

-- We have stated a Parallel Postulate, so we have the full power of Euclidean geometry.
-- We want to focus on the coordinate plane.

To emphasize the coordinate plane, I've titled this post "Translations on the Coordinate Plane" rather than just "Translations." Yesterday I warned you that another translation lesson was coming up soon, and as it turns out, that day is today.

I said that there are several ways to prove that the transformation mapping (x, y) to (x + h, y + k) really is a translation. First is a coordinate proof using the Slope and Distance Formulas, but I mentioned that we can't use those until after dilations in the second semester. Second is to manipulate the transformations until various mirrors cancel, but that requires arcane symbols such as:

T = r_U(n) o r_U(m)

and I said that it would only confuse students. Fortunately, there's a third way to complete the proof that I alluded to a few weeks back -- and I've decided to use this form of the proof in today's post.

We begin by noting that the two transformations (x, y) -> (x + h, y) and (x, y) -> (x, y + k) are already proved to be translations. The first translation slides points along the x-axis h units, and the second slides points along the y-axis k units. But how are points not on either axis transformed? We found out that in neutral geometry, we can prove that each point moves in the same direction, but not necessarily the same distance.

Now that we're in Euclidean geometry, we can prove that translations slide every point the same distance -- which is what we expect translations to do. We consider the points (0, 0), (a, 0), (a, b), and (0, b), and these points are the vertices of some quadrilateral. We know by definition, the side along the x-axis has length and also the side along y-axis has length b. We also know that three of the angles are right angles -- the angle at (0, 0) since the axes are perpendicular, the angle at (a, 0) because x = a is perpendicular to the x-axis, and the angle at (0, b) because y = b is perpendicular to the y-axis. In Euclidean geometry, we know that the sum of the angles of a quadrilateral is 360, so if three of the angles are right angles, so is the fourth, and the quadrilateral is a rectangle.

But now we know the lengths of two sides of this rectangle, a and b, and we want to find the lengths of the other two sides. Of course, the other two sides must also have length a and b as opposite sides of a rectangle are congruent. Yet how do we know this? Most Geometry texts would now say that every rectangle is a parallelogram and the opposite sides of a parallelogram are congruent, therefore the opposite sides of a rectangle are congruent. But unfortunately, the U of Chicago text doesn't give the Parallelogram Tests and Consequences until the last part of Chapter 7. We've already covered Lessons 7-1 to 7-5 on triangle congruence, and implied the Parallelogram Test for opposite sides to prove that every rhombus is a parallelogram, but we've certainly never proved the Parallelogram Consequence that every pgram has opposite sides congruent.

But let's consider the Quadrilateral Hierarchy yet again. Now only is every rectangle a parallelogram, but every rectangle is an isosceles trapezoid in two different ways -- and we've indeed proved that the opposite sides (legs) of an isosceles trapezoid are congruent. Therefore, we really do know that the opposite sides of a rectangle are congruent.

What we've shown is that the distance from (a, 0) to (a, b) is b units, and that the distance from (0, b) to (a, b) is a units. So horizontal and vertical distance work exactly as we expect them too -- that is, we now know that the distance from (x, y) to (x + h, y) is h units, and the distance from (x + h, y) to (x + h, y + k) is k units.

So (x, y) -> (x + h, y + k) moves points h units horizontally and then k units vertically. But that still doesn't tell us that the composite of the horizontal and vertical translations is itself a translation. Let's instead try the following -- let P(a, b) and Q(c, d) be two points, P' and Q' be the images of P and Q under the first translation, and then P" and Q" be the images of P' and Q' under the second one.

Then as the first translation slides h units, PP' = QQ' = h, and as the second translation slides k units, P'P" = Q'Q" = k, and as all horizontal and vertical lines are perpendicular (by the same rectangle argument given earlier), both angles PP'P" and QQ'Q" have measure 90. Thus triangles PP'P" and QQ'Q" are congruent by SAS.

So by CPCTC, PP" = QQ" without requiring the Distance Formula. Also, we have that the angles P'PP" and Q'QQ" -- the angles PP" and QQ" make with the horizontal -- are congruent, meaning that PP" and QQ" are in the same direction. So the transformation maps every point the same distance in the same direction. Therefore the transformation is a translation. QED

Notice that this proof essentially assumes a sort of "Converse to the Two Reflections Theorem for Translations" -- the forward theorem asserts that if a transformation is a translation, then it moves every point the same distance and direction, and so the converse would say that if a transformation moves every point the same distance and direction, then it's a translation. But this converse is trivial to prove -- as soon as we have a point P and its image P' and say that every point is moved the same distance and direction as P, then it's easy to find two mirrors to set up the translation. One of the mirrors can be placed at P and the other at the midpoint of PP', both mirrors perpendicular to PP".

Also, notice we assume that just because the two angles are equal, P and Q must be moved in the same direction. (And technically, we assume that any horizontal line must be perpendicular to any vertical line.) Both of these can be proved using corresponding angles. In a way, the proof would be similar to that on the old worksheet/activity from last month (Day 54) where we give two lines and two transversals and have students prove that the lines are parallel. Notice that on that old worksheet, if we're given that angles 5 and 9 are congruent, we can't conclude that any lines are parallel, but if we're given that one pair is parallel, we can combine the Parallel Consequences and Tests to conclude that the other pair is parallel, which is what we want.

Still, this proof should be more intuitive than the other versions which require symbolic manipulation or formulas we haven't covered yet to prove.