33. Triangle

*JKL*will undergo a transformation to create triangle

*J'K'L'*in the

*xy*-coordinate plane. Which transformations will result in triangle

*JKL*congruent to triangle

*J'K'L'*?

Select

**all**that apply:

A. (

*x*,

*y*) -> (-

*x*, -

*y*)

B. (

*x*,

*y*) -> (-

*x*,

*y*)

C. (

*x*,

*y*) -> (

*x*,

*y*- 5)

D. (

*x*,

*y*) -> (

*x*+ 3,

*y*- 5)

E. (

*x*,

*y*) -> (2

*x*, 3

*y*)

F. (

*x*,

*y*) -> (-

*x*,

*y*+ 3)

In other words, we are asking, which of these six transformations are

*isometries*? As it turns out, we can show five of these transformations to be isometries:

-- (A) is the rotation of 180 degrees centered at the origin.

-- (B) is the reflection in the

*y*-axis

-- (C) is the translation five units down.

-- (D) is the translation five units down and three units right.

-- (F) is a glide reflection. It is the composite of reflection in the

*y*-axis and translation three units up.

The only transformation that isn't an isometry is choice (E). This transformation is a "dilation" -- actually, it's not quite a dilation because it stretches figures by a horizontal factor of two and a vertical factor of three (while a true dilation would stretch by the same factor in all directions). So there are five correct answers -- (A), (B), (C), (D), and (F).

As usual, we expect the most common student error to be to mark fewer than five answers. Hopefully no student will try to mark (E) as correct -- but what if a student misreads the question as "which transformation will

*not*result in the triangles being congruent"? Otherwise, students are likely to miss choice (F), as glide reflections are always the most difficult isometries to understand.

Now here's an interesting question -- is it possible, just by looking at the coordinates, to tell which functions are isometries and which ones aren't? For example, we see from the above choices if a transformation maps

*x*to

*x*, -

*x*, or

*x*+

*h*and

*y*to

*y*, -

*y*, or

*y*+

*k*, then it is an isometry.

So what we're asking is for what sorts of functions

*f*(

*x*,

*y*) and

*g*(

*x*,

*y*) is the transformation:

(

*x*,

*y*) -> (

*f*(

*x*,

*y*),

*g*(

*x*,

*y*))

an isometry? It's obvious that

*f*and

*g*must be

*linear*functions. Looking at the answer choices again, we see that

*f*being

*x*, -

*x*, or

*x*+

*h*works, but not 2

*x*. So it appears our answer is that

*f*should be a linear function in

*x*with slope 1 or -1, and

*g*must be the same sort of function in

*y*:

(

*x*,

*y*) -> (

*f*(

*x*),

*g*(

*y*)),

*f*,

*g*linear functions with slope 1 or -1.

But as it turns out, this condition is sufficient, but not necessary, to produce an isometry. The only isometries that are of this form are translations, rotations of 180 degrees, and reflections (including glide reflections) whose mirrors are parallel to one of the axes.

Indeed, the complete answer to this question is quite complicated -- much too complicated for a high school Geometry student to figure out. It should be obvious though that our answer so far isn't complete when we look at reflection in the line

*y*=

*x*: (

*x*,

*y*) -> (

*y*,

*x*), or rotations of 90 degrees about the origin, which map (

*x*,

*y*) to either (

*y*, -

*x*) or (-

*y*,

*x*).

Rotations of angles that aren't multiples of 90 degrees and reflections in oblique mirrors are even more complicated than these. In Algebra II or Pre-Calc we sometimes use matrices to perform some of these rotations. With matrices, for example, we can prove that the rotation of angle

*theta*centered at the origin is:

(

*x*,

*y*) -> (

*x*cos

*theta*-

*y*sin

*theta*,

*x*sin

*theta*+

*y*cos

*theta*)

which, as we can see, is still linear in

*x*and

*y*. These transformations can be written in the form:

**v'**

**=**

**Av**+

**b**

**where**

**v**is the preimage written as a ordered pair (vector) and

**v'**is the image. Notice that this looks just like our crude solution using

*f*(

*x*) and

*g*(

*y*) earlier -- except that instead of having

*slope*1 or -1, the matrix

**A**must have

*determinant*1 or -1.

As it turns out,

**A**can be

*any*real 2x2 matrix with determinant 1 or -1 and we'll get a transformation that preserves

*area*, but this isn't quite the same as an isometry. The following transformation:

(

*x*,

*y*) = (2

*x*,

*y*/2)

preserves area, yet is not an isometry. It's just like choice (E) above, except that it doubles lengths in one direction and halves them in another, so that area is preserved.

As it turns out, the final complete answer requires knowledge of something called

*eigenvalues*, which are studied in Linear Algebra, a college-level class beyond Calculus. In order to produce an isometry, the matrix

**A**must have two eigenvalues, and both eigenvalues must have absolute value 1 (which sounds like 1 or -1 again, but they could also be complex numbers with absolute value 1).

But Geometry students don't have to know anything about eigenvalues to get this question right. All the right choices are recognizable as isometries, and the one that isn't has giveaway factors of 2 and 3 to show that it's not an isometry.

**PARCC Practice EOY Question 33**

**U of Chicago Correspondence: Lesson 6-5, Congruent Figures**

**Key Theorem: Definition of congruence**

**Two figures, F and G, are congruent figures if and only if G is the image of F under a translation, a reflection, a rotation, or any composite of these.**

**In a circle, the measure of an inscribed arc is one-half the measure of its intercepted arc.**

**Common Core Standard:**

CCSS.MATH.CONTENT.HSG.CO.B.6

Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.

**Commentary: Students need more practice with transformations on the coordinate plane, since the plane is not emphasized in the U of Chicago text. Thus students are given a triangle and asked to perform each of the six transformations on a given triangle**

*ABC*.

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