Geometry (B): READ: Translation and Reflection Reading

READ: Translation and Reflection Reading

Translations

Learning Objectives

Graph a translation in a coordinate plane.
Recognize that a translation is an isometry.
Use vectors to represent a translation.

Introduction

Translations are familiar to you from earlier lessons. In this lesson, we restate our earlier learning in terms of motions in a coordinate plane. We’ll use coordinates and vectors to express the results of translations.

Translations

Remember that a translation moves every point a given horizontal distance and/or a given vertical distance. For example, if a translation moves the point $A(3, 7)$ $2 \;\mathrm{units}$ to the right and $4\;\mathrm{units}$ up, to $A'(5, 11)$ then this translation moves every point the same way.

The original point (or figure) is called the preimage, in this case $A(3, 7)$ . The translated point (or figure) is called the image, in this case $A'(5, 11)$ , and is designated with the prime symbol.

Example 1

The point $A(3, 7)$ in a translation becomes the point $A'(2,4)$ . What is the image of $B(-6,1)$ in the same translation?

Point $A$ moved $1 \;\mathrm{unit}$ to the left and $3\;\mathrm{units}$ down to reach $A'$ . $B$ will also move $1\;\mathrm{unit}$ to the left and $3\;\mathrm{units}$ down.

$B' = (-6 - 1,1 - 3) = (-7, -2)$

$B'(-8, -2)$ is the image of $B(-6, 1)$ .

Notice the following:

$AB = \sqrt{(-6 - 3)^2 + (1 - 7)^2} = \sqrt{(-9)^2 + (-6)^2} = \sqrt{117}$
$A'B' = \sqrt{(-7 - 2)^2 + (-2 - 4)^2} = \sqrt{(-9)^2 + (-6)^2} = \sqrt{117}$

Since the endpoints of $\overline{AB}$ and $\overline{A'B'}$ moved the same distance horizontally and vertically, both segments have the same length.

Translation is an Isometry

An isometry is a transformation in which distance is “preserved.” This means that the distance between any two points is the same as the distance between the images of the points.

Did you notice this in example 1 above?

$AB = A'B'$ (since they are both equal to $\sqrt{117}$ )

Would we get the same result for any other point in this translation? Yes. It’s clear that for any point $X$ , the distance from $X$ to $X'$ will be $\sqrt{117}$ . Every point moves $\sqrt{117}\;\mathrm{units}$ to its image.

This is true in general.

Translation Isometry Theorem

Every translation in the coordinate plane is an isometry.

You will prove this theorem in the Lesson Exercises.

Vectors

Let’s look at the translation in example 1 in a slightly different way.

Example 2

The point $A(3, 7)$ in a translation is the point $A'(5, 11)$ . What is the image of $B(-6, 1)$ in the same translation?

The arrow from $A$ to $A'$ is called a vector, because it has a length and a direction. The horizontal and vertical components of the vector are $2$ and $4$ respectively.

To find the image of $B$ , we can apply the same transformation vector to point $B$ . The arrowhead of the vector is at $B'(-4, 5)$ .

The vector in example 2 is often represented with a boldface single letter $v$ .

The horizontal component of vector $v$ is $2$ .
The vertical component of vector $v$ is $4$ .
The vector can also be represented as a number pair made up of the horizontal and vertical components.

The vector for this transformation is $v = (2, 4).$

Example 3

A triangle has vertices $A(-2, -5), B(0, 2)$ , and $C(2, -5)$ . The vector for a translation is $v = (0, 5)$ . What are the vertices of the image of the triangle?

Add the horizontal and vertical components to the $x-$ and $y-$ coordinates of the vertices.

$A' & = (-2 + 0, -5 + 5) = (-2, 0)\\ B' & = (0 + 0, 2 + 5) = (0, 7)\\ C' & = (2 + 0, -5 + 5) = (2, 0)$

Challenge: Can you describe what this transformation does to the original triangle?

Lesson Summary

You can think of a translation as a way to move points in a coordinate plane. And you can be sure that the shape and size of a figure stays the same in a translation. For that reason a translation is called an isometry. (Note: Isometry is a compound word with two roots in Greek, “iso” and “metry.” You may know other words with these same roots, in addition to “isosceles” and “geometry.”)

Vectors provide an alternative way to represent a translation. A vector has a direction and a length—the exact features that are involved in moving a point in a translation.

Points to Consider

Think about some special transformation vectors. Can you picture what each one does to a figure in a coordinate plane?

$v = (0, 0)$

$v = (5, 0)$

$v = (0, -5)$

$v = (5, 5)$

This lesson was about two-dimensional space represented by a coordinate grid. But we know there are more than two dimensions. The real world is actually multidimensional. Vectors are well suited to describe motion in that world. What would transformation vectors look like there?

Reflections

Learning Objectives
Find the reflection of a point in a line on a coordinate plane.
Multiply matrices.
Apply matrix multiplication to reflections.
Verify that a reflection is an isometry.

Introduction

You studied translations earlier, and saw that matrix addition can be used to represent a translation in a coordinate plane. You also learned that a translation is an isometry.

In this lesson, we will analyze reflections in the same way. This time we will use a new operation, matrix multiplication, to represent a reflection in a coordinate plane. We will see that reflections, like translations, are isometries.

You will have an opportunity to discover one surprising—or even shocking!—fact of matrix arithmetic.

Reflection in a Line

A reflection in a line is as if the line were a mirror.

An object reflects in the mirror, and we see the image of the object.

The image is the same distance behind the mirror as the object is in front of the mirror.

The “line of sight” from the object to the mirror is perpendicular to the mirror itself.

The “line of sight” from the image to the mirror is also perpendicular to the mirror.

Technology Note - Geometry Software

Use your geometry software to experiment with reflections.

Try this.

Draw a line.
Draw a triangle.
Reflect the triangle in the line.
Look at your results.
Repeat with different lines, and figures other than a triangle.

Now try this.

Draw a line
Draw a point.
Reflect the point in the line.
Connect the point and its reflection with a segment.
Measure the distance of the original point to the line, and the distance of the reflected point to the line.
Measure the angle formed by the original line and the segment connecting the original point and its reflection.

Let’s put this information in more precise terms.

Reflection of a Point in a Line:

Point $P'$ is the reflection of point $P$ in line $k$ if and only if line $k$ is the perpendicular bisector of $\overline{PP'}$ .

Reflections in Special Lines

In a coordinate plane there are some “special” lines for which it is relatively easy to create reflections.

the $x-$ axis
the $y-$ axis
the line $y = x$ (this line makes a $45^\circ$ angle between the $x-$ and $y-$ axes)

We can develop simple formulas for reflections in these lines.

Let $P(x, y)$ be a point in the coordinate plane.

We now have these reflections of $P(x, y)$ :

Reflection of $P$ in the $x-$ axis is $Q(x, -y)$ .

: [ $x-$ coordinate stays the same, $y-$ coordinate becomes opposite]

Reflection of $P$ in the $y-$ axis is $R(-x, y)$ .

: [ $x-$ coordinate becomes opposite, $y-$ coordinate stays the same]

Reflection of $P$ in the line $y = x$ is $S(y, x)$ .

: [switch the $x-$ and $y-$ coordinates]

A look is enough to convince us of the first two reflections. We’ll prove the third one.

Example 1

Prove that the reflection of point $P(h, k)$ in the line $y = x$ is the point $S(k, h)$ .

Here is an “outline” proof.

First, we know the slope of $y=x$

Slope of $y = x$ is $1y = 1x + 0$ .

Next, let's assume investigate the slope of $\overline{PS}$ .

Slope of $\overline{PS}$ is $\frac{k - h} {h - k} = \frac{-1 (h - k)} {h - k} = -1$ .

Therefore, we have just shown that $\overline{PS}$ and $y=x$ are perpendicular.

$\overline{PS}$ is perpendicular to $y = x$ (product of slopes is $-1$ ).

Finally, we can show that $y=x$ is the perpendicular bisector of $\overline{PS}$ .

Midpoint of $\overline{PS}$ is $\left(\frac{h + k} {2}, \frac{h + k} {2}\right)$ .
Midpoint of $\overline{PS}$ is on $y = x$ ( $x-$ and $y-$ coordinates of $\overline{PS}$ are the same).
$y = x$ is the perpendicular bisector of $\overline{PS}$ .

Conclusion: $P$ and $S$ are reflections in the line $y = x$ . $\blacklozenge$

Example 2

Point $P(5, 2)$ is reflected in the line $y = x$ . The image is $P'$ . $P'$ is then reflected in the $y-$ axis. The image is $P''$ . What are the coordinates of $P''$ ?

We find one reflection at a time.

Reflect $P$ in $y = x. \ P'$ is $( 2, 5)$ .
Reflect $P'$ in the $y-$ axis. $P''$ is $(-2, 5)$ .

Reflections Are Isometries

A reflection in a line is an isometry. Distance between points is “preserved” (stays the same).

We will verify the isometry for reflection in the $x-$ axis. The story is very similar for reflection in the $y-$ axis. You can write a proof that reflection in $y = x$ is an isometry in the Lesson Exercises.

The diagram below shows $\overline{PQ}$ and its reflection in the $x-$ axis, $\overline{P'Q'}$ .

$PQ & = \sqrt{(m - h)^2 + (n - k)^2}\\ P'Q' & = \sqrt{(m - h)^2 + (- n -(-k))^2} = \sqrt{(m - h)^3 + (k - n)^2} = \sqrt{(m - h)^2 + (n - k)^2}\\ PQ & = P'Q'$

Conclusion: When a segment is reflected in the $x-$ axis, the image segment has the same length as the original segment. This is the meaning of isometry. You can see that a similar argument would apply to reflection in any line.

Last modified: Thursday, 8 July 2010, 9:59 AM