Geometry (B): READ: Rotations and Dilations Reading

Rotations

Learning Objectives

Find the image of a point in a rotation in a coordinate plane.
Recognize that a rotation is an isometry.

Sample Rotations

In this lesson we limit our study to rotations centered at the origin of a coordinate plane. We begin with some specific examples of rotations. Later we’ll see how these rotations fit into a general formula.

Remember how a rotation is defined. In a rotation centered at the origin with an angle of rotation of $n^\circ,$ a point moves counterclockwise along an arc of a circle. The central angle of the circle measures $n^\circ.$ The original point is one endpoint of the arc, and the image of the original point is the other endpoint of the arc.

180° Rotation

Our first example is rotation through an angle of $180^ \circ$ .

In a $180^ \circ$ rotation, the image of $P(h, k)$ is the point $P'(-h, -k)$ .

Notice:

$P$ and $P'$ are the endpoints of a diameter.
The rotation is the same as a “reflection in the origin.”

A $180^ \circ$ rotation is an isometry. The image of a segment is a congruent segment.

$PQ & = \sqrt{(k - t)^2 + (h - r)^2}\\ P'Q' & = \sqrt{(-k - -t)^2 + (-h - -r)^2} = \sqrt{(-k + t)^2 + (-h + r)^2}\\ & = \sqrt{(t - k)^2 + (r - h)^2} = \sqrt{(k - t)^2 + (h - r)^2}\\ PQ & = P'Q'$

If $M$ is a polygon matrix, then the matrix for the image of the polygon in a $180^ \circ$ rotation is the product $M$ $\begin{bmatrix} -1 &0\\ 0 &-1\\ \end{bmatrix}$ . The Lesson Exercises include exploration of this matrix for a $180^ \circ$ rotation.

90° Rotation

The next example is a rotation through an angle of $90^ \circ$ .

In a $90^ \circ$ rotation, the image of $P(h, k)$ is the point $P' (-k, h)$ .

Notice:

$\overline{P O}$ and $\overline{P' O}$ are radii of the same circle, so $PO = P'O$ .
$\angle{POP'}$ is a right angle.
The acute angle formed by $\overline{P O}$ and the $x-$ axis and the acute angle formed by $\overline{P' O}$ and the $x-$ axis are complementary angles.

A $90^ \circ$ rotation is an isometry. The image of a segment is a congruent segment.

$PQ & = \sqrt{(k - t)^2 + (h - r)^2}\\ P'Q' & = \sqrt{(h - r)^2 + (-k - -t)^2} = \sqrt{(h - r)^2 + (t - k)^2}\\ & = \sqrt{(k - t)^2 + (h - r)^2}\\ PQ & = P'Q'$

Dilations

Learning Objectives

Use the language of dilations.
Calculate and apply scalar products.
Use scalar products to represent dilations.

Introduction

We begin the lesson with a review of dilations, which were introduced in an earlier chapter. Like the other transformations, dilations can be expressed using matrices. Before we can do that, though, you will learn about a second kind of multiplication with matrices called scalar multiplication.

Dilation Refresher

The image of point $(h, k)$ in a dilation centered at the origin, with a scale factor $r$ , is the point $(rh, rk)$ .

For $r > 1$ , the dilation is an enlargement.

For $r < 1$ , the dilation is a reduction.

Any linear feature of an image is $r$ times as long as the length in the original figure.

Areas in the image are $r^2$ times the corresponding area in the original figure.

Scalar Products for Dilations

Recall from the Dilation Refresher above:

The image of point $(h, k)$ in a dilation centered at the origin, with a scale factor $r$ , is the point $(rh, rk)$ .

This is exactly the tool we need in order to use matrices for dilations.

Example 3

The following rectangle is dilated with a scale factor of $\frac{2} {3}$ .

Compositions with Dilations

Dilations can be one of the transformations in a composition, just as translations, reflections, and rotations can.

Example 4

We will use two transformations to move the circle below.

First we will dilate the circle with scale factor $4$ . Then, we will translate the new image $3 \;\mathrm{units}$ right and $5 \;\mathrm{units}$ up.