Rotation and Dilations: Understanding Point Rotation and Coordinate Rules

$P A$		$P A^{'}$
$P B$		$P B^{'}$
$P C$		$P C^{'}$
$P D$		$P D^{'}$

$m ∠ A P A^{'}$
$m ∠ B P B^{'}$
$m ∠ C P C^{'}$

Counterclockwise Rotations Around the Origin
Angle of Rotation	Rule
$9 0^{\circ}$	$(x, y) \to (- y, x)$
$18 0^{\circ}$	$(x, y) \to (- x, - y)$
$27 0^{\circ}$	$(x, y) \to (y, - x)$

Counterclockwise Rotations About the Origin
$9 0^{\circ}$ Rotation	$18 0^{\circ}$ Rotation	$27 0^{\circ}$ Rotation
$Preimage (x, y) \to Image (- y, x)$	$Preimage (x, y) \to Image (- x, - y)$	$Preimage (x, y) \to Image (y, - x)$

Dilation With Scale Factor $k = 3$
Preimage	Multiply by $k$	Image
$A (0, 2)$	$(3 \cdot 0, 3 \cdot 2)$	$A^{'} (0, 6)$
$B (3, 1)$	$(3 \cdot 3, 3 \cdot 1)$	$B^{'} (9, 3)$
$C (2, - 1)$	$(3 \cdot 2, 3 \cdot (- 1))$	$C^{'} (6, - 3)$

A rotation about the origin maps $△ A B C$ to $△ A^{'} B^{'} C^{'} .$ Which graph shows an angle that could be measured to determine the angle of rotation about the origin?

We know that a rotation about the origin maps △ ABC to △ A'B'C'. We want to determine which of the provided graphs shows an angle that we could measure to find the angle of rotation about the origin. Let's recall how to find an angle of rotation.

Draw a ray from the center of rotation — in this case, the origin — through a vertex in the preimage.
Draw a matching ray from the center of rotation through the vertex's image.
Measure the angle formed by the rays.

With this in mind, let's consider the graphs one by one. Instead of rays, we will consider the dotted segments to determine the angle of rotation.

Graph A

Let's take a look at the first graph.

In the graph we see a segment that connects A and B'. These two points do not correspond to a preimage-image pair because the image of point A is A', not B'. Additionally, the segment does not pass through the origin, which is the center of rotation. Therefore, we cannot use this graph to find the angle of rotation.

Graph B

Now let's apply the same reasoning as we did before to Graph B.

The point B' is not the image of the point A, so we cannot use this graph to determine the angle of rotation. The angle formed by points A, the origin, and B' is not the angle of rotation.

Graph C

Here is another graph to consider.

In this graph, we see a segment that connects B and its image B' and that passes through the origin. This means that the angle formed by points B, the origin, and B' is the angle of rotation. It is 180^(∘) as they form a straight angle. The answer is C.

Graph D

Let's consider our last graph.

Again, the point C' is not the image of the point B, so we cannot use this graph to find the angle of rotation.

The vertices of isosceles right triangle

D E F

are

D (- 5, 4),

E (- 5, - 3),

and

F (a, b) .

The sides

D E

and

D F

of this triangle are equal to each other. What are the possible coordinates of the image

F^{'}

after a

9 0^{\circ}

clockwise rotation about the origin?

We know that the isosceles right triangle DEF has vertices D(- 5,4), E(- 5,- 3), and F(a,b). We will start by finding the coordinates of F. Let's graph points D and E.

We can see that the distance between the points is equal to 7 units. Since triangle DEF is isosceles, DE=DF. This means that the length of the second leg DF must also be 7. We can plot F in two different positions, either to the left of D or to the right of D.

We have two possible locations for F. Triangle I & Triangle II F(2,4) & F(- 12,4) Now we can perform the 90^(∘) clockwise rotation of △ DEF. Note that a 90^(∘) clockwise rotation is equivalent to a 270^(∘) counterclockwise rotation. Let's recall the formula for a 270^(∘) counterclockwise rotation about the origin and apply it to each vertex of △ DEF.

Triangle I		Triangle II
(x,y)	(y ,- x)	(x,y)	(y ,- x)
D(- 5,4)	D'(4,5)	D(- 5,4)	D'(4,5)
E(- 5, - 3)	E'(- 3,5)	E(- 5, - 3)	E'(- 3,5)
F(2,4)	F'(4,- 2)	F(- 12,4)	F'(4,12)

Therefore, the image of vertex F after a 90^(∘) clockwise rotation about the origin can be either (4,- 2) or (4,12). Let's draw the possible triangles and perform the rotation.

When a counterclockwise rotation is performed about the origin, the coordinates of the image can be written in relation to the coordinates of the preimage.

Counterclockwise Rotations About the Origin
90^(∘) Rotation	180^(∘) Rotation	270^(∘) Rotation
ccc Preimage & & Image [0.5em] (x,y) & → & (- y,x)	ccc Preimage & & Image [0.5em] (x,y) & → & (- x,- y)	ccc Preimage & & Image [0.5em] (x,y) & → & (y,- x)

Davontay draws a triangle with vertices

F (3, 3),

G (7, 2),

and

H (8, - 5) .

He then rotates the triangle

18 0^{\circ}

about the origin. Which transformation creates an image identical to the one Davontay made by rotating the triangle?

Let's start by finding the coordinates of the image of Davontay's triangle after the 180^(∘) rotation about the origin. This rotation changes both the x- and y-coordinates into their opposites. ( x, y) ⟶ ( - x, - y) Let's apply this rule to the vertices of the triangle.

Preimage (x,y)	Image (- x, - y)	Simplify
F( -6, -2)	F'( - 3, - 3 )	F'(- 3,- 3)
G( 7, 2)	G'( - 7 , - 2 )	G'(- 7,- 2)
H( 8, -5)	H'( - 8 , -( -5) )	H'(- 8,5)

The rotated triangle has the following coordinates. F'(- 3,- 3), G'(- 7, - 2), H'(- 8, 5) Now let's draw the image of the triangle after the rotation!

We want to get this image using alternative transformations. First, let's remember how the rotation changed the coordinates of points. ( x, y) ⟶ ( - x, - y) Recall that reflections across the axes change the sign of coordinates, but only one at a time.

Transformation	Effect on Coordinates
Reflection in the x-axis	Changes the y-coordinate to its opposite
Reflection in the y-axis	Changes the x-coordinate to its opposite

If we perform only one of these transformations, we will not get to the desired result. However, if we perform both of them one after the other, we will be able to recreate the rotation! ( x, y) ⇓ c Reflection in thex-axis ( x, - y) ⇓ c Reflection in they-axis ( - x, - y) Performing two reflections leaves us with the same result as performing one 180^(∘) rotation about the origin. Since this works for any point, it will also work for the vertices the triangle!

When we reflect the preimage first in the x-axis and then in the y-axis, we get the same image as the rotation 180^(∘) around the origin.

A square is dilated using a scale factor of

5 .

The resulting image is then dilated again using a scale factor of

\frac{1}{2 0} .

What scale factor should be used to dilate the original square to get the final image?

We are told that a square is dilated using a scale factor of 5. After that, the image is dilated using a scale factor of 120. We want to find a scale factor that dilates the original preimage square into the final image. Recall that to find the image of a vertex after a dilation with scale factor k, we multiply its coordinates by k. ccc Preimage & & Image [0.5em] (x,y)& ⇒ & ( kx, ky) This means that if we apply two dilations, we actually apply two scale factors. ccc Preimage & & Image [0.5em] (x,y)& ⇒ & ( k_1* x, k_1* y) ( k_1x, k_1y)& ⇒ & ( k_2 * k_1x, k_2 * k_1 y) This means that the scale factor of the final image will be the product of the first scale factor multiplied by the second scale factor. In our case, the first scale factor is 5 and the second is 120. Let's multiply them!

Therefore, dilating the original square by the scale tfrac14 will result in the same image as the original process.

A line segment has endpoints

U (- 8, - 9)

and

V (- 8, 1) .

Which of the following pairs of points could be the endpoints of the image after a dilation?

We know that a line segment has endpoints U(-8,-9) and V(-8,1). We want to determine which one of the given answers represents a figure that is an image of the given segment after a dilation. A. & U'(- 8, 9) andV'(- 8,- 1) B. & U'(8, 9) and V'(8,1) C. & U'(- 16, - 18) and V'(- 16,2) D. & U'(80, 90) and V'(- 80,10) Recall that when a dilation by a scale factor of k is applied to a point, both of its coordinates are multiplied by the scale factor, k. ccc Preimage & & Image (x,y) & ⇒ & ( kx, ky) To check which of the pairs of points could be the endpoints of the image after a dilation, we need to check if there is a scale factor k that could produce this figure. Recall that the scale factor of dilation is a ratio between the coordinates in the image and the preimage. k=Image Coordinate/Corresponding Preimage Coordinate In other words, to check if a figure is the image of the given segment, we can divide each coordinate of the potential image's endpoints by the corresponding preimage coordinates. If the scale factors from each coordinate are equal, the figure is the image. Let's do it!

	Point U'		Point V'
Option	x-coordinate	y-coordinate	x-coordinate	y-coordinate
A	- 8/- 8=1	9/- 9=-1	-8/- 8=1	- 1/1=-1
B	8/- 8=-1	9/-9=- 1	8/- 8=-1	1/1=1
C	-16/-8=2	-18/-9=2	-16/-8=2	2/1=2
D	80/-8=- 10	90/-9= - 10	- 80/- 8= 10	10/1=10

The scale factor is the same for every coordinate only for the points from option C, so the segment with endpoints U'(- 16,- 18) and V'(- 16,2) is the image of the line segment UV after a dilation. The answer is C.

Diego and Jordan are creating shadows on the wall using a flashlight and their hands. Jordan traces the outline of Diego's shadow on the wall.

Which statements are true? Select all that apply.

I . II . III . The type of dilation here is an enlargement . As the distance between Diego and the flashlight decreases, the shadow on the wall becomes smaller . The flashlight represents the center of dilation .

If Diego's index finger is

6

inches long and the length of the index finger's shadow is

15

inches, what is the scale factor?

We know that we can use a flashlight to project shadow puppets on a wall. Let's take a look at the visualization.

As we can see, the shadow cast on the wall is larger than our hand. This means that the shadow is the image of our hand after a dilation. This type of dilation is an enlargement because the image is larger than the preimage, which in our case is our hand. Therefore, Statement I is true. ✓ I. & The type of dilation here is an enlargement. The second statement is false because the shadow on the wall becomes larger as the flashlight gets closer to our hand. * II. & As the distance between Diego and the & flashlight decreases, the shadow on the & wall becomes smaller. Remember that if we draw lines connecting corresponding vertices of the image and preimage, the lines will meet at a point called the center of dilation. For shadow puppets, the light rays meet at the flashlight, so the flashlight is the center of dilation. ✓ III. & The flashlight represents the center & of dilation. As a result, Statements I and III are true.

We know that Diego's index finger is 6 inches long. We want to find the scale factor of the dilation when the index finger's shadow is 15 inches long. Remember that the scale factor is the ratio of the length in the image to the corresponding length in the preimage. Scale Factor= Image Length/Corresponding Preimage Length In our case, the measurement of the image is 15 inches and the measurement of the preimage is 6 inches. Let's substitute these values into the formula and simplify.

The scale factor is 52, or 2.5.

Rotations and Dilations

Catch-Up and Review

Rotating Points About a Fixed Point

Rotating a Triangle About a Fixed Point

Rotation

Rotation of a Point Around Another Point

Cyrptawheel

Hint

Solution

Extra

Practice Rotating Triangles

Rotating a Point Around the Origin at $9 0^{\circ},$ $18 0^{\circ},$ and $27 0^{\circ}$

A Gateway to a New Realm

Answer

Hint

Solution

Extra

Exploring Dilations

Dilating a Drawing

How to Dilate?

Dilation

Coordinate Rule for Dilations

Practicing Dilatius's Teachings

Answer

Hint

Solution

Checking Our Answer

Dilatius's Final Lesson

Hint

Solution

Checking Our Answer

Comparing Rotations and Dilations

Graph A

Graph B

Graph C

Graph D

Rotations and Dilations

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