Math / Geometry

QuestionWrite the coordinates of the vertices after a rotation $270^{\circ}$ counterclockwise around the origin. $s^{\prime}(I$ , $\square$ $\square$ ) $T^{\prime}($ $\square$ , ) $\square$ $u^{\prime}$ , $\square$ ) $\square$

Studdy Solution

STEP 1

What is this asking? We need to spin the triangle $STU$ around the origin $270^\circ$ counter-clockwise and find the new locations of the corners! Watch out! Don't mix up clockwise and counter-clockwise rotations!
Also, remember that the origin is our center of rotation.

STEP 2

1. Rotation Rule
2. Rotate Point S
3. Rotate Point T
4. Rotate Point U

STEP 3

Let's remember the super cool rule for rotating a point $(x, y)$ counter-clockwise around the origin.
For a $270^\circ$ rotation, the new point becomes $(y, -x)$ .
Basically, the $y$ becomes the new $x$ , and the $x$ becomes the negative of the new $y$ !

STEP 4

Our original point $S$ is at $(8, 2)$ .
Let's apply our rotation rule.
The $y$ coordinate, $2$ , becomes the new $x$ .
The $x$ coordinate, $8$ , becomes the negative of the new $y$ , which is $-8$ .
So, $S'$ is at $(2, -8)$ !

STEP 5

$T$ starts at $(6, 6)$ .
Using our rotation rule, the $y$ coordinate, $6$ , becomes the new $x$ .
The $x$ coordinate, $6$ , becomes the negative of the new $y$ , which is $-6$ .
Therefore, $T'$ lands at $(6, -6)$ !

STEP 6

$U$ is originally at $(2, 2)$ .
Applying the rotation rule, the $y$ coordinate, $2$ , becomes the new $x$ .
The $x$ coordinate, $2$ , becomes the negative of the new $y$ , giving us $-2$ .
So, $U'$ is located at $(2, -2)$ !

STEP 7

$S'$ is at $(2, -8)$ . $T'$ is at $(6, -6)$ . $U'$ is at $(2, -2)$ .

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QuestionWrite the coordinates of the vertices after a rotation $270^{\circ}$ counterclockwise around the origin. $s^{\prime}(I$ , $\square$ $\square$ ) $T^{\prime}($ $\square$ , ) $\square$ $u^{\prime}$ , $\square$ ) $\square$

Studdy Solution

STEP 1

What is this asking? We need to spin the triangle $STU$ around the origin $270^\circ$ counter-clockwise and find the new locations of the corners! Watch out! Don't mix up clockwise and counter-clockwise rotations!
Also, remember that the origin is our center of rotation.

STEP 2

1. Rotation Rule
2. Rotate Point S
3. Rotate Point T
4. Rotate Point U

STEP 3

Let's remember the super cool rule for rotating a point $(x, y)$ counter-clockwise around the origin.
For a $270^\circ$ rotation, the new point becomes $(y, -x)$ .
Basically, the $y$ becomes the new $x$ , and the $x$ becomes the negative of the new $y$ !

STEP 4

Our original point $S$ is at $(8, 2)$ .
Let's apply our rotation rule.
The $y$ coordinate, $2$ , becomes the new $x$ .
The $x$ coordinate, $8$ , becomes the negative of the new $y$ , which is $-8$ .
So, $S'$ is at $(2, -8)$ !

STEP 5

$T$ starts at $(6, 6)$ .
Using our rotation rule, the $y$ coordinate, $6$ , becomes the new $x$ .
The $x$ coordinate, $6$ , becomes the negative of the new $y$ , which is $-6$ .
Therefore, $T'$ lands at $(6, -6)$ !

STEP 6

$U$ is originally at $(2, 2)$ .
Applying the rotation rule, the $y$ coordinate, $2$ , becomes the new $x$ .
The $x$ coordinate, $2$ , becomes the negative of the new $y$ , giving us $-2$ .
So, $U'$ is located at $(2, -2)$ !

STEP 7

$S'$ is at $(2, -8)$ . $T'$ is at $(6, -6)$ . $U'$ is at $(2, -2)$ .

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QuestionWrite the coordinates of the vertices after a rotation 270∘270^{\circ}270∘ counterclockwise around the origin. s′(Is^{\prime}(Is′(I , □\square□ □\square□ ) T′(T^{\prime}(T′( □\square□ , ) □\square□ u′u^{\prime}u′ , □\square□ ) □\square□

Studdy Solution

STEP 1

What is this asking? We need to spin the triangle STUSTUSTU around the origin 270∘270^\circ270∘ counter-clockwise and find the new locations of the corners! Watch out! Don't mix up clockwise and counter-clockwise rotations!Also, remember that the origin is our center of rotation.

STEP 2

1. Rotation Rule2. Rotate Point S3. Rotate Point T4. Rotate Point U

STEP 3

Let's remember the super cool rule for rotating a point (x,y)(x, y)(x,y) counter-clockwise around the origin.For a 270∘270^\circ270∘ rotation, the new point becomes (y,−x)(y, -x)(y,−x).Basically, the yyy becomes the new xxx, and the xxx becomes the negative of the new yyy!

STEP 4

Our **original point** SSS is at (8,2)(8, 2)(8,2).Let's apply our **rotation rule**.The yyy **coordinate**, 222, becomes the new xxx.The xxx **coordinate**, 888, becomes the negative of the new yyy, which is −8-8−8.So, S′S'S′ is at (2,−8)(2, -8)(2,−8)!

STEP 5

TTT starts at (6,6)(6, 6)(6,6).Using our **rotation rule**, the yyy **coordinate**, 666, becomes the new xxx.The xxx **coordinate**, 666, becomes the negative of the new yyy, which is −6-6−6.Therefore, T′T'T′ lands at (6,−6)(6, -6)(6,−6)!

STEP 6

UUU is originally at (2,2)(2, 2)(2,2).Applying the **rotation rule**, the yyy **coordinate**, 222, becomes the new xxx.The xxx **coordinate**, 222, becomes the negative of the new yyy, giving us −2-2−2.So, U′U'U′ is located at (2,−2)(2, -2)(2,−2)!

STEP 7

S′S'S′ is at (2,−8)(2, -8)(2,−8). T′T'T′ is at (6,−6)(6, -6)(6,−6). U′U'U′ is at (2,−2)(2, -2)(2,−2).

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionWrite the coordinates of the vertices after a rotation $270^{\circ}$ counterclockwise around the origin. $s^{\prime}(I$ , $\square$ $\square$ ) $T^{\prime}($ $\square$ , ) $\square$ $u^{\prime}$ , $\square$ ) $\square$

What is this asking? We need to spin the triangle $STU$ around the origin $270^\circ$ counter-clockwise and find the new locations of the corners! Watch out! Don't mix up clockwise and counter-clockwise rotations!
Also, remember that the origin is our center of rotation.

1. Rotation Rule
2. Rotate Point S
3. Rotate Point T
4. Rotate Point U

Let's remember the super cool rule for rotating a point $(x, y)$ counter-clockwise around the origin.
For a $270^\circ$ rotation, the new point becomes $(y, -x)$ .
Basically, the $y$ becomes the new $x$ , and the $x$ becomes the negative of the new $y$ !

Our original point $S$ is at $(8, 2)$ .
Let's apply our rotation rule.
The $y$ coordinate, $2$ , becomes the new $x$ .
The $x$ coordinate, $8$ , becomes the negative of the new $y$ , which is $-8$ .
So, $S'$ is at $(2, -8)$ !

$T$ starts at $(6, 6)$ .
Using our rotation rule, the $y$ coordinate, $6$ , becomes the new $x$ .
The $x$ coordinate, $6$ , becomes the negative of the new $y$ , which is $-6$ .
Therefore, $T'$ lands at $(6, -6)$ !

$U$ is originally at $(2, 2)$ .
Applying the rotation rule, the $y$ coordinate, $2$ , becomes the new $x$ .
The $x$ coordinate, $2$ , becomes the negative of the new $y$ , giving us $-2$ .
So, $U'$ is located at $(2, -2)$ !

$S'$ is at $(2, -8)$ . $T'$ is at $(6, -6)$ . $U'$ is at $(2, -2)$ .