Math / Geometry

QuestionWrite the coordinates of the vertices after a dilation with a scale factor of 3 , centered at the origin. $\begin{array}{l} P^{\prime}(\square, \square) \\ Q^{\prime}(\square, \square) \\ R^{\prime}(\square, \square) \\ S^{\prime}(\square, \square) \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to stretch a shape by making it 3 times bigger, keeping the origin as the center of the stretching action! Watch out! Don't forget that dilation means multiplying both the $x$ and $y$ coordinates by the scale factor.

STEP 2

1. Dilate point P
2. Dilate point Q
3. Dilate point R
4. Dilate point S

STEP 3

Our initial point P has coordinates $(0, -3)$ .
We're dilating with a scale factor of $3$ , centered at the origin.
This means we multiply both the $x$ and $y$ coordinates of P by $3$ .

STEP 4

So, $P'$ will have coordinates $(0 \cdot 3, -3 \cdot 3) = (0, -9)$ .
See how the origin stays put, and P' moves further away?
That's dilation, baby!

STEP 5

Point Q starts at $(3, 0)$ .
We apply the same dilation with a scale factor of $3$ , centered at the origin.

STEP 6

Multiplying both coordinates by $3$ , we get $Q'$ at $(3 \cdot 3, 0 \cdot 3) = (9, 0)$ .
Q just got super-sized!

STEP 7

R's original coordinates are $(0, 3)$ .
Let's stretch it by our scale factor of $3$ , centered at the origin.

STEP 8

Multiplying both coordinates by $3$ , we find $R'$ at $(0 \cdot 3, 3 \cdot 3) = (0, 9)$ .
R is reaching for the stars!

STEP 9

S begins its journey at $(-3, 0)$ .
We're going to dilate it with a scale factor of $3$ , centered at the origin.

STEP 10

Multiplying both coordinates by $3$ gives us $S'$ at $(-3 \cdot 3, 0 \cdot 3) = (-9, 0)$ .
S just took a giant leap outwards!

STEP 11

$\begin{array}{l} P^{\prime}(0, -9) \\ Q^{\prime}(9, 0) \\ R^{\prime}(0, 9) \\ S^{\prime}(-9, 0) \end{array}$

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QuestionWrite the coordinates of the vertices after a dilation with a scale factor of 3 , centered at the origin. $\begin{array}{l} P^{\prime}(\square, \square) \\ Q^{\prime}(\square, \square) \\ R^{\prime}(\square, \square) \\ S^{\prime}(\square, \square) \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to stretch a shape by making it 3 times bigger, keeping the origin as the center of the stretching action! Watch out! Don't forget that dilation means multiplying both the $x$ and $y$ coordinates by the scale factor.

STEP 2

1. Dilate point P
2. Dilate point Q
3. Dilate point R
4. Dilate point S

STEP 3

Our initial point P has coordinates $(0, -3)$ .
We're dilating with a scale factor of $3$ , centered at the origin.
This means we multiply both the $x$ and $y$ coordinates of P by $3$ .

STEP 4

So, $P'$ will have coordinates $(0 \cdot 3, -3 \cdot 3) = (0, -9)$ .
See how the origin stays put, and P' moves further away?
That's dilation, baby!

STEP 5

Point Q starts at $(3, 0)$ .
We apply the same dilation with a scale factor of $3$ , centered at the origin.

STEP 6

Multiplying both coordinates by $3$ , we get $Q'$ at $(3 \cdot 3, 0 \cdot 3) = (9, 0)$ .
Q just got super-sized!

STEP 7

R's original coordinates are $(0, 3)$ .
Let's stretch it by our scale factor of $3$ , centered at the origin.

STEP 8

Multiplying both coordinates by $3$ , we find $R'$ at $(0 \cdot 3, 3 \cdot 3) = (0, 9)$ .
R is reaching for the stars!

STEP 9

S begins its journey at $(-3, 0)$ .
We're going to dilate it with a scale factor of $3$ , centered at the origin.

STEP 10

Multiplying both coordinates by $3$ gives us $S'$ at $(-3 \cdot 3, 0 \cdot 3) = (-9, 0)$ .
S just took a giant leap outwards!

STEP 11

$\begin{array}{l} P^{\prime}(0, -9) \\ Q^{\prime}(9, 0) \\ R^{\prime}(0, 9) \\ S^{\prime}(-9, 0) \end{array}$

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Studdy Solution

STEP 1

What is this asking? We need to stretch a shape by making it **3 times bigger**, keeping the origin as the center of the stretching action! Watch out! Don't forget that dilation means multiplying *both* the xxx and yyy coordinates by the scale factor.

STEP 2

1. Dilate point P2. Dilate point Q3. Dilate point R4. Dilate point S

STEP 3

Our **initial point** P has coordinates (0,−3)(0, -3)(0,−3).We're dilating with a **scale factor** of 333, centered at the origin.This means we **multiply** both the xxx and yyy coordinates of P by 333.

STEP 4

So, P′P'P′ will have coordinates (0⋅3,−3⋅3)=(0,−9)(0 \cdot 3, -3 \cdot 3) = (0, -9)(0⋅3,−3⋅3)=(0,−9).See how the origin stays put, and P' moves further away?That's dilation, baby!

STEP 5

Point Q starts at (3,0)(3, 0)(3,0).We apply the same **dilation** with a **scale factor** of 333, centered at the origin.

STEP 6

Multiplying both coordinates by 333, we get Q′Q'Q′ at (3⋅3,0⋅3)=(9,0)(3 \cdot 3, 0 \cdot 3) = (9, 0)(3⋅3,0⋅3)=(9,0).Q just got super-sized!

STEP 7

R's original coordinates are (0,3)(0, 3)(0,3).Let's **stretch** it by our **scale factor** of 333, centered at the origin.

STEP 8

Multiplying both coordinates by 333, we find R′R'R′ at (0⋅3,3⋅3)=(0,9)(0 \cdot 3, 3 \cdot 3) = (0, 9)(0⋅3,3⋅3)=(0,9).R is reaching for the stars!

STEP 9

S begins its journey at (−3,0)(-3, 0)(−3,0).We're going to **dilate** it with a **scale factor** of 333, centered at the origin.

STEP 10

Multiplying both coordinates by 333 gives us S′S'S′ at (−3⋅3,0⋅3)=(−9,0)(-3 \cdot 3, 0 \cdot 3) = (-9, 0)(−3⋅3,0⋅3)=(−9,0).S just took a giant leap outwards!

STEP 11

P′(0,−9)Q′(9,0)R′(0,9)S′(−9,0)\begin{array}{l} P^{\prime}(0, -9) \\ Q^{\prime}(9, 0) \\ R^{\prime}(0, 9) \\ S^{\prime}(-9, 0) \end{array}P′(0,−9)Q′(9,0)R′(0,9)S′(−9,0)​

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What is this asking? We need to stretch a shape by making it 3 times bigger, keeping the origin as the center of the stretching action! Watch out! Don't forget that dilation means multiplying both the $x$ and $y$ coordinates by the scale factor.

1. Dilate point P
2. Dilate point Q
3. Dilate point R
4. Dilate point S

Our initial point P has coordinates $(0, -3)$ .
We're dilating with a scale factor of $3$ , centered at the origin.
This means we multiply both the $x$ and $y$ coordinates of P by $3$ .

So, $P'$ will have coordinates $(0 \cdot 3, -3 \cdot 3) = (0, -9)$ .
See how the origin stays put, and P' moves further away?
That's dilation, baby!

Point Q starts at $(3, 0)$ .
We apply the same dilation with a scale factor of $3$ , centered at the origin.

Multiplying both coordinates by $3$ , we get $Q'$ at $(3 \cdot 3, 0 \cdot 3) = (9, 0)$ .
Q just got super-sized!

R's original coordinates are $(0, 3)$ .
Let's stretch it by our scale factor of $3$ , centered at the origin.

Multiplying both coordinates by $3$ , we find $R'$ at $(0 \cdot 3, 3 \cdot 3) = (0, 9)$ .
R is reaching for the stars!

S begins its journey at $(-3, 0)$ .
We're going to dilate it with a scale factor of $3$ , centered at the origin.

Multiplying both coordinates by $3$ gives us $S'$ at $(-3 \cdot 3, 0 \cdot 3) = (-9, 0)$ .
S just took a giant leap outwards!

$\begin{array}{l} P^{\prime}(0, -9) \\ Q^{\prime}(9, 0) \\ R^{\prime}(0, 9) \\ S^{\prime}(-9, 0) \end{array}$