a -1,2 b 2,2 c 1,-1 d -2,-1 A -2,4 B 4,4 C 2,-2 D-4,-2 In the image, parallelogram ABCD

Question

is dilated in reference to the origin to form parallelogram A′B′C′D′
. What is the magnitude of dilation?

(1 point)
Responses

1
1

12
Start Fraction 1 over 2 End Fraction

13
Start Fraction 1 over 3 End Fraction

2

Answer 1

To find the magnitude of dilation for the parallelogram ABCD transforming to A'B'C'D', we can use the coordinates of the points before and after the dilation.

The given points of parallelogram ABCD are:

• A(-2, 4)
• B(4, 4)
• C(2, -2)
• D(-4, -2)

And the points of A'B'C'D' after dilation (as provided):

• A'(-1, 2)
• B'(2, 2)
• C'(1, -1)
• D'(-2, -1)

To find the dilation factor (k), we can take one pair of corresponding points. Let's use point A and A' for this calculation.

The formula for dilation from the origin is:

\[ A' = k \cdot A \]

Using point A(-2, 4):

\[ A' = (x', y') = (k \cdot x, k \cdot y) \]

So for point A:

\[ (-1, 2) = (k \cdot -2, k \cdot 4) \]

This gives us two equations:

1. For the x-coordinates: \[ -1 = -2k \] \[ k = \frac{1}{2} \]

2. For the y-coordinates: \[ 2 = 4k \] \[ k = \frac{1}{2} \]

Since we calculated the same “k” from both coordinates, the dilation factor is consistent.

Thus, the magnitude of the dilation is:

\[ \frac{1}{2} \]

The correct response is:

1/2

Answer 2

Triangle XYZ
is drawn on a coordinate plane with vertices X(0,4)
, Y(2,0)
, and Z(4,7)
. If you dilate the triangle to form triangle X′Y′Z′
using a magnitude of 12
and a center of dilation at (4,7)
, what are the coordinate points of the new vertices?(1 point)
Responses

X′(4,11)
, Y′(6,7)
, Z′(8,14)
upper X prime left parenthesis negative 4 comma 11 right parenthesis , upper Y prime left parenthesis 6 comma 7 right parenthesis , upper Z prime left parenthesis 8 comma 14 right parenthesis

X′(2,5.5)
, Y′(3,3.5)
, Z′(4,7)
upper X prime left parenthesis 2 comma 5.5 right parenthesis , upper Y prime left parenthesis 3 comma 3.5 right parenthesis , upper Z prime left parenthesis 4 comma 7 right parenthesis

X′(−4,−3)
, Y′(−2,−7)
, Z′(0,0)
upper X prime left parenthesis negative 4 comma negative 3 right parenthesis , upper Y prime left parenthesis negative 2 comma negative 7 right parenthesis , upper Z prime left parenthesis 0 comma 0 right parenthesis

X′(−4,1)
, Y′(0,−7)
, Z′(4,7)

Answer 3

To find the new coordinates of the vertices of triangle XYZ after dilation with a center at (4, 7) and a magnitude of \( \frac{1}{2} \), we can follow these steps:

1. Identify the original coordinates of the vertices:

   • \( X(0, 4) \)
   • \( Y(2, 0) \)
   • \( Z(4, 7) \)

2. Determine the coordinates after dilation: The formula for dilating a point \( (x, y) \) around a center \( (h, k) \) with a scale factor \( m \) is given by: \[ (x', y') = (h + m(x - h), k + m(y - k)) \]

   In this case, the center of dilation \( (h, k) = (4, 7) \) and the scale factor \( m = \frac{1}{2} \).

3. Calculate the new coordinates for each vertex:

   • For \( X(0, 4) \): \[ x' = 4 + \frac{1}{2}(0 - 4) = 4 - 2 = 2 \] \[ y' = 7 + \frac{1}{2}(4 - 7) = 7 - \frac{3}{2} = 7 - 1.5 = 5.5 \] So, \( X' = (2, 5.5) \).

   • For \( Y(2, 0) \): \[ x' = 4 + \frac{1}{2}(2 - 4) = 4 + \frac{1}{2}(-2) = 4 - 1 = 3 \] \[ y' = 7 + \frac{1}{2}(0 - 7) = 7 - \frac{7}{2} = 7 - 3.5 = 3.5 \] So, \( Y' = (3, 3.5) \).

   • For \( Z(4, 7) \): \[ x' = 4 + \frac{1}{2}(4 - 4) = 4 + 0 = 4 \] \[ y' = 7 + \frac{1}{2}(7 - 7) = 7 + 0 = 7 \] So, \( Z' = (4, 7) \).

4. Compile the new coordinates:

   • \( X' = (2, 5.5) \)
   • \( Y' = (3, 3.5) \)
   • \( Z' = (4, 7) \)

Now, checking the responses, the correct answer is:

X′(2,5.5), Y′(3,3.5), Z′(4,7).