To solve this problem, we need to determine the original positions of the points [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] before the dilation occurred. We know the dilation had a scale factor of 2 with a center at [tex]\((0, 2)\)[/tex].
1. Identify Given Points and Center of Dilation:
- [tex]\(X'(0,8)\)[/tex]
- [tex]\(Y'(0,2)\)[/tex]
- Center of dilation: [tex]\((0,2)\)[/tex]
- Scale factor: 2
2. Find the Pre-Image Coordinates:
For a point [tex]\(P'(x', y')\)[/tex] that is dilated at the center [tex]\((0, 2)\)[/tex] with scale factor 2, the pre-image point [tex]\(P(x, y)\)[/tex] can be found using the formula:
[tex]\[
y - 2 = \frac{1}{2}(y' - 2)
\][/tex]
Solving this for each coordinate:
- For [tex]\(X'(0,8)\)[/tex]:
[tex]\[
y - 2 = \frac{1}{2}(8 - 2) \implies y - 2 = 3 \implies y = 5
\][/tex]
So [tex]\(X(0, 5)\)[/tex].
- For [tex]\(Y'(0,2)\)[/tex]:
[tex]\[
y - 2 = \frac{1}{2}(2 - 2) \implies y - 2 = 0 \implies y = 2
\][/tex]
So [tex]\(Y(0, 2)\)[/tex].
3. Determine Lengths of Segments:
- Length of [tex]\(\overline{X'Y'}\)[/tex]:
[tex]\[
|8 - 2| = 6
\][/tex]
- Length of [tex]\(\overline{XY}\)[/tex]:
[tex]\[
|5 - 2| = 3
\][/tex]
Since the dilation scale factor was 2, the length of the pre-image [tex]\(\overline{XY}\)[/tex] should be half of the length of [tex]\(\overline{X'Y'}\)[/tex].
[tex]\[
\frac{6}{2} = 3
\][/tex]
4. Match the Pre-Image Points and Length with the Given Statements:
- The points [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] after dilation are [tex]\(X(0, 5)\)[/tex] and [tex]\(Y(0, 2)\)[/tex].
- The length of [tex]\(\overline{XY}\)[/tex] is half the length of [tex]\(\overline{X'Y'}\)[/tex].
The statement that matches this description is:
[tex]\[
\overline{XY} \text{ is located at } X(0, 5) \text{ and } Y(0, 2) \text{ and is half the length of } \overline{X'Y'}
\][/tex]
Thus, the correct statement is:
[tex]\[
\text{Option 3: } \overline{X Y} \text{ is located at } X(0,5) \text{ and } Y(0,2) \text{ and is half the length of } \overline{X^{\prime} Y^{\prime}}.
\][/tex]
