To determine the pre-image of the segment [tex]\(\overline{A'B'}\)[/tex] when it is dilated at a scale factor of [tex]\(\frac{1}{2}\)[/tex] from the center [tex]\((2,0)\)[/tex], follow these steps:
1. Identify Given Points and Data:
- [tex]\(A' = (0,0)\)[/tex]
- [tex]\(B' = (2,0)\)[/tex]
- Scale Factor = [tex]\(\frac{1}{2}\)[/tex]
- Center of Dilation = [tex]\((2, 0)\)[/tex]
2. Use the Dilation Formula to Find Pre-Image Coordinates:
The dilation formula for a point [tex]\((x', y')\)[/tex] with respect to a center [tex]\((cx, cy)\)[/tex] and a scale factor [tex]\(k\)[/tex] is:
[tex]\[
x = cx + \frac{x' - cx}{k}
\][/tex]
[tex]\[
y = cy + \frac{y' - cy}{k}
\][/tex]
3. Calculate Coordinates of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] in the Pre-Image:
- For [tex]\(A\)[/tex]:
[tex]\[
A'_x = 0, \quad A'_y = 0
\][/tex]
From center [tex]\((2,0)\)[/tex], we have:
[tex]\[
A_x = 2 + \frac{0 - 2}{0.5} = 2 + \frac{-2}{0.5} = 2 - 4 = -2
\][/tex]
[tex]\[
A_y = 0 + \frac{0 - 0}{0.5} = 0
\][/tex]
Therefore, [tex]\(A = (-2,0)\)[/tex].
- For [tex]\(B\)[/tex]:
[tex]\[
B'_x = 2, \quad B'_y = 0
\][/tex]
Again from center [tex]\((2,0)\)[/tex], we have:
[tex]\[
B_x = 2 + \frac{2 - 2}{0.5} = 2 + \frac{0}{0.5} = 2
\][/tex]
[tex]\[
B_y = 0 + \frac{0 - 0}{0.5} = 0
\][/tex]
Therefore, [tex]\(B = (2,0)\)[/tex].
4. Verify Length of the Segment:
- Distance of [tex]\(\overline{A'B'}\)[/tex]:
[tex]\[
\overline{A'B'} = \sqrt{(2 - 0)^2 + (0 - 0)^2} = 2
\][/tex]
- Distance of [tex]\(\overline{AB}\)[/tex]:
[tex]\[
\overline{AB} = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4
\][/tex]
Here, the length of [tex]\(\overline{AB}\)[/tex] is twice the length of [tex]\(\overline{A'B'}\)[/tex].
5. Conclusion:
The pre-image [tex]\(\overline{A B}\)[/tex] of the segment [tex]\(\overline{A' B'}\)[/tex] is located at the points [tex]\(A(-2, 0)\)[/tex] and [tex]\(B(2, 0)\)[/tex] and is twice the length of [tex]\(\overline{A'B'}\)[/tex].
Therefore, the correct statement is:
[tex]\[
\overline{A B} \text{ is located at } A(-2,0) \text{ and } B(2,0) \text{ and is twice the length of } \overline{A^{\prime} B^{\prime}}.
\][/tex]
