The given endpoints of [tex]\(\overline{B^{\prime} C^{\prime}}\)[/tex] are [tex]\(B^{\prime}(2,0)\)[/tex] and [tex]\(C^{\prime}(2,6)\)[/tex].
We are told that [tex]\(\overline{BC}\)[/tex] was created by dilating [tex]\(\overline{B^{\prime} C^{\prime}}\)[/tex] from the origin with a scale factor of 2.
To find the pre-image points [tex]\(B\)[/tex] and [tex]\(C\)[/tex], we need to divide the coordinates of [tex]\(B^{\prime}\)[/tex] and [tex]\(C^{\prime}\)[/tex] by the scale factor. Therefore:
1. For point [tex]\(B\)[/tex]:
[tex]\[
B = \left( \frac{B^{\prime}_x}{2}, \frac{B^{\prime}_y}{2} \right) = \left( \frac{2}{2}, \frac{0}{2} \right) = (1, 0)
\][/tex]
2. For point [tex]\(C\)[/tex]:
[tex]\[
C = \left( \frac{C^{\prime}_x}{2}, \frac{C^{\prime}_y}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)
\][/tex]
Next, we verify their sizes to ensure that [tex]\( \overline{BC} \)[/tex] is correctly described.
The distance between [tex]\(B^{\prime}\)[/tex] and [tex]\(C^{\prime}\)[/tex] is calculated as:
[tex]\[
B^{\prime}C^{\prime} = \sqrt{(C^{\prime}_x - B^{\prime}_x)^2 + (C^{\prime}_y - B^{\prime}_y)^2} = \sqrt{(2 - 2)^2 + (6 - 0)^2} = \sqrt{0 + 36} = 6
\][/tex]
The distance between [tex]\(B\)[/tex] and [tex]\(C\)[/tex] is:
[tex]\[
BC = \sqrt{(C_x - B_x)^2 + (C_y - B_y)^2} = \sqrt{(1 - 1)^2 + (3 - 0)^2} = \sqrt{0 + 9} = 3
\][/tex]
Since [tex]\( BC \)[/tex] is half the size of [tex]\( B^{\prime}C^{\prime} \)[/tex], we conclude that:
[tex]\[
\overline{BC} \text{ is located at } B(1,0) \text{ and } C(1,3) \text{ and is one-half the size of } \overline{B^{\prime}C^{\prime}}.
\][/tex]
Thus, the correct statement is:
[tex]\[
\overline{BC} \text{ is located at } B(1,0) \text{ and } C(1,3) \text{ and is one-half the size of } \overline{B^{\prime}C^{\prime}}.
\][/tex]
