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[tex]$\overline{X^{\prime} Y^{\prime}}$[/tex] has endpoints located at [tex]$X^{\prime}(0,8)$[/tex] and [tex]$Y^{\prime}(0,2)$[/tex]. It was dilated at a scale factor of 2 from center [tex]$(0,2)$[/tex]. Which statement describes the pre-image?

A. [tex]$\overline{X Y}$[/tex] is located at [tex]$X(0,1)$[/tex] and [tex]$Y(0,5)$[/tex] and is half the length of [tex]$\overline{X^{\prime} Y^{\prime}}$[/tex].
B. [tex]$\overline{X Y}$[/tex] is located at [tex]$X(0,1)$[/tex] and [tex]$Y(0,5)$[/tex] and is twice the length of [tex]$\overline{X^{\prime} Y^{\prime}}$[/tex].
C. [tex]$\overline{X Y}$[/tex] is located at [tex]$X(0,5)$[/tex] and [tex]$Y(0,2)$[/tex] and is half the length of [tex]$\overline{X^{\prime} Y}$[/tex].
D. [tex]$\overline{X Y}$[/tex] is located at [tex]$X(0,5)$[/tex] and [tex]$Y(0,2)$[/tex] and is twice the length of [tex]$\overline{X^{\prime} Y^{\prime}}$[/tex].


Sagot :

To solve this problem, we need to determine the pre-image coordinates and the properties of the segment [tex]\(\overline{XY}\)[/tex] before dilation.

The given endpoints of the dilated segment [tex]\(\overline{X'Y'}\)[/tex] are [tex]\(X'(0, 8)\)[/tex] and [tex]\(Y'(0, 2)\)[/tex]. We are told that the dilation was centered at [tex]\((0, 2)\)[/tex] with a scale factor of [tex]\(2\)[/tex].

Here's a step-by-step solution to determine the pre-image coordinates:

1. Determine the coordinates of point [tex]\(X\)[/tex]:
The coordinates of [tex]\(X'\)[/tex] are [tex]\((0, 8)\)[/tex]. Given the dilation center [tex]\((0, 2)\)[/tex] and the scale factor [tex]\(2\)[/tex], the coordinates of [tex]\(X\)[/tex] can be found as follows:

[tex]\[ X = \left( \frac{0 - 0}{2} + 0, \frac{8 - 2}{2} + 2 \right) = (0, \frac{6}{2} + 2) = (0, 3 + 2) = (0, 5) \][/tex]

2. Determine the coordinates of point [tex]\(Y\)[/tex]:
The coordinates of [tex]\(Y'\)[/tex] are [tex]\((0, 2)\)[/tex]. Given the dilation center [tex]\((0, 2)\)[/tex] and the scale factor [tex]\(2\)[/tex], the coordinates of [tex]\(Y\)[/tex] can be found as follows:

[tex]\[ Y = \left( \frac{0 - 0}{2} + 0, \frac{2 - 2}{2} + 2 \right) = (0, \frac{0}{2} + 2) = (0, 0 + 2) = (0, 2) \][/tex]

3. Verify the length and coordinates of segment [tex]\(\overline{XY}\)[/tex]:
The length of the original segment [tex]\(\overline{X'Y'}\)[/tex] can be calculated by the distance formula (though in this case it's simply vertical distance):

[tex]\[ \text{Length of } \overline{X'Y'} = |8 - 2| = 6 \][/tex]

For the pre-image segment [tex]\(\overline{XY}\)[/tex]:

[tex]\[ \text{Length of } \overline{XY} = |5 - 2| = 3 \][/tex]

Since the pre-image segment is half the length of the dilated segment:

[tex]\[ \text{Length of } \overline{XY} = \frac{\text{Length of } \overline{X'Y'}}{2} \][/tex]

4. Identify the correct statement:
Comparing with the given choices:

- [tex]\(\overline{XY}\)[/tex] is located at [tex]\(X(0, 5)\)[/tex] and [tex]\(Y(0, 2)\)[/tex] and is half the length of [tex]\(\overline{X'Y'}\)[/tex].

The correct statement 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 corresponding choice is:
3.
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