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[tex]$\overline{AB}$[/tex] is dilated from the origin to create [tex]$\overline{A^{\prime}B^{\prime}}$[/tex] at [tex]$A^{\prime}(0, 8)$[/tex] and [tex]$B^{\prime}(8, 12)$[/tex]. What scale factor was [tex]$\overline{AB}$[/tex] dilated by?

A. [tex]$\frac{1}{2}$[/tex]
B. 2
C. 3
D. 4


Sagot :

To find the scale factor of the dilation that maps [tex]\(\overline{AB}\)[/tex] to [tex]\(\overline{A'B'}\)[/tex] with the given coordinates [tex]\(A'(0,8)\)[/tex] and [tex]\(B'(8,12)\)[/tex], we need to compare the distances of the original points from the origin and the distances of the transformed points from the origin.

Since the dilation is centered at the origin [tex]\((0, 0)\)[/tex], the coordinates of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are scaled by the same factor [tex]\(k\)[/tex] to produce [tex]\(A'\)[/tex] and [tex]\(B'\)[/tex].

1. Assume the original coordinates of [tex]\(A\)[/tex] are [tex]\((x_1 = 0, y_1)\)[/tex]. Given that after dilation, [tex]\(A' = (0,8)\)[/tex]:
- This indicates that [tex]\(y_1\)[/tex] is scaled to 8.

2. Next, assume the original coordinates of [tex]\(B\)[/tex] are [tex]\((x_2, y_2)\)[/tex]. Given [tex]\(B' = (8,12)\)[/tex]:
- Coordinates [tex]\((x_2, y_2)\)[/tex] are scaled to [tex]\((8, 12)\)[/tex].

Given that dilation is uniform, the same scale factor [tex]\(k\)[/tex] applies to both the x and y coordinates. Let's determine this scale factor starting with [tex]\(A\)[/tex]:

- Since [tex]\(A\)[/tex] scales to [tex]\(A' = (0,8)\)[/tex] and the x-coordinate is 0,
[tex]\[ k \cdot y_1 = 8 \][/tex]
but we do not have the specific value of [tex]\(y_1$. Any non-zero original coordinate when multiplied by k should give 8. We need to also consider \(B'\)[/tex].

For [tex]\(B\)[/tex] with [tex]\(B' = (8, 12)\)[/tex]:
- Particularly for [tex]\(x_2\)[/tex]:
[tex]\[ k \cdot x_2 = 8 \][/tex]
- And for [tex]\(y_2\)[/tex]:
[tex]\[ k \cdot y_2 = 12 \][/tex]

Finding consistent [tex]\(k\)[/tex]:
- Solving [tex]\(k \cdot x_2 = 8\)[/tex] for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{8}{x_2} \][/tex]

- Solving [tex]\(k \cdot y_2 = 12\)[/tex] for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{12}{y_2} \][/tex]

If coordinates scale uniformly, then:
[tex]\[ \frac{8}{x_2} = \frac{12}{y_2} \rightarrow 8 y_2 = 12 x_2 \rightarrow \frac{y_2}{x_2} = \frac{12}{8} = \frac{3}{2} \][/tex]

Since [tex]\(k = 2\)[/tex], we check:
[tex]\[ x_2 = \frac{8}{2} = 4 y_2 = \frac{12}{2} = 6 \][/tex]

Hence the coordinates [tex]\(A = (0, 4)\)[/tex] or equivalents coincide if k is checked as unitary:
Overall, the consistent scale factor [tex]\(k\)[/tex] verifying dilation:
[tex]\( \boxed{2} \)[/tex]

The scale factor was [tex]\( \boxed{2} \)[/tex].