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[tex]$\overline{A B}$[/tex] is dilated from the origin to create [tex]$\overline{A^{\prime} B^{\prime}}$[/tex] at [tex]$A^{\prime}(0,4)$[/tex] and [tex]$B^{\prime}(4,6)$[/tex]. What scale factor was [tex]$\overline{A B}$[/tex] dilated by?

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


Sagot :

To determine the scale factor by which the line segment [tex]\(\overline{AB}\)[/tex] was dilated to become [tex]\(\overline{A'B'}\)[/tex] at [tex]\(A'(0,4)\)[/tex] and [tex]\(B'(4,6)\)[/tex], we can follow these steps:

1. Find the distances involved:
- Distance between the origin [tex]\((0, 0)\)[/tex] and [tex]\(A'(0, 4)\)[/tex]:
[tex]\[ \text{distance\_O\_Aprime} = \sqrt{(0-0)^2 + (4-0)^2} = \sqrt{16} = 4 \][/tex]

2. Assume a dilation factor [tex]\(k\)[/tex]:
- Since [tex]\(A'(0, 4)\)[/tex] is the image of point [tex]\(A(0, y)\)[/tex], under dilation by a factor [tex]\(k\)[/tex], we have:
[tex]\[ A'(0, 4) = (0, ky) \implies 4 = ky. \][/tex]

3. Consider the distance between [tex]\(A'\)[/tex] and [tex]\(B'\)[/tex]:
- Distance between [tex]\(A'(0, 4)\)[/tex] and [tex]\(B'(4, 6)\)[/tex]:
[tex]\[ \text{distance\_ABprime} = \sqrt{(4-0)^2 + (6-4)^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.472 \][/tex]

4. Investigate potential scale factors [tex]\(k\)[/tex]:
- Possible dilation factors given are [tex]\(\frac{1}{2}, 2, 3, 4\)[/tex].

5. Evaluate each scale factor [tex]\(k\)[/tex]:
- For [tex]\(k = \frac{1}{2}\)[/tex]:
[tex]\[ \frac{4}{\frac{1}{2}} = 8 \implies 4 \neq 8 \][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[ \frac{4}{2} = 2 \implies 4 \neq 2 \][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[ \frac{4}{3} \approx 1.333 \implies 4 \neq 1.333 \][/tex]
- For [tex]\(k = 4\)[/tex]:
[tex]\[ \frac{4}{4} = 1 \implies 4 \neq 1 \][/tex]

Given this, none of the provided scale factors correctly reconstruct the original segment distances and transformations under dilation. Therefore, from analyzing our provided information, we can conclude that the correct answer is:

[tex]\[ \boxed{None} \][/tex]