To prove that the length of segment \( A'B' \) is half the length of segment \( AB \) after the dilation with a scale factor of \( \frac{1}{2} \) and a center at the origin, we need to follow these steps:
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Understand the Dilation Process:
For a dilation centered at the origin with a scale factor \( k \), the coordinates of any point \( (x, y) \) transform to \( (kx, ky) \). Since the scale factor is \( \frac{1}{2} \), the new coordinates become: \[ (x', y') = \left(\frac{1}{2}x, \frac{1}{2}y\right) \] -
Analyze Each Case:
We will apply the dilation transformation to the given points \( A \) and \( B \) to find \( A' \) and \( B' \) in each option:a. \( A' = (1, 5) \), \( B' = (3, 6) \)
\[ \text{Length of } A'B' = \sqrt{(3 - 1)^2 + (6 - 5)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]b. \( A' = (-1, -2) \), \( B' = (-3, -3) \)
\[ \text{Length of } A'B' = \sqrt{(-3 + 1)^2 + (-3 + 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]c. \( A' = (4, 8) \), \( B' = (12, 12) \)
\[ \text{Length of } A'B' = \sqrt{(12 - 4)^2 + (12 - 8)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \]d. \( A' = (1, 2) \), \( B' = (3, 3) \)
\[ \text{Length of } A'B' = \sqrt{(3 - 1)^2 + (3 - 2)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \] -
Finding the Length of \( AB \) for each option:
We need to find the original points \( A \) and \( B \) based on \( A' \) and \( B' \). Since \( (A', B') \) are obtained by multiplying \( (A, B) \) by \( \frac{1}{2} \):- \( A = (2x', 2y') \)
- \( B = (2x'', 2y'') \)
Now let’s recalculate the original lengths for each:
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For option a: \[ A = (2, 10), B = (6, 12) \implies AB = \sqrt{(6 - 2)^2 + (12 - 10)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \]
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For option b: \[ A = (-2, -4), B = (-6, -6) \implies AB = \sqrt{(-6 + 2)^2 + (-6 + 4)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} \]
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For option c: This is clearly incorrect since \( A' \) and \( B' \) return point values beyond original points: \[ A = (8, 16), B = (24, 24) \implies AB = \sqrt{(24 - 8)^2 + (24 - 16)^2} = \sqrt{16^2 + 8^2} = \sqrt{256 + 64} = \sqrt{320} \]
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For option d: \[ A = (2, 4), B = (6, 6) \implies AB = \sqrt{(6 - 2)^2 + (6 - 4)^2} = \sqrt{4^2 + 2^2} = \sqrt{20} \]
- Conclusion:
The length of \( A'B' \) is \( \sqrt{5} \) and \( AB\) is \( \sqrt{20} \). Thus, indeed, \( A'B' \) is half the length of \( AB \).
The options a, b, and d yields proper results as is valid for proving dilation with appropriate conditions met. Thus confirming:
The lengths that fit the dilations and confirmations are options: - a, b, and d are correct, as their lengths of the dilated segments \( A'B' \) conform to half that of \( AB \).
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