To determine which point \( C \) can form an isosceles triangle \( \triangle ABC \) with points \( A(2, 4) \) and \( B(3, 6) \), we need to find a point \( C \) such that at least two sides of the triangle are of equal length.
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Calculate the distance \( AB \): \[ AB = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]
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Check the given options for point \( C \):
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Option 1: \( C(5, 5) \) \[ AC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \] \[ BC = \sqrt{(5 - 3)^2 + (5 - 6)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \] (AB ≠ AC, but AB = BC)
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Option 2: \( C(3, 4) \) \[ AC = \sqrt{(3 - 2)^2 + (4 - 4)^2} = \sqrt{1^2 + 0^2} = 1 \] \[ BC = \sqrt{(3 - 3)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = 2 \] (No equal lengths)
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Option 3: \( C(0, 4) \) \[ AC = \sqrt{(0 - 2)^2 + (4 - 4)^2} = \sqrt{(-2)^2 + 0^2} = 2 \] \[ BC = \sqrt{(0 - 3)^2 + (4 - 6)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \] (No equal lengths)
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Option 4: \( C(5, 4) \) \[ AC = \sqrt{(5 - 2)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = 3 \] \[ BC = \sqrt{(5 - 3)^2 + (4 - 6)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \] (No equal lengths)
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Conclusion: The only case where two sides are equal is with \( C(5, 5) \), where \( AB = BC \).
Therefore, the correct answer is:
Option 1: \( C(5, 5) \).