To determine which point C makes triangle ABC isosceles, we need to check the distances between the points A (2, 4), B (3, 6), and each candidate point C provided.
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Calculate the distance between A and B: \[ AB = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]
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Now, let's check each option:
Option A: C (3, 4)
- Distance AC: \[ AC = \sqrt{(3 - 2)^2 + (4 - 4)^2} = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]
- Distance BC: \[ BC = \sqrt{(3 - 3)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2 \]
- Not isosceles (1, 2, \(\sqrt{5}\))
Option B: C (0, 4)
- Distance AC: \[ AC = \sqrt{(0 - 2)^2 + (4 - 4)^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2 \]
- Distance BC: \[ BC = \sqrt{(0 - 3)^2 + (4 - 6)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \]
- Not isosceles (2, \(\sqrt{5}\), \(\sqrt{13}\))
Option C: C (5, 5)
- Distance AC: \[ AC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]
- Distance BC: \[ BC = \sqrt{(5 - 3)^2 + (5 - 6)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
- Not isosceles (\(\sqrt{10}\), \(\sqrt{5}\), \(\sqrt{5}\))
Option D: C (5, 4)
- Distance AC: \[ AC = \sqrt{(5 - 2)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \]
- Distance BC: \[ BC = \sqrt{(5 - 3)^2 + (4 - 6)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
- Not isosceles (3, 2\(\sqrt{2}\), \(\sqrt{5}\))
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The only option where two sides are equal:
Option C plugs well, where \( AC \) is \( 1 \) and \( BC \) are equal. Thus, point \( C(5, 5) \) creates triangle ABC to be isosceles with two equal lengths (both being \( \sqrt{5} \)).
So, the answer is: C. C (5, 5).
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