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Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.

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Vertices of an isosceles are: A(2, 0) and B(2, 5). 

Let the third vertex is P(x, y)

On squaring both sides, we get

(x - 2)2 + (y)2 = (x - 2)2 + (y - 5)2

x2 -  4x + 4 + y2 = x2 - 4x + 4 + y2 - 10y + 25

y = \(\frac{5}2,\)

Also, PA = 3

\(\sqrt{(x - 2)^2 + (y)^2} = 3\)

On squaring both sides, we get

(x - 2)2 + (y)2 = 9

x2 - 4x + 4 + y2 = 9

x2 - 4x + y2 = 5

On substituting \(y = \frac{5}2,\)

\(x^2 - 4x + \frac{25}4 = 5\)

\(x^2 - 4x + \frac{25} 4 - 5 = 0\)

\(x^2 - 4x + \frac{5}4 = 0\)

Using quadratic formula:

Therefore coordinates of third vertex are:

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