# 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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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 $y = \frac{5}2,$

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

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

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

Therefore coordinates of third vertex are: