Find the third vertex of an equilateral triangle whose two vertices are (2, 4) and (2, 6).

Given:
Vertices A(2,4) and B( 2,6) 

Let the coordinate of C be (x, y) in an equilateral triangle all sides are of equal length.

 AB = AC = BC

Here AB = $$\sqrt {(2-2)² + (6-4)^2}$$​ = 2 units

and AC = $$\sqrt {(2-x)² + (4-y)^2}$$​ = 2units........(1)

BC = $$\sqrt {(2-x)² + (6-y)^2}$$​ = 2units.......(2)

also,

AC = BC

$$\sqrt {(2-x)² + (4-y)^2}$$ = $$\sqrt {(2-x)² + (6-y)^2}$$​…………(3)

squaring both sides we get,

(2 - x)²+(4 - y)² = (2 - x)² + (6- y)²

16+ y² - 8y = 36 + y² - 12y

4y = 20

 y = 5...... (4)

Now, putting y = 5 in equation (1) we have,

​$$\sqrt {(2-x)² + (4-5)^2}$$​ = 2

squaring both sides we have,

(2 - x)² + 1 = 4

= (2-x)² = 3

2-x = ±√3

x = 2 + √3

Thus the third vertex is  (x, y) = $$( 2 + \sqrt3 ,5)$$​​