<PRE><B>Question 1058558</B>
Since the line connecting (1, 2) and (4, -1) is the base of the isosceles triangle, the vertex 
must be along the line x = 6, with coordinates (6, y).


Since the triangle is isosceles, the distance from (1, 2) to (6, y) must be equal to the distance from (4, -1)
to (6, y).


To make things simpler, and avoid using radicals, the square of the distance between the points will
be compared:


Between (1, 2) and (6, y):


d^2 = (6 - 1)^2 + (y - 2)^2 = 25 + (y - 2)^2


Between (4, -1) and (6, y):


d^2 = (6 - 4)^2 + (y + 1)^2 = 4 + (y + 1)^2


Equating the two squares gives:


25 + (y - 2)^2 = 4 + (y + 1)^2


25 + y^2 - 4y + 4 = 4 + y^2 + 2y + 1


Subtracting y^2 from both sides leaves:


25 - 4y + 4 = 4 + 2y + 1


29 - 4y = 5 + 2y


2y + 4y = 29 - 5


6y = 24


y = 24/6


y = 4


Solution: The ordinate of the third vertex is 4.
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