Posted by shirsh on Saturday, December 1, 2012 at 9:12am.
the two diagonals of a square right-bisect each other at the point (2,2) , the midpoint of the given points .
let the other vertex be (x,y) , the endpoint of the other diagonal
since the two diagonals must be perpendiculars, their slopes must be negative reciprocals of each other
(y-2)/x-2) = -(2-0)/(2-1)
(y-2)/(x-2) = -2/1
-2x+4 = y-2
y = 6-2x
we also know that the segments of the diagonals are equal
√((x-2)^2 + (y-2)^2 ) = √((2-0)^2 + (2-1)^2 )
square both sides
(x-2)^2 + (y-2)^2 = 5 , ahhh, the equation of the circumscribed circle
subbing in our y = 6-2x
(x-2)^2 + (4-2x)^2 = 5
x^2 - 4x + 4 + 16 - 16x + 4x^2 - 5 = 0
5x^2 -20x +15 = 0
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
x = 1 or x = 3
if x = 1 then y = 6-2 = 4 ---> point (1,4)
if x = 3 then y = 6-6 = 0 ---> point (3,0)
I have a feeling there should be an easier way, but I can't see it right now. Sometimes I tend to overthink the question.