geometry
posted by shirsh .
if two opposites vertices of a square are (0,1)and(4,3)find the other vertices

the two diagonals of a square rightbisect 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
(y2)/x2) = (20)/(21)
(y2)/(x2) = 2/1
2x+4 = y2
y = 62x
we also know that the segments of the diagonals are equal
√((x2)^2 + (y2)^2 ) = √((20)^2 + (21)^2 )
square both sides
(x2)^2 + (y2)^2 = 5 , ahhh, the equation of the circumscribed circle
subbing in our y = 62x
(x2)^2 + (42x)^2 = 5
x^2  4x + 4 + 16  16x + 4x^2  5 = 0
5x^2 20x +15 = 0
x^2  4x + 3 = 0
(x1)(x3) = 0
x = 1 or x = 3
if x = 1 then y = 62 = 4 > point (1,4)
if x = 3 then y = 66 = 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.