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November 25, 2014

November 25, 2014

Posted by **shirsh** on Saturday, December 1, 2012 at 9:12am.

- geometry -
**Reiny**, Saturday, December 1, 2012 at 10:31amthe 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.

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