geometry

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if two opposites vertices of a square are (0,1)and(4,3)find the other vertices

  • geometry -

    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.

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