show that the lines x^-4xy+y^=0 and x+y=1 form an equilateral triangle and find its area

asked by prateek
  1. If that's x^2 - 4xy + y^2 = 0, that means that

    y = (2+√3)x and y = (2-√3)x

    Find where those lines intersect y = -x + 1

    A: (2+√3)x = 1-x
    x = 1/(3+√3)
    y = 1-x = (2+√3)/(3+√3)

    B:(2-√3)x = 1-x
    x = 1/(3-√3)
    y = 1 - x = (2-√3)/(3-√3)

    After a little algebra, we find that

    |OA|2 = 2/3
    |OB|2 = 2/3

    |AB|^2 = √6/3

    The altitude = 1/√2, so the area of OAB = 1/2 * 1/√2 * √6/3 = √3/6

    posted by Steve
  2. That's |AB| = √6/3

    posted by Steve

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