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The value of y that minimizes the sum of the two distances from (3,5) to (1,y) and from (1,y) to (4,9) can be written as \frac{a}{b} where a and b are coprime positive integers. Find a + b

  • Algebra -

    d = √(4+(y-5)^2) + √(9+(y-9)^2)
    dd/dy = (y-5)/√(4+(y-5)^2) + (y-9)/√(9+(y-9)^2)
    = [(y-9)√(4+(y-5)^2) + (y-5)√(9+(y-9)^2)]/(√(4+(y-5)^2) * √(9+(y-9)^2))
    dd/dy=0 when
    (y-9)√(4+(y-5)^2) + (y-5)√(9+(y-9)^2) = 0

    y = 33/5

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