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
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
a+b=38