Algebra
posted by Anonymous .
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+(y5)^2) + √(9+(y9)^2)
dd/dy = (y5)/√(4+(y5)^2) + (y9)/√(9+(y9)^2)
= [(y9)√(4+(y5)^2) + (y5)√(9+(y9)^2)]/(√(4+(y5)^2) * √(9+(y9)^2))
dd/dy=0 when
(y9)√(4+(y5)^2) + (y5)√(9+(y9)^2) = 0
y = 33/5
a+b=38
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