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Mathematics 27 Online
OpenStudy (anonymous):

find the point on the line y=x nearest to the point (4,1). THe ans. is (5/2,5/2)

OpenStudy (anonymous):

geometry or calculus?

OpenStudy (anonymous):

calculus

OpenStudy (anonymous):

whew, then it is easy

OpenStudy (anonymous):

ok show to me :D

OpenStudy (anonymous):

any point on the line \(y=x\) will look like \((x,x)\) and the square of the distance between any point \((x,x)\) and \((4,1)\) is \[f(x)=(x-4)^2+(x-1)^2\]

OpenStudy (anonymous):

multiply out, combine like terms, take the derivative, set it equal to zero and solve for \(x\)

OpenStudy (anonymous):

in fact you don't need calculus because \(f\) is a quadratic, and it will have a minimum at the vertex

OpenStudy (anonymous):

you good from there?

OpenStudy (anonymous):

ok :D thats all master ???

OpenStudy (anonymous):

that is it, unless you need help with the algebra

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