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tmartin
 3 years ago
Find the point on the line 6x+y=9 that is closest to the point (3,1).
tmartin
 3 years ago
Find the point on the line 6x+y=9 that is closest to the point (3,1).

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ASAAD123
 3 years ago
Best ResponseYou've already chosen the best response.06x + y = 9 y = 6x + 9 slope is 6 perpendicular would have slope 1/6 (y  1) = (1/6) (x + 3) y  1 = (1/6) x + (1/2) multiplying both sides by 6 6y  6 = x + 3 x  6y = 9 _______1 From above we know given equation is y = 6x + 9 substitute this equation in 1 x  6(6x + 9) = 9 x + 36x  54 = 9 37 x = 45 x = 45/37 y = 6(45/37) + 9 y = 270/37 + (333/37) y = 63/37 point is (45/37, 63/37)

Mathmuse
 3 years ago
Best ResponseYou've already chosen the best response.0@ASAAD123 http://answers.yahoo.com/question/index?qid=20081106202035AAlEt5X The above solution is algebraic, but since this is a calculus section lets explore calculus type solutions: dw:1353354005150:dw

Mathmuse
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353354087858:dw To find the values of point x,y will will have to minimize the length of the dotted line segment. first we need the length of this line segment as a function of x: \[Length = \sqrt{\Delta x^2+\Delta y^2}\] \[Length = \sqrt{(x(3))^2+(y1)^2}\] from the equation of the line we know that: \[y=6x+9\] so we can sub this into the length formula \[Length = \sqrt{(x(3))^2+((6x+9)1)^2}\] \[Length = ({37x^290x+73})^{\frac{1}{2}}\] Set the derivative of this to zero to arrive at the minimum value for length, that is the closest point

Mathmuse
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: when finding zeros of a complicated function, just concentrate on the numerator. Any value that causes the numerator to go to zero will push the whole function to zero.
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