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

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ASAAD123 Group TitleBest ResponseYou've already chosen the best response.0
6x + 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)
 one year ago

Mathmuse Group TitleBest 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
 one year ago

Mathmuse Group TitleBest ResponseYou've already chosen the best response.0
dw: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
 one year ago

Mathmuse Group TitleBest ResponseYou've already chosen the best response.0
Hint: 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.
 one year ago
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