## 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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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)

2. Mathmuse

@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|

3. Mathmuse

|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: $Length = \sqrt{\Delta x^2+\Delta y^2}$ $Length = \sqrt{(x-(-3))^2+(y-1)^2}$ from the equation of the line we know that: $y=-6x+9$ so we can sub this into the length formula $Length = \sqrt{(x-(-3))^2+((-6x+9)-1)^2}$ $Length = ({37x^2-90x+73})^{\frac{1}{2}}$ Set the derivative of this to zero to arrive at the minimum value for length, that is the closest point

4. Mathmuse

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.