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Math Help - find a point

  1. #1
    Member M670's Avatar
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    find a point

    Find the point on the line 4x+4y+3=0 which is closet to P (1, 2)?

    So I re wrote my line as y=-x-(3/4)
    Then i take d^2=(x-1)^2+(-x-(3/4)-2)^2=2x^2+(7x/2)+(137/16)
    Once I take the dirivative i get 4x+(7/2) soove for x i get -(7/8) and then y = 1/8
    But its wrong. What did i do wrong?
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  2. #2
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    Re: find a point

    M670

    The slope intercept form of the equation of the given line L is : y = -1x-(3/4)
    the closest point on the line L from the point P(1,2) is the point that the perpendicular line to the line L , passing from the point P, intecects the line L.
    the equation of the line that passes from the point P and is perpendicular to L is y = x+1

    to find the coordinates of the point of intercection of these two lines solve the system of the two simultaneous equations
    y = -1x-3/4
    y=x+1

    the answer is x=-7/8 and y= 1/8 therefore the point is Q(-7/8 , 1/8)

    Pls revise the equation of the line in the plane .
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  3. #3
    Member M670's Avatar
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    Re: find a point

    Quote Originally Posted by MINOANMAN View Post
    M670

    The slope intercept form of the equation of the given line L is : y = -1x-(3/4)
    the closest point on the line L from the point P(1,2) is the point that the perpendicular line to the line L , passing from the point P, intecects the line L.
    the equation of the line that passes from the point P and is perpendicular to L is y = x+1

    to find the coordinates of the point of intercection of these two lines solve the system of the two simultaneous equations
    y = -1x-3/4
    y=x+1

    the answer is x=-7/8 and y= 1/8 therefore the point is Q(-7/8 , 1/8)

    Pls revise the equation of the line in the plane .

    Point Q = (-7/8,1/8) is incorrect when I submitted that...
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  4. #4
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    Re: find a point

    Quote Originally Posted by M670 View Post
    Find the point on the line 4x+4y+3=0 which is closet to P (1, 2)?
    So I re wrote my line as y=-x-(3/4)
    Then i take d^2=(x-1)^2+(-x-(3/4)-2)^2=2x^2+(7x/2)+(137/16)
    Once I take the dirivative i get 4x+(7/2) soove for x i get -(7/8) and then y = 1/8
    But its wrong. What did i do wrong?

    No calculus needed. Where does the line 4x-4y+4=0 intersect the given line?
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  5. #5
    Member M670's Avatar
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    Re: find a point

    The question is asking me to find the point closest to (1,2) on the line y=-x-\frac{3}{4}
    Then I get d^2=(x-1)^2+(-x-\frac{3}{4}-2)^2 which is = 2x^2+\frac{7x}{2}+\frac{137}{16}
    Then once I get d^2'=4x+\frac{7}{2} I set it equal to 0 and solve for x
    Thus getting x=-\frac{7}{8} and y = \frac{1}{8}

    Shoot I think I found my mistake my original formula is -4x+4y+3=0 which would make it 4y=4x-3then I should have y=x-\frac{3}{4}
    Last edited by M670; April 7th 2013 at 09:56 AM.
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  6. #6
    Member M670's Avatar
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    Re: find a point

    All this time it was the Minus sign that messed me up........
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  7. #7
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    Re: find a point

    OK MAN
    Next time give us correct information....
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