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Math Help - Distance from a point to a line

  1. #1
    Member VitaX's Avatar
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    Distance from a point to a line

    Find the distance from P to L:

    A. P(2,1) L: y = x + 2
    B. P(4,6) L: 4x + 3y = 12

    If I can get some guidance on these and steps to follow I should be able to do the rest of the problems. The books explanation on how to do these is a little fuzzy to me.
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  2. #2
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    Quote Originally Posted by VitaX View Post
    Find the distance from P to L:

    A. P(2,1) L: y = x + 2
    B. P(4,6) L: 4x + 3y = 12

    If I can get some guidance on these and steps to follow I should be able to do the rest of the problems. The books explanation on how to do these is a little fuzzy to me.
    There is a long way to get the result - and of course a shortcut.

    A. Long way:
    1. Calculate the equation of the perpendicular line to the line L: y = -x+3
    2. Calculate the coordinates of the point of intersection of L and the normal to L: Q\left(\dfrac12\ ,\ \dfrac52\right)
    3. Calculate the distance \overline{QP} = \dfrac32 \sqrt{2}


    B. Shortcut:

    Let P(x_P, y_P) denote a point and L:Ax+By+C=0 a straight line then the distance d of P from L is calculated by:

    d(P,L)=\dfrac{Ax_P + By_P + C}{\sqrt{A^2+B^2}}

    With your problem: The line L:y = x+2~\implies~-x+y-2=0

    Plug in the values of A, B, C and the coordinates of P and you'll get the same result.

    I'll leave the second example for you.
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  3. #3
    Member VitaX's Avatar
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    Thanks for the help. I got the same answer as you did using the long method. I'll try the other one now.

    Edit: For B. I get d = 4.4 units. Correct?
    Last edited by VitaX; August 29th 2009 at 02:00 AM.
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  4. #4
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    Quote Originally Posted by VitaX View Post
    Thanks for the help. I got the same answer as you did using the long method. I'll try the other one now.

    Edit: For B. I get d = 4.4 units. <<<< perfect!
    ...
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  5. #5
    Member VitaX's Avatar
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    I ran into some difficult problems in this section of the homework. The reason they are difficult is because the problems are all variables. So trying to solve them is more difficult (to me atleast)

    Find the distance from P to L:

    P(a,b) L: x = -1 Edit : I get d = a + 1 ... correct?
    P(x1,y1) L: Ax + By = C

    How would I go about solving these two?
    Last edited by VitaX; August 29th 2009 at 07:07 PM.
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  6. #6
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    Quote Originally Posted by VitaX View Post
    I ran into some difficult problems in this section of the homework. The reason they are difficult is because the problems are all variables. So trying to solve them is more difficult (to me atleast)

    Find the distance from P to L:

    P(a,b) L: x = -1 Edit : I get d = a + 1 ... correct? <<<<<<< Yes
    P(x1,y1) L: Ax + By = C

    How would I go about solving these two?
    In both cases you can use the formula I've posted in my previous post:

    L: x = -1~\implies~1\cdot x + 0\cdot y + 1 = 0

    Thus:

    d(P(a,b),L)=\dfrac{1 \cdot a+0\cdot b + 1}{\sqrt{1^2+0^2}}=a+1

    And the 2nd problem yields nearly the formula which I posted in post#2. (There is a slight difference with the constant C)
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