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Math Help - Finding the minimum distance from a point to a line-Having trouble

  1. #1
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    Finding the minimum distance from a point to a line-Having trouble

    Hey!
    I am having trouble finding the minimum distance from a point to a line. Here is the question:

    For each of the following, find the distance from Q(3,-2) to each line.
    a) 3x - 2y - 6 = 0

    Solution
    First I put into the form y = mx + b (because I am more familiar with this)
    y = \frac{3}{2}x - 3

    I now know that the slope of the line is  \frac {3}{2} , which means that the line perpendicular to it has the slope of  \frac {-2}{3} .
    With this information I found the equation of a line with the slope  \frac{3}{-2} . Which is  y = \frac{3}{-2}x + \frac{5}{2} .

    I then found where the equations  y = \frac{3}{-2}x + \frac{5}{2} and y = \frac{3}{2}x - 3 intersect by solving the system of equations. I got the point  (\frac{11}{6}, \frac{-1}{4}) .

    So I now know the point on the equation closest to the point Q. So all I need to do is find the distance between them. So:
    \sqrt{((\frac{11}{6})(3))^2 + (-2)(\frac{1}{-4})^2}
     = \sqrt{\frac{61}{2}}
    This is my final answer.

    The book says the answer should be  \frac{7\sqrt{13}}{13} .

    I am doing exactly what the textbook is telling me to do. What am I doing wrong?
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  2. #2
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    Quote Originally Posted by Kakariki View Post
    Hey!
    I am having trouble finding the minimum distance from a point to a line. Here is the question:

    For each of the following, find the distance from Q(3,-2) to each line.
    a) 3x - 2y - 6 = 0

    Solution
    First I put into the form y = mx + b (because I am more familiar with this)
    y = \frac{3}{2}x - 3

    I now know that the slope of the line is  \frac {3}{2} , which means that the line perpendicular to it has the slope of  \frac {-2}{3} . correct
    With this information I found the equation of a line with the slope  \frac{3}{-2} . Which is  y = \frac{3}{-2}x + \frac{5}{2} . you changed your mind here though,

    I then found where the equations  y = \frac{3}{-2}x + \frac{5}{2} and y = \frac{3}{2}x - 3 intersect by solving the system of equations. I got the point  (\frac{11}{6}, \frac{-1}{4}) .

    So I now know the point on the equation closest to the point Q. So all I need to do is find the distance between them. So:
    \sqrt{((\frac{11}{6})(3))^2 + (-2)(\frac{1}{-4})^2}
     = \sqrt{\frac{61}{2}}
    This is my final answer.

    The book says the answer should be  \frac{7\sqrt{13}}{13} .

    I am doing exactly what the textbook is telling me to do. What am I doing wrong?

    Bobak
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  3. #3
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    Quote Originally Posted by bobak View Post
    Bobak
    Oh dear! What a silly mistake I made!!! Thank you for pointing it out.
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  4. #4
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    Quote Originally Posted by Kakariki View Post
    Hey!
    I am having trouble finding the minimum distance from a point to a line. Here is the question:

    For each of the following, find the distance from Q(3,-2) to each line.
    a) 3x - 2y - 6 = 0

    Solution
    First I put into the form y = mx + b (because I am more familiar with this)
    y = \frac{3}{2}x - 3

    I now know that the slope of the line is  \frac {3}{2} , which means that the line perpendicular to it has the slope of  \frac {-2}{3} .
    With this information I found the equation of a line with the slope  \frac{3}{-2} . Which is  y = \frac{3}{-2}x + \frac{5}{2} .

    I then found where the equations  y = \frac{3}{-2}x + \frac{5}{2} and y = \frac{3}{2}x - 3 intersect by solving the system of equations. I got the point  (\frac{11}{6}, \frac{-1}{4}) .

    So I now know the point on the equation closest to the point Q. So all I need to do is find the distance between them. So:
    \sqrt{((\frac{11}{6})(3))^2 + (-2)(\frac{1}{-4})^2}
     = \sqrt{\frac{61}{2}}
    This is my final answer.

    The book says the answer should be  \frac{7\sqrt{13}}{13} .

    I am doing exactly what the textbook is telling me to do. What am I doing wrong?
    3x - 2y - 6 = 0

    2y = 3x - 6

    y = \frac{3}{2}x - 3.


    So on this line, any point can be written as

    \left(x, \frac{3}{2}x - 3\right).


    The distance from this arbitrary point to Q(3, -2) is


    D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

     = \sqrt{\left(x - 3\right)^2 + \left[\frac{3}{2}x - 3 - (-2)\right]^2}

     = \sqrt{(x - 3)^2 + \left(\frac{3}{2}x - 1\right)^2}

     = \sqrt{x^2 - 6x + 9 + \frac{9}{4}x^2 - 3x + 1}

     = \sqrt{\frac{13}{4}x^2 - 9x + 10}

     = \left(\frac{13}{4}x^2 - 9x + 10\right)^{\frac{1}{2}}


    So to find the minimum distance, you need to differentiate the function, set it  = 0 and solve for x.


    Let u = \frac{13}{4}x^2 - 9x + 10 so that D = u^{\frac{1}{2}}.

    \frac{du}{dx} = \frac{13}{2}x - 9.

    \frac{dD}{du} = \frac{1}{2}u^{-\frac{1}{2}}

     = \frac{1}{2\sqrt{\frac{13}{4}x^2 - 9x + 10}}.

    So \frac{dD}{dx} = \frac{\frac{13}{2}x - 9}{2\sqrt{\frac{13}{4}x^2 - 9x + 10}}

     = \frac{13x - 18}{4\sqrt{\frac{13}{4}x^2 - 9x + 10}}

     = \frac{13x - 18}{2\sqrt{13x^2 - 36x + 40}}.


    Setting this equal to 0...

    \frac{13x - 18}{2\sqrt{13x^2 - 36x + 40}} = 0

    13x - 18 = 0

    13x = 18

    x = \frac{18}{13}.


    When x = \frac{18}{13}, y = \frac{3}{2}\left(\frac{18}{13}\right) - 3

     = \frac{27}{13} - 3

     = -\frac{12}{13}.



    So the point where the distance between the point Q(3, -2) and the line is minimised is at (x, y) = \left(\frac{18}{13}, -\frac{12}{13}\right). The distance between them is

    D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

     = \sqrt{\left(\frac{18}{13} - 3\right)^2 + \left[-\frac{12}{13} - (-2)\right]^2}

     = \sqrt{\left(-\frac{21}{13}\right)^2 + \left(\frac{14}{13}\right)^2}

     = \sqrt{\frac{441}{169} + \frac{196}{169}}

     = \sqrt{\frac{637}{169}}

     = \frac{7\sqrt{13}}{13}.
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