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Shortest distance between a line and a point?

  1. Jul 20, 2017 #1
    1. The problem statement, all variables and given/known data
    find the shortest distance from (0,0) to the line passing A(2,3) and B(3,5)

    2. Relevant equations
    ## \frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1} ##
    y-y1 = m (x-x1)
    m1 * m2 = -1 (m1 perpendicular to m2)

    3. The attempt at a solution
    line passing A and B points
    ## \frac{y-3}{5-3} ## = ## \frac{x-2}{3-2} ##
    y-3 = 2(x-2)
    y = 2x - 1

    m1 = 2
    m2 = -1/2

    the line perpendicular to line 1 and passing (0,0)
    y -y1 = m(x-x1)
    y-0=-1/2(x-0)
    y' = -1/2x'

    line 1 and line 2 intersect at (x,y)
    y = y', x = x'
    2x - 1 = -1/2x'
    3/2x = 1
    x = 2/3

    y' = -1/2'x = -1/2(2/3) = -1/3

    x,y = 2/3 , -1/3

    distance is from (0,0) to (2/3 , -1/3)
    distance = ## \sqrt{(2/3-0)^2 + (-1/3-0)^2} ##
    distance = ## \sqrt{4/9 + 1/9} ##
    distance = ## \sqrt{5}/3 ##

    what's wrong? I got wrong answer
     
  2. jcsd
  3. Jul 20, 2017 #2

    ehild

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    The red line.
     
  4. Jul 20, 2017 #3

    SammyS

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    Simply an Algebra mistake. Highlighted in red.

    If you graph the line ##\ y=2x-1\ ##, you will notice that the point (2/3, −1/3) does not lie on the line.
     
    Last edited: Jul 20, 2017
  5. Jul 20, 2017 #4

    Mark44

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    @ehild already pointed out the mistake in the last line above. In addition, the three lines I picked out above are ambiguous, as -1/2x' might be interpreted as ##-\frac{1}{2x'}## by some when you probably meant ##-\frac{1}2 x'##. The same goes for 3/2x.
     
  6. Jul 20, 2017 #5
    thank you for all the corrections... :)
     
  7. Jul 20, 2017 #6

    SammyS

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    What's your answer now?
     
  8. Jul 20, 2017 #7
    (1/5) ##\sqrt5##
     
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