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What is the distance between two points

  1. Jul 23, 2014 #1
    Definition/Summary

    In the plane with coordinate system [itex](O, \vec{i}, \vec{j})[/itex] are given the points [itex]A(x_1,y_1)[/itex] and [itex]B(x_2,y_2)[/itex] (see the picture). We want to determine the distance d between the points A and B.

    Equations

    distance between two points [itex](x_1,y_1)[/itex] and [itex](x_2,y_2)[/itex]:

    [tex]d=|\vec{AB}|=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

    distance from (0,0) to point M(x,y):

    [tex]d=|\vec{OM}|=\sqrt{x^2+y^2}[/tex]

    Extended explanation

    Because of the fact that the distance is equal of the module of [itex]\vec{AB}[/itex],

    d=|AB|

    we need to find the module of the vector [itex]\vec{AB}=(x,y)[/itex]. Because of the rectangular triangle ACB (see the picture), satisfying the Pythagorean theorem, we have:

    [tex]|\vec{AB}|^2=|\vec{AC}|^2+|\vec{CB}|^2 \ \ \ \ (1)[/tex]

    Because of:

    [tex]|\vec{AC}|=|\vec{A'B'}|=|x_2-x_1|[/tex] and [tex]|\vec{CB}|=|\vec{A''B''}|=|y_2-y_1|,[/tex]

    substituting in (1) we have:

    [tex]|\vec{AB}|^2 = |x_2-x_1|^2 + |y_2 - y_1|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2[/tex]

    So, for finding the distance d between two points we have the formula:

    [tex]d=|\vec{AB}|=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

    If the distance d, is from the point (0,0) to arbitary point M(x,y), which [itex]d=|\vec{OM}|[/itex], then we have:

    [tex]d=|\vec{OM}|=\sqrt{x^2+y^2}[/tex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
  2. jcsd
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