What is the distance between two points

[Greg Bernhardt]

Definition/Summary

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

Equations

distance between two points (x_1,y_1) and (x_2,y_2):

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

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

d=|\vec{OM}|=\sqrt{x^2+y^2}

Extended explanation

Because of the fact that the distance is equal of the module of \vec{AB},

d=|AB|

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

|\vec{AB}|^2=|\vec{AC}|^2+|\vec{CB}|^2 \ \ \ \ (1)

Because of:

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

substituting in (1) we have:

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

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

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

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

d=|\vec{OM}|=\sqrt{x^2+y^2}

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[fresh_42]

For a more general view one could do a forum and / or insight search on the keyword "metric". The definition above is basically the theorem of Pythagoras, but distances can also be defined on non Euclidean spaces which is e.g. necessary in cosmology!