no it does not
as long as you subtract the x's from the x's and y's from the y's
consider
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
and
[tex]D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]
alright, lets say x2-x1=z and x1-x2=-z
also y2-y1=t and y1-y2=-t
if we subsituted then we get
[tex]D=\sqrt{(z)^2+(t)^2}[/tex]
and
[tex]D=\sqrt{(-z)^2+(-t)^2}[/tex]
which simplifies to
[tex]D=\sqrt{z^2+t^2}[/tex]
and
[tex]D=\sqrt{z^2+t^2}[/tex]
the same thing
because the square of a negative becomes positive
so it doesn't matter
