Quick math question

sqrt((x2-x1)^2(y2-y1)^2))

>>47230
Imagine there's a right-triangle between the line and one axis.

Apply Pythagoras.

Depends on metric you use

File: 1384720783067.jpg (3MB, 1924x1080px) Image search: [Google]

3MB, 1924x1080px

OP here, that's all the data we were presented with, could I possibly see an example?

>>47230
Euclidean distance between two points.

If you have two points (x1,y1) and (x2,y2) with x1 = x2 then the two points are connected by a segment
parallel to the y-axis and the distance is |y2 - y1|.

If instead y1 = y2 then analogously the distance is |x2 - x1|.

In the other cases the distance d is the length of an hypothenuse of a right triangle with short sides parallel to the x- and to the y-axis, of length |x2 - x1| and |y2 - y1|, respectively.

Now use the theorem of Pythagoras and you obtain

d^2 = |x2 -x1|^2 + |y2 - y1|^2.
Note that because of the squares you can drop the absolute values here, and by taking the square root you get

d = sqrt((x2 -x1)^2 + (y2 - y1)^2)

(>>47238 forgot the +)

Note also that this formula covers also the first two cases with y1 = y2 and x1 = x2, that is, you need only to remember it (or simply the Theorem of Pythagoras).

>>47248
yes. also: http://lmgtfy.com/?q=distance+between+two+points