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Quick math question
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You are currently reading a thread in /wsr/ - Worksafe Requests

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Alright, basic ass shit but someone left me ignorant. I'm a sophmore taking college algebra at my school. I understand linear equations easily enough for the most part, but our teacher neglected to cover how you calculate distance between two point when they're formatted as such (A,B) and (0,0). A quick pointer would do me a world of good, cause as far as I've been shown a line with this arrangement shouldn't even exist. Got some cool wallpapers on top of this one to share in exchange.
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sqrt((x2-x1)^2(y2-y1)^2))
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>>47230
Imagine there's a right-triangle between the line and one axis.

Apply Pythagoras.
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Depends on metric you use
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1384720783067.jpg
3 MB, 1924x1080
OP here, that's all the data we were presented with, could I possibly see an example?
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>>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).
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>>47248
yes. also: http://lmgtfy.com/?q=distance+between+two+points
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