From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the ''x''- and ''y''-axes, respectively.
-	[[Bild:4_1_4_a-1(2)_1.gif|center]]	+
-	[[Bild:4_1_4_a-1(2)_2.gif|center]]
-	<center> [[Bild:4_1_4a-1(2).gif]] </center>	+	[[Image:4_1_4_a-1(2)_1.gif|center]]
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_1_4a-2(2).gif]] </center>	+	In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the ''x''- and ''y''-directions, respectively.
-	{{NAVCONTENT_STOP}}	+
		+
		+	{| align="center"
		+	|align="center"|[[Image:4_1_4_a-1(2)_2.gif|center]]
		+	|-
		+	|align="center"|<small>∆''x''&nbsp;=&nbsp;5&nbsp;-&nbsp;1&nbsp;=&nbsp;4 &nbsp;and&nbsp; ∆''y''&nbsp;=&nbsp;4&nbsp;-&nbsp;1&nbsp;=&nbsp;3</small>
		+	|}
		+
		+
		+	Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
		+
		+	{{Displayed math||<math>\begin{align}
		+	d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt]
		+	&= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.}
		+	\end{align}</math>}}
		+
		+
		+	Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula
		+
		+	{{Displayed math||<math>d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.}</math>}}

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Current revision

If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the x- and y-axes, respectively.

In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the x- and y-directions, respectively.

∆x = 5 - 1 = 4  and  ∆y = 4 - 1 = 3

Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:

\displaystyle \begin{align} d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt] &= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.} \end{align}

Note: In general, the distance between two points \displaystyle (x,y) and \displaystyle (a,b) is given by the formula

\displaystyle d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.}