Solution 4.1:4a - Revision history

Tek at 10:40, 3 October 2008

Tek — Fri, 03 Oct 2008 10:40:50 GMT

←Older revision		Revision as of 10:40, 3 October 2008
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-	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	+	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.
-	~~<math>~~x~~</math>~~	+
-	- and	+
-	~~<math>~~y~~</math>~~	+
-	-axes.	+

-	{{NAVCONTENT_START}}

	[[Image:4_1_4_a-1(2)_1.gif\|center]]		[[Image:4_1_4_a-1(2)_1.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
-	<math>x</math>
-	-
-	<math>y</math>
-	-directions.

-	[[Image:4_1_4_a-1(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}}

-	~~Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points~~:	+	{\| align="center"
		+	\|align="center"\|[[Image:4_1_4_a-1(2)_2.gif\|center]]
		+	\|-
		+	\|align="center"\|<small>∆''x'' = 5 - 1 = 4  and  ∆''y'' = 4 - 1 = 3</small>
		+	\|}


-	~~<math>\begin{align}~~	+	Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
-	~~& d=\sqrt{\left( \Delta x \right)^{2}+\left( \Delta y \right)^{2}}=\sqrt{4^{2}+3^{2}} \\~~	+
-	~~& =\sqrt{16+9}=\sqrt{25}=5 \\~~	+
-	~~\end{align}</math>~~	+

		+	{{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>\left( x \right.,\left. y \right)</math>
-	and
-	<math>\left( a \right.,\left. b \right)</math>
-	is given by the formula

		+	Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula

-	<math>d=\sqrt{~~\left~~( x-a ~~\right~~)^{2}+~~\left~~( y-b ~~\right~~)^{2}}</math>	+	{{Displayed math\|\|<math>d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.}</math>}}

Ian at 09:56, 27 September 2008

Ian — Sat, 27 Sep 2008 09:56:10 GMT

←Older revision		Revision as of 09:56, 27 September 2008
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		+	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
		+	<math>x</math>
		+	- and
		+	<math>y</math>
		+	-axes.
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
		+
	[[Image:4_1_4_a-1(2)_1.gif\|center]]		[[Image:4_1_4_a-1(2)_1.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
		+	<math>x</math>
		+	-
		+	<math>y</math>
		+	-directions.

	[[Image:4_1_4_a-1(2)_2.gif\|center]]		[[Image:4_1_4_a-1(2)_2.gif\|center]]

-	~~<center> [[Image:4_1_4a-1(2).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Image:4_1_4a-2(2).gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
		+
		+
		+	<math>\begin{align}
		+	& d=\sqrt{\left( \Delta x \right)^{2}+\left( \Delta y \right)^{2}}=\sqrt{4^{2}+3^{2}} \\
		+	& =\sqrt{16+9}=\sqrt{25}=5 \\
		+	\end{align}</math>
		+
		+
		+	NOTE: In general, the distance between two points
		+	<math>\left( x \right.,\left. y \right)</math>
		+	and
		+	<math>\left( a \right.,\left. b \right)</math>
		+	is given by the formula
		+
		+
		+	<math>d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}</math>

Tekbot: Lösning 4.1:4a moved to Solution 4.1:4a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:35:03 GMT

Lösning 4.1:4a moved to Solution 4.1:4a: Robot: moved page

←Older revision

Revision as of 09:35, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:04:39 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Martin at 13:21, 15 August 2008

Martin — Fri, 15 Aug 2008 13:21:31 GMT

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Oskar — Wed, 02 Apr 2008 12:57:58 GMT

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