Solution 2.2:5d - Revision history

Tek at 12:44, 24 September 2008

2008-09-24T12:44:43Z

←Older revision		Revision as of 12:44, 24 September 2008
Line 1:		Line 1:
-	If two non-vertical lines are perpendicular to each other, their ~~gradients~~	+	If two non-vertical lines are perpendicular to each other, their slopes <math>k_{1}</math> and <math>k_{2}</math> satisfy the relation <math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a slope that is given by
-	<math>k_{1}</math>	+
-	and	+
-	<math>k_{2}</math>	+
-	satisfy the relation	+

-	<math>k_{1}k_{2}=-1</math>~~, and from this we have that the line we are looking for must have a gradient that is given by~~	+	{{Displayed math\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}

-		+	since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>
-	~~<math>k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}</math>~~	+	(the coefficient in front of ''x'').
-		+
-		+
-	since the line	+
-	<math>y=2x+5</math>	+
-	has a ~~gradient~~	+
-	<math>k_{1}=2</math>	+
-	(the coefficient in front of	+
-	~~<math>~~x~~</math>~~	+
-	).	+

	The line we are looking for can thus be written in the form		The line we are looking for can thus be written in the form

		+	{{Displayed math\|\|<math>y=-\frac{1}{2}x+m</math>}}

-	~~<math>y=-\frac{1}{2}x+m</math>~~	+	with ''m'' as an unknown constant.
-		+
-		+
-	with	+
-	~~<math>~~m~~</math>~~	+
-	as an unknown constant.	+
-		+
-	~~Because the point~~	+
-	~~<math>\left( 2 \right.,\left. 4 \right)</math>~~	+
-	~~should lie on the line,~~	+
-	~~<math>\left( 2 \right.,\left. 4 \right)</math>~~	+
-	~~must satisfy the equation of the line,~~	+
-		+
-		+
-	~~<math>4=-\frac{1}{2}\centerdot 2+m</math>~~	+
-		+
-		+
-	~~i.e.~~	+
-	~~<math>m=5</math>. The equation of the line is .~~	+
-		+

-	~~<math>y=-\frac{1}{~~2~~}x+5</math>~~	+	Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,

		+	{{Displayed math\|\|<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}

		+	i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.


-	~~{{NAVCONTENT_START}}~~	+	<center>[[Image:2_2_5d-2(2).gif]]</center>
-	<center> [[Image:2_2_5d-2(2).gif]] </center>	+

Ian at 09:50, 18 September 2008

2008-09-18T09:50:10Z

←Older revision		Revision as of 09:50, 18 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	If two non-vertical lines are perpendicular to each other, their gradients
-	<~~center~~> ~~[[Image:2_2_5d~~-1(2).~~gif]]~~ </~~center~~>	+	<math>k_{1}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>k_{2}</math>
		+	satisfy the relation
		+
		+	<math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a gradient that is given by
		+
		+
		+	<math>k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}</math>
		+
		+
		+	since the line
		+	<math>y=2x+5</math>
		+	has a gradient
		+	<math>k_{1}=2</math>
		+	(the coefficient in front of
		+	<math>x</math>
		+	).
		+
		+	The line we are looking for can thus be written in the form
		+
		+
		+	<math>y=-\frac{1}{2}x+m</math>
		+
		+
		+	with
		+	<math>m</math>
		+	as an unknown constant.
		+
		+	Because the point
		+	<math>\left( 2 \right.,\left. 4 \right)</math>
		+	should lie on the line,
		+	<math>\left( 2 \right.,\left. 4 \right)</math>
		+	must satisfy the equation of the line,
		+
		+
		+	<math>4=-\frac{1}{2}\centerdot 2+m</math>
		+
		+
		+	i.e.
		+	<math>m=5</math>. The equation of the line is .
		+
		+
		+	<math>y=-\frac{1}{2}x+5</math>
		+
		+
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Image:2_2_5d-2(2).gif]] </center>		<center> [[Image:2_2_5d-2(2).gif]] </center>
-	{{NAVCONTENT_STOP}}
-	[[Image:2_2_5_d.jpg\|center\|300px]]

Tekbot: Lösning 2.2:5d moved to Solution 2.2:5d: Robot: moved page

2008-09-09T08:48:03Z

Lösning 2.2:5d moved to Solution 2.2:5d: Robot: moved page

←Older revision

Revision as of 08:48, 9 September 2008

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

2008-08-21T06:41:09Z

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

←Older revision		Revision as of 06:41, 21 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_5d-1(2).gif]] </center>	+	<center> [[Image:2_2_5d-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_5d-2(2).gif]] </center>	+	<center> [[Image:2_2_5d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[~~Bild~~:2_2_5_d.jpg\|center\|300px]]	+	[[Image:2_2_5_d.jpg\|center\|300px]]

Madde at 16:24, 14 August 2008

2008-08-14T16:24:49Z

←Older revision		Revision as of 16:24, 14 August 2008
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	<center> [[Bild:2_2_5d-2(2).gif]] </center>		<center> [[Bild:2_2_5d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[Bild:2_2_5_d.jpg\|center\|~~px300~~]]	+	[[Bild:2_2_5_d.jpg\|center\|300px]]

Madde at 16:23, 14 August 2008

2008-08-14T16:23:30Z

←Older revision		Revision as of 16:23, 14 August 2008
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	<center> [[Bild:2_2_5d-2(2).gif]] </center>		<center> [[Bild:2_2_5d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	[[Bild:2_2_5_d.jpg\|center\|px300]]

Lina at 14:49, 31 March 2008

2008-03-31T14:49:54Z

←Older revision		Revision as of 14:49, 31 March 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:~~2_2_5c~~-1(2).gif]] </center>	+	<center> [[Bild:2_2_5d-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:~~2_2_5c~~-2(2).gif]] </center>	+	<center> [[Bild:2_2_5d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_2_5c-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:2_2_5c-2(2).gif {{NAVCONTENT_STOP}}

2008-03-31T14:49:39Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_5c-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:2_2_5c-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_2_5c-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:2_2_5c-2(2).gif]] </center>
{{NAVCONTENT_STOP}}