Solution 2.2:5a - Revision history

Tek at 11:55, 24 September 2008

2008-09-24T11:55:49Z

←Older revision		Revision as of 11:55, 24 September 2008
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	Let's write down the equation for a straight line as		Let's write down the equation for a straight line as

		+	{{Displayed math\|\|<math>y=kx+m\,,</math>}}

-	~~<math>y=kx+~~m~~</math>~~	+	where ''k'' and ''m'' are constants which we shall determine.

		+	Since the points (2,3) and (3,0) should lie on the line, they must also satisfy the equation of the line,

-	~~where~~	+	{{Displayed math\|\|<math>3=k\cdot 2+m\qquad\text{and}\qquad 0=k\cdot 3+m\,\textrm{.}</math>}}
-	<math>k~~</math>~~	+
-	and	+
-	~~<math>~~m</math>	+
-	~~are constants which we shall determine.~~	+

-	~~Since~~ the ~~points~~	+	If we take the difference between the equations, ''m'' disappears and we can work out the slope ''k'',
-	~~<math>\left( 2 \right.~~, ~~\left. 3 \right)</math>~~	+
-	and	+
-	~~<math>\left( 3 \right., \left. 0 \right)</math>~~	+
-	~~should lie on the line, they must also satisfy the equation of~~ the ~~line~~,	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	3-0 &= k\cdot 2+m-(k\cdot 3+m)\,,\\[5pt]
		+	3 &= -k\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>3=k\centerdot 2+m</math>~~	+	Substituting this into the equation <math>0=k\centerdot 3+m</math> then gives us a value for ''m'',
-	~~and~~	+
-	<math>0=k\centerdot 3+m</math>	+

		+	{{Displayed math\|\|<math>m=-3k=-3\cdot (-3)=9\,\textrm{.}</math>}}

-	~~If we take~~ the ~~difference between the equations,~~	+	The equation of the line is thus <math>y=-3x+9</math>.
-	~~<math>m</math>~~	+
-	~~disappears and we can work out the gradient~~	+
-	<math>k</math>,	+


-	<~~math~~>~~3-0=k\centerdot 2+m-\left( k\centerdot 3+m \right)~~</~~math~~>	+	<center>[[Image:S1_2_2_5_a.jpg]]</center>


-	~~<math>~~3~~=-k</math>~~	+	Note. To be completely certain that we have calculated correctly, we check that the points (2,3) and (3,0) satisfy the equation of the line:

-	~~Substituting this into the equation~~	+	:*(''x'',''y'') = (2,3): <math>\text{LHS} = 3\ </math> and <math>\ \text{RHS} = -3\cdot 2+9 = 3\,</math>.
-	~~<math>0=k\centerdot 3+m</math>~~	+	:*(''x'',''y'') = (3,0): <math>\text{LHS} = 0\ </math> and <math>\ \text{LHS} = -3\cdot 3+9 = 0\,</math>.
-	~~then gives us a value for~~	+
-	~~<math>m</math>,~~	+
-		+
-		+
-	~~<math>m=-3k=-3\centerdot \left( -3 \right)=9</math>~~	+
-		+
-		+
-	~~The equation of the line is thus~~	+
-	~~<math>y=-3x+9</math>.~~	+
-		+
-		+
-		+
-	~~NOTE~~: ~~To be completely certain that we have calculated correctly, we check that the points~~	+
-	~~<math>\left( 2 \right., \left. 3 \right)</math>~~	+
-	~~and~~	+
-	~~<math>\left( 3 \right., \left. 0 \right)</math>~~	+
-	~~satisfy the equation of the line:~~	+
-		+
-	~~<math>\left~~( x ~~\right.~~, ~~\left.~~ y ~~\right~~)=~~\left~~( 2 ~~\right.~~, ~~\left.~~ 3 ~~\right~~)</math>: LHS=	+
-	~~<math>~~3</math>	+
-	and ~~RHS=~~	+
-	<math>-3\~~centerdot~~ 2+9=3</math>	+
-		+
-		+
-	~~<math>\left~~( x ~~\right.~~, ~~\left.~~ y ~~\right~~)=~~\left~~( 3 ~~\right.~~, ~~\left.~~ 0 ~~\right~~)</math>: LHS=	+
-	~~<math>~~0</math>	+
-	and ~~LHS=~~	+
-	<math>-3\~~centerdot~~ 3+9=0</math>	+
-		+
-		+
-		+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<!--<center> [[Image:2_2_5a-1(2)~~.~~gif]] </center>-->~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-		+
-	~~[[Image:S1_2_2_5_a.jpg]]~~	+
-	~~<!--<center> [[Image:2_2_5a-2(2).gif]] </center>-->~~	+
-	~~{{NAVCONTENT_STOP}}~~	+

Ian at 09:02, 18 September 2008

2008-09-18T09:02:26Z

←Older revision		Revision as of 09:02, 18 September 2008
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		+	Let's write down the equation for a straight line as
		+
		+
		+	<math>y=kx+m</math>
		+
		+
		+	where
		+	<math>k</math>
		+	and
		+	<math>m</math>
		+	are constants which we shall determine.
		+
		+	Since the points
		+	<math>\left( 2 \right., \left. 3 \right)</math>
		+	and
		+	<math>\left( 3 \right., \left. 0 \right)</math>
		+	should lie on the line, they must also satisfy the equation of the line,
		+
		+
		+	<math>3=k\centerdot 2+m</math>
		+	and
		+	<math>0=k\centerdot 3+m</math>
		+
		+
		+	If we take the difference between the equations,
		+	<math>m</math>
		+	disappears and we can work out the gradient
		+	<math>k</math>,
		+
		+
		+	<math>3-0=k\centerdot 2+m-\left( k\centerdot 3+m \right)</math>
		+
		+
		+	<math>3=-k</math>
		+
		+	Substituting this into the equation
		+	<math>0=k\centerdot 3+m</math>
		+	then gives us a value for
		+	<math>m</math>,
		+
		+
		+	<math>m=-3k=-3\centerdot \left( -3 \right)=9</math>
		+
		+
		+	The equation of the line is thus
		+	<math>y=-3x+9</math>.
		+
		+
		+
		+	NOTE: To be completely certain that we have calculated correctly, we check that the points
		+	<math>\left( 2 \right., \left. 3 \right)</math>
		+	and
		+	<math>\left( 3 \right., \left. 0 \right)</math>
		+	satisfy the equation of the line:
		+
		+	<math>\left( x \right., \left. y \right)=\left( 2 \right., \left. 3 \right)</math>: LHS=
		+	<math>3</math>
		+	and RHS=
		+	<math>-3\centerdot 2+9=3</math>
		+
		+
		+	<math>\left( x \right., \left. y \right)=\left( 3 \right., \left. 0 \right)</math>: LHS=
		+	<math>0</math>
		+	and LHS=
		+	<math>-3\centerdot 3+9=0</math>
		+
		+
		+
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Tekbot: Lösning 2.2:5a moved to Solution 2.2:5a: Robot: moved page

2008-09-09T08:47:03Z

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

←Older revision

Revision as of 08:47, 9 September 2008

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

2008-08-21T06:40:39Z

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

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Martin at 15:17, 14 August 2008

2008-08-14T15:17:16Z

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Martin at 15:13, 14 August 2008

2008-08-14T15:13:31Z

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Martin at 15:12, 14 August 2008

2008-08-14T15:12:41Z

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Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_2_5a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:2_2_5a-2(2).gif {{NAVCONTENT_STOP}}

2008-03-31T14:46:10Z

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