Aus Online Mathematik Brückenkurs 1

Version vom 08:48, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.2:5d moved to Solution 2.2:5d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 09:50, 18. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{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]]

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Version vom 09:50, 18. Sep. 2008

If two non-vertical lines are perpendicular to each other, their gradients \displaystyle k_{1} and \displaystyle k_{2} satisfy the relation

\displaystyle k_{1}k_{2}=-1, and from this we have that the line we are looking for must have a gradient that is given by

\displaystyle k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}

since the line \displaystyle y=2x+5 has a gradient \displaystyle k_{1}=2 (the coefficient in front of \displaystyle x ).

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

\displaystyle y=-\frac{1}{2}x+m

with \displaystyle m as an unknown constant.

Because the point \displaystyle \left( 2 \right.,\left. 4 \right) should lie on the line, \displaystyle \left( 2 \right.,\left. 4 \right) must satisfy the equation of the line,

\displaystyle 4=-\frac{1}{2}\centerdot 2+m

i.e. \displaystyle m=5. The equation of the line is .

\displaystyle y=-\frac{1}{2}x+5