Aus Online Mathematik Brückenkurs 1

Version vom 11:55, 24. Sep. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:28, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	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>}}	+	{{Abgesetzte Formel||<math>y=kx+m\,,</math>}}
	where ''k'' and ''m'' are constants which we shall determine.		where ''k'' and ''m'' are constants which we shall determine.
Zeile 7:		Zeile 7:
	Since the points (2,3) and (3,0) should lie on the line, they must also satisfy the equation of the line,		Since the points (2,3) and (3,0) should lie on the line, they must also satisfy the equation of the line,
-	{{Displayed math||<math>3=k\cdot 2+m\qquad\text{and}\qquad 0=k\cdot 3+m\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>3=k\cdot 2+m\qquad\text{and}\qquad 0=k\cdot 3+m\,\textrm{.}</math>}}
	If we take the difference between the equations, ''m'' disappears and we can work out the slope ''k'',		If we take the difference between the equations, ''m'' disappears and we can work out the slope ''k'',
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	3-0 &= k\cdot 2+m-(k\cdot 3+m)\,,\\[5pt]		3-0 &= k\cdot 2+m-(k\cdot 3+m)\,,\\[5pt]
	3 &= -k\,\textrm{.}		3 &= -k\,\textrm{.}
Zeile 18:		Zeile 18:
	Substituting this into the equation <math>0=k\centerdot 3+m</math> then gives us a value for ''m'',		Substituting this into the equation <math>0=k\centerdot 3+m</math> then gives us a value for ''m'',
-	{{Displayed math||<math>m=-3k=-3\cdot (-3)=9\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>m=-3k=-3\cdot (-3)=9\,\textrm{.}</math>}}
	The equation of the line is thus <math>y=-3x+9</math>.		The equation of the line is thus <math>y=-3x+9</math>.

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Version vom 08:28, 22. Okt. 2008

Let's write down the equation for a straight line as

\displaystyle y=kx+m\,,

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,

\displaystyle 3=k\cdot 2+m\qquad\text{and}\qquad 0=k\cdot 3+m\,\textrm{.}

If we take the difference between the equations, m disappears and we can work out the slope k,

\displaystyle \begin{align} 3-0 &= k\cdot 2+m-(k\cdot 3+m)\,,\\[5pt] 3 &= -k\,\textrm{.} \end{align}

Substituting this into the equation \displaystyle 0=k\centerdot 3+m then gives us a value for m,

\displaystyle m=-3k=-3\cdot (-3)=9\,\textrm{.}

The equation of the line is thus \displaystyle y=-3x+9.

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:

• (x,y) = (2,3): \displaystyle \text{LHS} = 3\ and \displaystyle \ \text{RHS} = -3\cdot 2+9 = 3\,.
• (x,y) = (3,0): \displaystyle \text{LHS} = 0\ and \displaystyle \ \text{LHS} = -3\cdot 3+9 = 0\,.