Lösung 2.2:5a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.2:5a moved to Solution 2.2:5a: Robot: moved page)
Zeile 1: Zeile 1:
 +
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>
 +
 +
 +
{{NAVCONTENT_START}}
{{NAVCONTENT_START}}
<!--<center> [[Image:2_2_5a-1(2).gif]] </center>-->
<!--<center> [[Image:2_2_5a-1(2).gif]] </center>-->

Version vom 09:02, 18. Sep. 2008

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


\displaystyle y=kx+m


where \displaystyle k and \displaystyle m are constants which we shall determine.

Since the points \displaystyle \left( 2 \right., \left. 3 \right) and \displaystyle \left( 3 \right., \left. 0 \right) should lie on the line, they must also satisfy the equation of the line,


\displaystyle 3=k\centerdot 2+m and \displaystyle 0=k\centerdot 3+m


If we take the difference between the equations, \displaystyle m disappears and we can work out the gradient \displaystyle k,


\displaystyle 3-0=k\centerdot 2+m-\left( k\centerdot 3+m \right)


\displaystyle 3=-k

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


\displaystyle m=-3k=-3\centerdot \left( -3 \right)=9


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 \displaystyle \left( 2 \right., \left. 3 \right) and \displaystyle \left( 3 \right., \left. 0 \right) satisfy the equation of the line:

\displaystyle \left( x \right., \left. y \right)=\left( 2 \right., \left. 3 \right): LHS= \displaystyle 3 and RHS= \displaystyle -3\centerdot 2+9=3


\displaystyle \left( x \right., \left. y \right)=\left( 3 \right., \left. 0 \right): LHS= \displaystyle 0 and LHS= \displaystyle -3\centerdot 3+9=0


Image:S1_2_2_5_a.jpg

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:5a“