Lösung 4.1:4c
Aus Online Mathematik Brückenkurs 1
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| Zeile 2: | Zeile 2: | ||
<math>(x,0)</math> to <math>(3,3)</math> and <math>(5,1)</math> is given by | <math>(x,0)</math> to <math>(3,3)</math> and <math>(5,1)</math> is given by | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{and}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}} |
respectively. Because these two distances should be the same, we get the equation | respectively. Because these two distances should be the same, we get the equation | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sqrt{(x-3)^2+9} = \sqrt{(x-5)^2+1}</math>}} |
or, if we take the square, so as to get rid of the square root sign, | or, if we take the square, so as to get rid of the square root sign, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(x-3)^2 + 9 = (x-5)^2+1\,\textrm{.}</math>}} |
Now, expand the squares and collect together all the terms onto one side, | Now, expand the squares and collect together all the terms onto one side, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
& x^2-6x+9+9 = x^2-10x+25+1\\[5pt] | & x^2-6x+9+9 = x^2-10x+25+1\\[5pt] | ||
&\quad\Leftrightarrow\quad 4x-8=0\,\textrm{.} | &\quad\Leftrightarrow\quad 4x-8=0\,\textrm{.} | ||
| Zeile 28: | Zeile 28: | ||
is | is | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10}</math>}} |
and the distance between <math>(2,0)</math> and <math>(5,1)</math> is | and the distance between <math>(2,0)</math> and <math>(5,1)</math> is | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}} |
Note: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots. | Note: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots. | ||
Version vom 08:48, 22. Okt. 2008
Let the point on the x-axis have coordinates \displaystyle (x,0), where x is an unknown number. Then, using the distance formula, the distance from the point \displaystyle (x,0) to \displaystyle (3,3) and \displaystyle (5,1) is given by
| \displaystyle \sqrt{(x-3)^2+(0-3)^2}\qquad\text{and}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,} |
respectively. Because these two distances should be the same, we get the equation
| \displaystyle \sqrt{(x-3)^2+9} = \sqrt{(x-5)^2+1} |
or, if we take the square, so as to get rid of the square root sign,
| \displaystyle (x-3)^2 + 9 = (x-5)^2+1\,\textrm{.} |
Now, expand the squares and collect together all the terms onto one side,
| \displaystyle \begin{align}
& x^2-6x+9+9 = x^2-10x+25+1\\[5pt] &\quad\Leftrightarrow\quad 4x-8=0\,\textrm{.} \end{align} |
This gives that \displaystyle x=2, i.e. the point on the x-axis is \displaystyle (2,0)\,.
As a final step, we check that we have calculated correctly and that the distances really are the same. The distance between \displaystyle (2,0) and \displaystyle (3,3)
is
| \displaystyle \sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10} |
and the distance between \displaystyle (2,0) and \displaystyle (5,1) is
| \displaystyle \sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.} |
Note: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots.
