Wiki-Quellcode von Lösung Länge und Mittelpunkt einer Strecke 2
Zuletzt geändert von akukin am 2025/06/05 15:09
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| author | version | line-number | content |
|---|---|---|---|
| |
1.1 | 1 | (%class=abc%) |
| 2 | 1. ((({{formula}} | ||
| 3 | x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{-3+0}{2}=-1,5 | ||
| 4 | {{/formula}} | ||
| 5 | {{formula}} | ||
| 6 | y_{M}=\frac{y_{1}+y_{2}}{2}=\frac{2+0}{2}=1 | ||
| 7 | {{/formula}} | ||
| 8 | → {{formula}}M(-1,5|1){{/formula}} | ||
| 9 | |||
| 10 | {{formula}} | ||
| 11 | x_{M}=\frac{4+(-2)}{2}=1 | ||
| 12 | {{/formula}} | ||
| 13 | {{formula}} | ||
| 14 | y_{M}=3,5=\frac{y_{1}+5}{2}\Leftrightarrow y_{1}=2 | ||
| 15 | {{/formula}} | ||
| 16 | |||
| 17 | ))) | ||
| 18 | 1. ((({{formula}} | ||
| 19 | x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{3+7}{2}=5 | ||
| 20 | {{/formula}} | ||
| 21 | {{formula}} | ||
| 22 | y_{M}=\frac{y_{1}+y_{2}}{2}=\frac{-5+2}{2}=-1,5 | ||
| 23 | {{/formula}} | ||
| 24 | → {{formula}}M(5|-1,5){{/formula}} | ||
| 25 | |||
| 26 | __Geradengleichung bestimmen:__ | ||
| 27 | Hauptform oder Punkt-Steigungs-Form liefert {{formula}}y=0,5x-4{{/formula}} | ||
| 28 | ))) | ||
| 29 | 1. Der Schnittpunkt der Geraden {{formula}}y=0,5x-4{{/formula}} mit der y-Achse lautet {{formula}}S_y(0|-4){{/formula}} | ||
| 30 | Länge der Strecke {{formula}}S_yA{{/formula}} berechnen: | ||
| 31 | {{formula}} | ||
| 32 | \overline{S_{y}A}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=\sqrt{(3-0)^{2}+(-5-(-4))^{2}}=\sqrt{10} | ||
| 33 | {{/formula}} |