Wiki-Quellcode von Lösung Seitenhalbierende im Dreieck
Zuletzt geändert von akukin am 2025/06/08 22:56
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author | version | line-number | content |
---|---|---|---|
1 | Dreieck {{formula}}A(-1|-2), B(5|3){{/formula}} und {{formula}}C(3|7){{/formula}} | ||
2 | (%class=abc%) | ||
3 | 1. (((Mittelpunkt von {{formula}}BC: \quad M_{BC}(4|5){{/formula}}. | ||
4 | |||
5 | Steigung {{formula}}AM_{BC}: m = \frac{5 - (-2)}{4 - (-1)} = 1,4{{/formula}} | ||
6 | |||
7 | Gerade durch {{formula}}A(-1|-2){{/formula}} und {{formula}}M_{BC}(4|5){{/formula}}: {{formula}}g(x) = 1,4x - 0,6{{/formula}}))) | ||
8 | 1. (((Mittelpunkt von {{formula}}AC: \quad M_{AC}(1|2,5){{/formula}}. | ||
9 | |||
10 | Steigung {{formula}}BM_{AC}: m = \frac{2,5 - 3}{1 - 5} = \frac{-0,5}{-4} = 0,125{{/formula}} | ||
11 | |||
12 | Gerade durch {{formula}}B(5|3){{/formula}} und {{formula}}M_{AC}(1|2,5){{/formula}}: {{formula}}h(x) = 0,125x + 2,375{{/formula}}))) | ||
13 | 1. (((Schnitt {{formula}}g(x) = h(x){{/formula}} | ||
14 | {{formula}}1,4x - 0,6 = 0,125x + 2,375{{/formula}} | ||
15 | |||
16 | {{formula}}x = \frac{7}{3} = 2,3333 \quad y = \frac{8}{3} = 2,6666{{/formula}} | ||
17 | |||
18 | Schwerpunkt {{formula}}S\left(\frac{7}{3}\bigl|\frac{8}{3}\right) \quad S(2,333|2,666){{/formula}}))) |