Предмет: Геометрия,
автор: Мil
Помогите,пожалуйста, решить задачку по геометрии..........Даны точки А(-1,2), В(1,-2), С(7,2). Найти основание биссектрисы АК и ее длину
Ответы
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Биссектриса внутреннего угла треугольника делит противоположную сторону в отношении, равном отношению двух прилежащих сторон.
![frac{BK}{KC}=frac{AB}{AC}; frac{BK}{KC}=frac{AB}{AC};](https://tex.z-dn.net/?f=frac%7BBK%7D%7BKC%7D%3Dfrac%7BAB%7D%7BAC%7D%3B)
![AB= sqrt{4+16}= sqrt{20}=2 sqrt{5}; AB= sqrt{4+16}= sqrt{20}=2 sqrt{5};](https://tex.z-dn.net/?f=AB%3D+sqrt%7B4%2B16%7D%3D+sqrt%7B20%7D%3D2+sqrt%7B5%7D%3B)
![AC= sqrt{64+0}= sqrt{64}=8; AC= sqrt{64+0}= sqrt{64}=8;](https://tex.z-dn.net/?f=AC%3D+sqrt%7B64%2B0%7D%3D+sqrt%7B64%7D%3D8%3B)
![frac{BK}{KC}= frac{ 2sqrt{5}}{8}=frac{ sqrt{5}}{4}; frac{BK}{KC}= frac{ 2sqrt{5}}{8}=frac{ sqrt{5}}{4};](https://tex.z-dn.net/?f=frac%7BBK%7D%7BKC%7D%3D+frac%7B+2sqrt%7B5%7D%7D%7B8%7D%3Dfrac%7B+sqrt%7B5%7D%7D%7B4%7D%3B)
координаты точки K, которая делит отрезок BC в отношении λ, выражаются формулами:
![x_K= frac{x_B+ lambda x_C}{1+ lambda};y_K= frac{y_B+ lambda y_C}{1+ lambda}; x_K= frac{x_B+ lambda x_C}{1+ lambda};y_K= frac{y_B+ lambda y_C}{1+ lambda};](https://tex.z-dn.net/?f=x_K%3D+frac%7Bx_B%2B+lambda+x_C%7D%7B1%2B+lambda%7D%3By_K%3D+frac%7By_B%2B+lambda+y_C%7D%7B1%2B+lambda%7D%3B++)
![lambda= frac{ sqrt{5}}{4}; x_K= frac{1+ 7*frac{ sqrt{5}}{4}}{1+ frac{ sqrt{5}}{4}};y_K= frac{-2+ 2*frac{ sqrt{5}}{4}}{1+ frac{ sqrt{5}}{4}}; lambda= frac{ sqrt{5}}{4}; x_K= frac{1+ 7*frac{ sqrt{5}}{4}}{1+ frac{ sqrt{5}}{4}};y_K= frac{-2+ 2*frac{ sqrt{5}}{4}}{1+ frac{ sqrt{5}}{4}};](https://tex.z-dn.net/?f=lambda%3D+frac%7B+sqrt%7B5%7D%7D%7B4%7D%3B+x_K%3D+frac%7B1%2B+7%2Afrac%7B+sqrt%7B5%7D%7D%7B4%7D%7D%7B1%2B+frac%7B+sqrt%7B5%7D%7D%7B4%7D%7D%3By_K%3D+frac%7B-2%2B+2%2Afrac%7B+sqrt%7B5%7D%7D%7B4%7D%7D%7B1%2B+frac%7B+sqrt%7B5%7D%7D%7B4%7D%7D%3B)
![x_K= frac{4+ 7sqrt{5}}{4+sqrt{5}}=frac{(4+ 7sqrt{5})(4-sqrt{5})}{(4+sqrt{5})(4-sqrt{5})}=frac{16+28sqrt{5}-35-4sqrt{5}}{11}= frac{24sqrt{5}-19}{11}; x_K= frac{4+ 7sqrt{5}}{4+sqrt{5}}=frac{(4+ 7sqrt{5})(4-sqrt{5})}{(4+sqrt{5})(4-sqrt{5})}=frac{16+28sqrt{5}-35-4sqrt{5}}{11}= frac{24sqrt{5}-19}{11};](https://tex.z-dn.net/?f=x_K%3D+frac%7B4%2B+7sqrt%7B5%7D%7D%7B4%2Bsqrt%7B5%7D%7D%3Dfrac%7B%284%2B+7sqrt%7B5%7D%29%284-sqrt%7B5%7D%29%7D%7B%284%2Bsqrt%7B5%7D%29%284-sqrt%7B5%7D%29%7D%3Dfrac%7B16%2B28sqrt%7B5%7D-35-4sqrt%7B5%7D%7D%7B11%7D%3D+frac%7B24sqrt%7B5%7D-19%7D%7B11%7D%3B)
![y_K= frac{-8+ 2 sqrt{5}}{4+sqrt{5}}=frac{(-8+ 2 sqrt{5)}(4-sqrt{5})}{(4+sqrt{5})(4-sqrt{5}}= frac{-32+ 8 sqrt{5}-10+8sqrt{5}}{11}= frac{16sqrt{5}-42}{11}; y_K= frac{-8+ 2 sqrt{5}}{4+sqrt{5}}=frac{(-8+ 2 sqrt{5)}(4-sqrt{5})}{(4+sqrt{5})(4-sqrt{5}}= frac{-32+ 8 sqrt{5}-10+8sqrt{5}}{11}= frac{16sqrt{5}-42}{11};](https://tex.z-dn.net/?f=y_K%3D+frac%7B-8%2B+2+sqrt%7B5%7D%7D%7B4%2Bsqrt%7B5%7D%7D%3Dfrac%7B%28-8%2B+2+sqrt%7B5%29%7D%284-sqrt%7B5%7D%29%7D%7B%284%2Bsqrt%7B5%7D%29%284-sqrt%7B5%7D%7D%3D+frac%7B-32%2B+8+sqrt%7B5%7D-10%2B8sqrt%7B5%7D%7D%7B11%7D%3D+frac%7B16sqrt%7B5%7D-42%7D%7B11%7D%3B)
![AK^2=(frac{24sqrt{5}-19}{11}+1)^2+(frac{16sqrt{5}-42}{11}-2)^2= AK^2=(frac{24sqrt{5}-19}{11}+1)^2+(frac{16sqrt{5}-42}{11}-2)^2=](https://tex.z-dn.net/?f=AK%5E2%3D%28frac%7B24sqrt%7B5%7D-19%7D%7B11%7D%2B1%29%5E2%2B%28frac%7B16sqrt%7B5%7D-42%7D%7B11%7D-2%29%5E2%3D)
![=(frac{24sqrt{5}-8}{11})^2+(frac{16sqrt{5}-64}{11})^2= =(frac{24sqrt{5}-8}{11})^2+(frac{16sqrt{5}-64}{11})^2=](https://tex.z-dn.net/?f=%3D%28frac%7B24sqrt%7B5%7D-8%7D%7B11%7D%29%5E2%2B%28frac%7B16sqrt%7B5%7D-64%7D%7B11%7D%29%5E2%3D)
![=(frac{8}{11})^2 (3sqrt{5}-1)^2+ (frac{16}{11})^2 (sqrt{5}-4)^2= =(frac{8}{11})^2 (3sqrt{5}-1)^2+ (frac{16}{11})^2 (sqrt{5}-4)^2=](https://tex.z-dn.net/?f=%3D%28frac%7B8%7D%7B11%7D%29%5E2+%283sqrt%7B5%7D-1%29%5E2%2B+%28frac%7B16%7D%7B11%7D%29%5E2+%28sqrt%7B5%7D-4%29%5E2%3D)
![=frac{64}{121} (45-6sqrt{5}+1)+ frac{256}{121}(5-8sqrt{5}+16)= =frac{64}{121} (45-6sqrt{5}+1)+ frac{256}{121}(5-8sqrt{5}+16)=](https://tex.z-dn.net/?f=%3Dfrac%7B64%7D%7B121%7D+%2845-6sqrt%7B5%7D%2B1%29%2B+frac%7B256%7D%7B121%7D%285-8sqrt%7B5%7D%2B16%29%3D)
![=frac{64}{121} (46-6sqrt{5})+ frac{256}{121}(21-8sqrt{5})= =frac{64}{121} (46-6sqrt{5})+ frac{256}{121}(21-8sqrt{5})=](https://tex.z-dn.net/?f=%3Dfrac%7B64%7D%7B121%7D+%2846-6sqrt%7B5%7D%29%2B+frac%7B256%7D%7B121%7D%2821-8sqrt%7B5%7D%29%3D)
![=frac{64}{121}(46-6sqrt{5}+84-32sqrt{5})=frac{64}{121}(130-38sqrt{5}) =frac{64}{121}(46-6sqrt{5}+84-32sqrt{5})=frac{64}{121}(130-38sqrt{5})](https://tex.z-dn.net/?f=%3Dfrac%7B64%7D%7B121%7D%2846-6sqrt%7B5%7D%2B84-32sqrt%7B5%7D%29%3Dfrac%7B64%7D%7B121%7D%28130-38sqrt%7B5%7D%29)
координаты точки K, которая делит отрезок BC в отношении λ, выражаются формулами:
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