tìm tọa độ khó

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khoai_lang_chay

vecto MN(1;3) MN=\sqrt[2]{10} =>BC=2\sqrt[2]{10}
vì BC//MN ; D thuoc BC nen viet duoc phuong trinh BC: 3x-y+1=0
=>B(a:3a+1) C(b;3b+1)
BC=2\sqrt[2]{10}
thay toa do B va C ta duoc: a=b+2 hoăc a=b-2
TH1 a=b+2
B(b+2;3b+7) C(b;3b+1)
vectoDB(b+2;3b+6) vecto MB(b+3;3b+8)
vectoDC(b;3b) vectoNC(b;3b-1)
AD la phan giac =>\frac{BD}{MB}=\frac{DC}{NC}
thay do dai cac vec to vao ta duoc phuong trinh bac 3: 5b mủ 3+14b bình + 5b -1=0
giai pt ta duoc toa do cac diem
TH2 tuong tu
 
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