trong mặt phẳng tọa độ Oxy cho tam giác ABC có phương trình chứa 2 cạnh AB,BC lần lượt là x+y-1=o và 3x-y+2=0. Tìm tọa độ trực tâm H của tam giác. Biết I(1;2) là tâm đường tròn ngoại tiếp tam giác ABC
 $(AB)\cap (BC)= B         \Rightarrow B(\frac{-1}{4};\frac{5}{4})$
   $\overrightarrow{BI}(\frac{5}{4};\frac{3}{4})           \rightarrow BI=\frac{\sqrt{34}}{4}$
 Gọi A(a;1-a)
Ta có :     $AI=BI\Leftrightarrow \sqrt{(a-1)^{2}+(a+1)^{2}}=\frac{\sqrt{34}}{4}\Leftrightarrow a=\frac{1}{4}\Rightarrow A(\frac{1}{4};\frac{3}{4})$
Tương tự với điểm C $\rightarrow C(\frac{9}{20};\frac{67}{20})$
Gọi AD , CK là các  cao trong tan giác
pt (AD) : $2x-2y+1=0$
pt (CK) : $2x+6y-21=0$
 (AD)  $\cap $ (CK)=H         $\Rightarrow H(\frac{9}{4};\frac{11}{4})$
Nếu mk trả lời đúng thì vote cho mk nhé –  Dân Nguyễn 29-04-14 09:44 PM

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