Cho tam giác $ABC$ có $3$ đỉnh $A(-3;0),B(3;0),C(2;6)$. Tìm tọa độ trọng tâm $G$ và trực tâm $H$ của tam giác.
muốn tính trọng tâm G thì lấy tọa độ của 3 đỉnh cộng lại rồi chia ba là được tọa độ của G nhé –  llelan53a.dhv 08-09-15 11:00 PM
Trọng tâm $G$ có tọa độ:
$x_G=\frac{x_A+x_B+x_C}{3}=\frac{2}{3},y_G=\frac{y_A+y_B+y_C}{3}=2$. Vậy $G(\frac{2}{3};2)$
Gọi $H(x;y)$ là trực tâm ta có:
$\left\{ \begin{array}{l} \overrightarrow{AH}.\overrightarrow{BC}=0\\\overrightarrow{BH}.\overrightarrow{CA}=0  \end{array} \right.  \Leftrightarrow  \left\{ \begin{array}{l} (x+3)(2-3)+(y-0)(6-0)=0\\(x-3)(-3-2)+(y-0)(0-6)=0 \end{array} \right. $
$ \Leftrightarrow  \left\{ \begin{array}{l} -x+6y=3\\ -5x-6y=-15 \end{array} \right.  \Leftrightarrow \left\{ \begin{array}{l} x=2\\ y=\frac{5}{6}  \end{array} \right. $. Vậy $H(2;\frac{5}{6})$
thêm 1 đáp án nhé, tham khảo nha –  hoàng anh thọ 05-07-12 10:09 PM

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