Cho tam giác $ABC$ với $A(-1,2);B(2,0);C(3,4)$
$a.$ Tìm tọa độ trọng tâm $G$, trực tâm $H$ và tâm $I$ đường tròn ngoại tiếp của tam giác $ABC$
$b.$ Chứng minh rằng $\overrightarrow {GH}=2\overrightarrow {GI}  $ có nhận xét gì về ba điểm $G,H,I$ ?
$a.$

Vì $G$ là trọng tâm của $\Delta ABC$ nên ta có:
$\begin{cases}x_G=\frac{x_A+x_B+x_C}{3}  \\ y_G=\frac{y_A+y_B+g_C}{3}  \end{cases} \Rightarrow  \begin{cases}x_G=\frac{-1+2+3}{3}=\frac{4}{3}   \\ y_g=\frac{2+0+4 }{3}=2 \end{cases} \Rightarrow  G(\frac{4}{3};2 )$
$H$ là trực tâm $\Delta ABC$
$\Rightarrow  \begin{cases}\overrightarrow {AH}\bot \overrightarrow {BC}   \\ \overrightarrow {BH}\bot \overrightarrow {CA}   \end{cases} \Leftrightarrow  \begin{cases}\overrightarrow {AH}.\overrightarrow {BC}=0   \\ \overrightarrow {BH}.\overrightarrow {CA}=0   \end{cases}   (*)$
Với :
$\overrightarrow {AH}=(x_H+1;y_H-2),     \overrightarrow {BC}=(3-2;4-0)=(1;4)  $
$\overrightarrow {BH}=(x_H-2;y_H)        \overrightarrow {CA}=(-1-3;2-4)=(-4;-2)  $
Vậy $(*)\Leftrightarrow  \begin{cases}x_H+1+4(y_H-2)=0 \\ -4(x_H-2)-2y_H=0 \end{cases} \Leftrightarrow  \begin{cases}x_H+4y_H=7 \\ 2x_H+y_H=4 \end{cases} \Leftrightarrow  \begin{cases}x_H=\frac{9}{7}  \\ y_H=\frac{10}{7}  \end{cases} $
Vậy $H(\frac{9}{7};\frac{10}{7}  )$
Gọi $I(x_I;y_I)$ là tâm đường tròn ngoại tiếp $\Delta ABC$
Ta có : $IA^2=IB^2=IC^2$
$\Leftrightarrow  \begin{cases}IA^2=IB^2 \\ IB^2=IC^2 \end{cases} \Leftrightarrow  \Leftrightarrow  \begin{cases}(-1-x_I)^2+(2-y_I)^2=(2-x_I)^2+(0-y_I)^2 \\ (2-x_I)^2+(0-y_I)^2=(3-x_I)^2+(4-y_I)^2 \end{cases} $
$\Leftrightarrow  \begin{cases}6x_I-4y_I=-1 \\ 2x_I+8y_I=21 \end{cases} \Leftrightarrow  \begin{cases}x_I=\frac{19}{14}  \\ y_I=\frac{16}{7}  \end{cases} \Rightarrow  I(\frac{19}{14} ;\frac{16}{7} )$
$b.$ Ta có : $\overrightarrow {HG}=(\frac{4}{3}-\frac{9}{7};2-\frac{10}{7}   )=(\frac{1}{21} ;\frac{4}{7} ) $
$\overrightarrow {GI}=(\frac{19}{14} -\frac{4}{3};\frac{16}{7}  -2)=(\frac{1}{42} ;\frac{2}{7} ) $
$\Rightarrow  \overrightarrow {HG}=2\overrightarrow {GI}  $ Vì $\Rightarrow  \overrightarrow {HG}=2\overrightarrow {GI}  $ suy ra $H,I,G$ thẳng hàng

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