\(\overrightarrow{AH}=\left(x-2;y-6\right)\)
\(\overrightarrow{BC}=\left(8;-4\right)\)
\(\overrightarrow{BH}=\left(x+3;y-4\right)\)
Vì H là chân đường cao kẻ từ A xuống BC
nên ta có:
\(\left\{{}\begin{matrix}8\left(x-2\right)+\left(-4\right)\left(y-6\right)=0\\\dfrac{x+3}{8}=\dfrac{y-4}{-4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-16-4y+24=0\\-4x-12=8y-32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-4y+8=0\\-4x-8y=-20\end{matrix}\right.\)
=>x=1/5; y=12/5