Gọi \(I\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{IA}=\left(2-x;1-y\right)\\\overrightarrow{BC}=\left(-2;4\right)\\\overrightarrow{CI}=\left(-3-x;2-y\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\overrightarrow{IA}-\overrightarrow{BC}=\left(2-2x;-2y-2\right)\\4\overrightarrow{CI}=\left(-12-4x;8-4y\right)\end{matrix}\right.\)
\(2\overrightarrow{IA}-\overrightarrow{BC}=4\overrightarrow{CI}\Rightarrow\left\{{}\begin{matrix}2-2x=-12-4x\\-2y-2=8-4y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-7\\y=5\end{matrix}\right.\) \(\Rightarrow I\left(-7;5\right)\)
Cho I(x;y)
Ta có \(\overrightarrow{IA}=\left(2-x;1-y\right)\)
\(\overrightarrow{BC}=\left(-2;4\right);\overrightarrow{CI}=\left(x+3;y-2\right)\)
Để tm bth trên ta có
\(\Leftrightarrow\left\{{}\begin{matrix}4-2x+2=4x+12\\1-y-4=4y-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
-> I(-1;1)
