a: AMBC là hình bình hành
=>\(\overrightarrow{AM}=\overrightarrow{CB}\)
\(\overrightarrow{AM}=\left(x-2;y-1\right);\overrightarrow{CB}=\left(-1;-21\right)\)
=>\(\left\{{}\begin{matrix}x-2=-1\\y-1=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-20\end{matrix}\right.\)
vậy: M(1;-20)
b: \(I\left(x;y\right);A\left(2;1\right);B\left(3;-4\right);C\left(4;17\right)\)
\(\overrightarrow{IA}=\left(2-x;1-y\right)\)
\(\overrightarrow{IB}=\left(3-x;-4-y\right)\)
\(\overrightarrow{BC}=\left(4-3;17+4\right)=\left(1;21\right)\)
\(\overrightarrow{IA}+2\overrightarrow{IB}=4\overrightarrow{BC}\)
=>\(\left\{{}\begin{matrix}2-x+2\left(3-x\right)=4\cdot1=4\\1-y+2\left(-4-y\right)=4\cdot21=84\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-x+6-2x=4\\1-y-8-2y=84\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=4-6-2=-4\\-3y=84+8-1=84+7=91\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{91}{3}\end{matrix}\right.\)
vậy: I(4/3;-91/3)
c: A là trọng tâm của ΔCAN
=>\(\left\{{}\begin{matrix}x_C+x_A+x_N=3\cdot x_A\\y_C+y_A+y_N=3\cdot y_A\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4+2+x_N=3\cdot2=6\\17+1+y_N=3\cdot1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_N=0\\y_N=3-18=-15\end{matrix}\right.\)
vậy: N(0;-15)
d: B là trung điểm của AE
=>\(\left\{{}\begin{matrix}x_A+x_E=2\cdot x_B\\y_A+y_E=2\cdot y_B\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_E=2\cdot3-2=6-2=4\\y_E=2\cdot\left(-4\right)-1=-8-1=-9\end{matrix}\right.\)
vậy: E(4;-9)
