§4. Hệ trục tọa độ

Câu 4: 

a: Tọa độ điểm I là:

\(\left\{{}\begin{matrix}x_I=\dfrac{1+2}{2}=\dfrac{3}{2}\\y_I=\dfrac{-2+3}{2}=\dfrac{1}{2}\end{matrix}\right.\)

Tọa độ điểm G là:

\(\left\{{}\begin{matrix}x_G=\dfrac{1+2-1}{3}=\dfrac{2}{3}\\y_g=\dfrac{-2+3-2}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

Câu 1: 

a: A={3;4;5;6}

b: B={-1}

a: \(\overrightarrow{AB}=\left(4;-5\right)\)

\(\overrightarrow{BC}=\left(-2;4\right)\)

\(\overrightarrow{AC}=\left(2;-1\right)\)

\(\overrightarrow{u}=\overrightarrow{AB}-3\overrightarrow{BC}+2\overrightarrow{AC}\)

\(=\left(4-3\cdot\left(-2\right)+2\cdot2;-5-3\cdot4+2\cdot\left(-1\right)\right)=\left(14;-19\right)\)

b: Để ABDC là hình bình hành thì 

\(\overrightarrow{AB}=\overrightarrow{CD}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_D-1=4\\y_D-3=-5\end{matrix}\right.\Leftrightarrow D\left(5;-2\right)\)

Chọn C

\(\overrightarrow{AM}=\left(x_M+3;y_M-2\right)\)

\(\overrightarrow{AB}=\left(4;3\right)\)

\(\overrightarrow{BC}=\left(3;2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_M+3=22\\y_M-2=16\end{matrix}\right.\Leftrightarrow M\left(19;18\right)\)

Chọn D

D

D

\(\left\{{}\begin{matrix}x_G=\dfrac{-1+\left(-2\right)+4}{3}=\dfrac{1}{3}\\y_G=\dfrac{1+3+\left(-5\right)}{3}=-\dfrac{1}{3}\end{matrix}\right.\)