1.
\(\overrightarrow{a}=\left(1;-2\right);\overrightarrow{b}=\left(3;0\right);\overrightarrow{c}=\left(0;-4\right)\)
\(\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{c}=\left(7;2\right)\)
2.
Đặt \(\overrightarrow{c}=x.\overrightarrow{a}+y\overrightarrow{b}\)
\(\Rightarrow\left\{{}\begin{matrix}3=2x-y\\6=x+4y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\overrightarrow{c}=2\overrightarrow{a}+\overrightarrow{b}\)
3.
\(\left\{{}\begin{matrix}x_M=\dfrac{x_A+x_B}{2}=1\\y_M=\dfrac{y_A+y_B}{2}=3\end{matrix}\right.\) \(\Rightarrow M\left(1;3\right)\)
\(\left\{{}\begin{matrix}x_N=\dfrac{x_B+x_C}{2}=\dfrac{3}{2}\\y_N=\dfrac{y_B+y_C}{2}=\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow N\left(\dfrac{3}{2};\dfrac{5}{2}\right)\)
Tương tự: \(P\left(-\dfrac{1}{2};\dfrac{7}{2}\right)\)
b. \(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{2}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=3\end{matrix}\right.\) \(\Rightarrow G\left(\dfrac{2}{3};3\right)\)
c. Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;-2\right)\\\overrightarrow{DC}=\left(-x;3-y\right)\end{matrix}\right.\)
ABCD là hbh khi \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}-x=4\\3-y=-2\end{matrix}\right.\) \(\Rightarrow D\left(-4;5\right)\)
