7.
\(\overrightarrow{AB}=\left(4;3\right)\); \(\overrightarrow{AC}=\left(6;1\right)\)
\(\overrightarrow{u}=2.\left(4;3\right)-\left(6;1\right)=\left(2;5\right)\)
b.
Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(-4-x;-y\right)\\\overrightarrow{MB}=\left(-x;3-y\right)\\\overrightarrow{MC}=\left(2-x;1-y\right)\end{matrix}\right.\)
\(\overrightarrow{MA}+2\overrightarrow{MB}+3\overrightarrow{MC}=\left(2-6x;9-6y\right)\)
\(\overrightarrow{MA}+2\overrightarrow{MB}+3\overrightarrow{MC}=\overrightarrow{0}\Rightarrow\left\{{}\begin{matrix}2-6x=0\\9-6y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow M\left(\dfrac{1}{3};\dfrac{3}{2}\right)\)
8.
a.
Áp dụng công thức trung điểm:
\(\left\{{}\begin{matrix}x_C=\dfrac{x_M+x_B}{2}\\y_C=\dfrac{y_M+y_B}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_M=2x_C-x_B=2.\left(-3\right)-\left(-1\right)=-5\\y_M=2y_C-y_B=2.2-\left(-2\right)=6\end{matrix}\right.\)
\(\Rightarrow M\left(-5;6\right)\)
b.
Gọi G là trọng tâm ABC, theo công thức trọng tâm:
\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{2-1-3}{3}=-\dfrac{2}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{1-2+2}{3}=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow G\left(-\dfrac{2}{3};\dfrac{1}{3}\right)\)
c.
Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-3;-3\right)\\\overrightarrow{DC}=\left(-3-x;2-y\right)\end{matrix}\right.\)
ABCD là hình bình hành \(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3-x=-3\\2-y=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\)
\(\Rightarrow D\left(0;5\right)\)
