Ôn tập chương I

\(AB^2+CD^2-\left(BC^2+DA^2\right)=\overrightarrow{AB}^2+\overrightarrow{CD}^2-\overrightarrow{BC}^2-\overrightarrow{AD}^2\)

\(=\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{AB}-\overrightarrow{AD}\right)+\left(\overrightarrow{CD}-\overrightarrow{BC}\right)\left(\overrightarrow{CD}+\overrightarrow{BC}\right)\)

\(=\overrightarrow{DB}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)+\overrightarrow{DB}\left(\overrightarrow{BC}+\overrightarrow{DC}\right)\)

\(=\overrightarrow{DB}\left(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{BC}+\overrightarrow{DC}\right)\)

\(=2\overrightarrow{AC}.\overrightarrow{DB}\) (đpcm)

???

Theo đề, ta có:

\(\left\{{}\begin{matrix}-3a+b=2\\5a+b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-8a=6\\5a+b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{4}\\b=-4-5a=-4-5\cdot\dfrac{-3}{4}=-4+\dfrac{15}{4}=-\dfrac{1}{4}\end{matrix}\right.\)

a) Gọi G(x_G;y_G)

x_G=\(\dfrac{X_A+X_B+X_C}{3}=\dfrac{3-2+1}{3}\)=\(\dfrac{2}{3}\)

y_G=\(\dfrac{Y_A+Y_B+Y_C}{3}=\dfrac{3+4+5}{3}=4\)

⇒G(\(\dfrac{2}{3};4\))

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 \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}-x=4\\3-y=-2\end{matrix}\right.\) \(\Rightarrow D\left(-4;5\right)\)