§4. Hệ trục tọa độ

a: \(\overrightarrow{u}=\left(0-2\cdot\left(-4\right);0-2\cdot5\right)=\left(8;-10\right)\)

\(\overrightarrow{AB}=\left(-1;-4\right)\)

Ta có: ABCD là hình bình hành

nên \(\overrightarrow{AB}=\overrightarrow{DC}\)

\(\Leftrightarrow\overrightarrow{DC}=\left(-1;-4\right)\)

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

?

a.

\(\overrightarrow{AB}=\left(-1;6\right)\) ; \(\overrightarrow{AC}=\left(2;4\right)\) ; \(\overrightarrow{BC}=\left(3;-2\right)\)

b.

Do \(\dfrac{-1}{2}\ne\dfrac{6}{4}\Rightarrow\) hai vecto \(\overrightarrow{AB};\overrightarrow{AC}\) ko cùng phương

\(\Rightarrow\) 3 điểm A;B;C không thẳng hàng

Hay A;B;C là 3 đỉnh của 1 tam giác

c.

\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=\dfrac{1+0}{2}=\dfrac{1}{2}\\y_I=\dfrac{y_A+y_B}{2}=\dfrac{-2+4}{2}=1\end{matrix}\right.\) \(\Rightarrow I\left(\dfrac{1}{2};1\right)\)

\(\Rightarrow\overrightarrow{CI}=\left(-\dfrac{5}{2};-1\right)\Rightarrow CI=\sqrt{\left(-\dfrac{5}{2}\right)^2+\left(-1\right)^2}=\dfrac{\sqrt{29}}{2}\)

d.

Gọi \(D\left(x;y\right)\Rightarrow\overrightarrow{DC}=\left(3-x;2-y\right)\)

ABCD là hbh khi và chỉ khi:

\(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}3-x=-1\\2-y=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\) \(\Rightarrow D\left(4;-4\right)\)

e.

\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{1+0+3}{3}=\dfrac{4}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{-2+4+2}{3}=\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow G\left(\dfrac{4}{3};\dfrac{4}{3}\right)\)

f.

D đối xứng A qua C khi và chỉ khi C là trung điểm AD

\(\Rightarrow\left\{{}\begin{matrix}x_C=\dfrac{x_A+x_D}{2}\\y_C=\dfrac{y_A+y_D}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_D=2x_C-x_A=3.2-1=5\\y_D=2y_C-y_A=2.2-\left(-2\right)=6\end{matrix}\right.\)

\(\Rightarrow D\left(5;6\right)\)

g.

A là trọng tâm tam giác MBC khi:

\(\left\{{}\begin{matrix}x_A=\dfrac{x_M+x_B+x_C}{3}\\y_A=\dfrac{y_M+y_B+y_C}{3}\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_M=3x_A-x_B-x_C=3.1-0-3=0\\y_M=3y_A-y_B-y_C=3.\left(-2\right)-4-2=-12\end{matrix}\right.\)

\(\Rightarrow M\left(0;-12\right)\)

Lời giải:
\(\overrightarrow{a}=-2\overrightarrow{i}+3\overrightarrow{j}=-2(1,0)+3(0,1)=(-2.1+3.0, -2.0+3.1)=(-2,3)\)

Fugg