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:
\(\left\{{}\begin{matrix}x_A=\dfrac{x_M+x_B+x_C}{3}\\y_A=\dfrac{y_M+y_B+y_C}{3}\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_M=3x_A-x_B-x_C=3.1-0-3=0\\y_M=3y_A-y_B-y_C=3.\left(-2\right)-4-2=-12\end{matrix}\right.\)
\(\Rightarrow M\left(0;-12\right)\)
h.
Gọi \(M\left(x;y\right)\Rightarrow\overrightarrow{CM}=\left(x-3;y-2\right)\)
\(2\overrightarrow{AB}-3\overrightarrow{AC}=2\left(-1;6\right)-3\left(2;4\right)=\left(-8;0\right)\)
\(\overrightarrow{CM}=2\overrightarrow{AB}-3\overrightarrow{AC}\Leftrightarrow\left\{{}\begin{matrix}x-3=-8\\y-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=2\end{matrix}\right.\)
\(\Rightarrow M\left(-5;2\right)\)
i.
Gọi \(N\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AN}=\left(x-1;y+2\right)\\\overrightarrow{BN}=\left(x;y-4\right)\\\overrightarrow{CN}=\left(x-3;y-2\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{AN}+2\overrightarrow{BN}-4\overrightarrow{CN}=\left(x-1;y+2\right)+2\left(x;y-4\right)-4\left(x-3;y-2\right)\)
\(\Rightarrow\overrightarrow{AN}+2\overrightarrow{BN}-4\overrightarrow{CN}=\left(-x+11;-y+2\right)\)
\(\overrightarrow{AN}+2\overrightarrow{BN}-4\overrightarrow{CN}=\overrightarrow{0}\Leftrightarrow\left\{{}\begin{matrix}-x+11=0\\-y+2=0\end{matrix}\right.\)
\(\Rightarrow N\left(11;2\right)\)
j.
\(\left\{{}\begin{matrix}2\overrightarrow{AB}=\left(-2;12\right)\\3\overrightarrow{AC}=\left(6;12\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{u}=2\overrightarrow{AB}-3\overrightarrow{AC}=\left(-2;12\right)-\left(6;12\right)\)
\(\Rightarrow\overrightarrow{u}=\left(-8;0\right)\)
k.
Do F thuộc Oy nên tọa độ có dạng: \(\left(0;y\right)\)
\(\Rightarrow\overrightarrow{AF}=\left(-1;y+2\right)\)
\(\Rightarrow AF=\sqrt{\left(-1\right)^2+\left(y+2\right)^2}=\sqrt{y^2+4y+5}\)
\(AF=\sqrt{10}\Rightarrow\sqrt{y^2+4y+5}=\sqrt{10}\)
\(\Rightarrow y^2+4y-5=0\Rightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}F\left(0;1\right)\\F\left(0;-5\right)\end{matrix}\right.\)
