a: \(\overrightarrow{AE}=\dfrac{2}{3}\overrightarrow{EC}\)

=>E nằm giữa A và C và AE=2/3EC

Ta có: AE+EC=AC(E nằm giữa A và C)

=>\(AC=\dfrac{2}{3}EC+EC=\dfrac{5}{3}EC\)

=>\(\dfrac{AE}{AC}=\dfrac{\dfrac{2}{3}EC}{\dfrac{5}{3}EC}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)

=>\(AE=\dfrac{2}{5}AC\)

=>\(\overrightarrow{AE}=\dfrac{2}{5}\cdot\overrightarrow{AC}\)

\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}\)

\(=-\overrightarrow{AB}+\dfrac{2}{5}\cdot\overrightarrow{AC}\)

b: \(\left|\overrightarrow{IA}+\overrightarrow{IG}\right|=\left|\overrightarrow{IA}-\overrightarrow{IG}\right|\)

=>\(\left[{}\begin{matrix}\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IA}-\overrightarrow{IG}\\\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IG}-\overrightarrow{IA}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2\cdot\overrightarrow{IG}=\overrightarrow{0}\\2\cdot\overrightarrow{IA}=\overrightarrow{0}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}I\equiv G\\I\equiv A\end{matrix}\right.\)

1.

\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=-\dfrac{3}{2}\\y_I=\dfrac{y_A+y_B}{2}=1\end{matrix}\right.\) \(\Rightarrow I\left(-\dfrac{3}{2};1\right)\)

\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=0\\y_G=\dfrac{y_A+y_B+y_C}{3}=0\end{matrix}\right.\) \(\Rightarrow G\left(0;0\right)\)

2.

\(\left\{{}\begin{matrix}\overrightarrow{CI}=\left(-\dfrac{9}{2};3\right)\\\overrightarrow{AG}=\left(-2;-3\right)\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}CI=\sqrt{\left(-\dfrac{9}{2}\right)^2+3^2}=\dfrac{3\sqrt{13}}{2}\\AG=\sqrt{\left(-2\right)^2+\left(-3\right)^2}=\sqrt{13}\end{matrix}\right.\)

3.

Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-7;-4\right)\\\overrightarrow{DC}=\left(3-x;-2-y\right)\end{matrix}\right.\)

\(ABCD\) là hbh \(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\)

\(\Leftrightarrow\left\{{}\begin{matrix}-7=3-x\\-4=-2-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=10\\y=2\end{matrix}\right.\) 

\(\Rightarrow D\left(10;2\right)\)

4. Gọi \(H\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{CH}=\left(x-3;y+2\right)\\\overrightarrow{AH}=\left(x-2;y-3\right)\\\overrightarrow{BC}=\left(8;-1\right)\end{matrix}\right.\)

H là trực tâm \(\Leftrightarrow\left\{{}\begin{matrix}AH\perp BC\\CH\perp AB\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AH}.\overrightarrow{BC}=0\\\overrightarrow{CH}.\overrightarrow{AB}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8\left(x-2\right)-1\left(y-3\right)=0\\-7\left(x-3\right)-4\left(y+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-y=13\\-7x-4y=-13\end{matrix}\right.\) \(\Rightarrow H\left(\dfrac{5}{3};\dfrac{1}{3}\right)\)

\(I\left(\frac{3-11}{2};\frac{2+0}{2}\right)\Rightarrow I\left(-4;1\right)\)

\(G\left(\frac{3+5-11}{3};\frac{2+4+0}{3}\right)\Rightarrow G\left(-1;2\right)\)

\(M\left(-22-5;0-4\right)\Rightarrow M\left(-27;-4\right)\)

\(D\left(3+5--11;2+4-0\right)\Rightarrow D\left(19;6\right)\)

 a) Ta thấy \(\overrightarrow{AB}\left(3;2\right)\) và \(\overrightarrow{AC}\left(4;-3\right)\). Vì \(\dfrac{3}{4}\ne\dfrac{2}{-3}\) nên A, B, C không thẳng hàng.

 b) Ta có \(\overrightarrow{BC}\left(1;-5\right)\) 

 Do vậy \(AB=\left|\overrightarrow{AB}\right|=\sqrt{3^2+2^2}=\sqrt{13}\)

\(AC=\left|\overrightarrow{AC}\right|=\sqrt{4^2+\left(-3\right)^2}=5\)

\(BC=\left|\overrightarrow{BC}\right|=\sqrt{1^2+\left(-5\right)^2}=\sqrt{26}\)

\(\Rightarrow C_{ABC}=AB+AC+BC=5+\sqrt{13}+\sqrt{26}\)

c) Gọi M, N, P lần lượt là trung điểm BC, CA, AB.

\(\Rightarrow P=\left(\dfrac{x_A+x_B}{2};\dfrac{y_A+y_B}{2}\right)=\left(-\dfrac{3}{2};3\right)\)

\(N=\left(\dfrac{x_A+x_C}{2};\dfrac{y_A+y_C}{2}\right)=\left(-1;\dfrac{1}{2}\right)\)

\(M=\left(\dfrac{x_B+x_C}{2};\dfrac{y_B+y_C}{2}\right)=\left(\dfrac{1}{2};\dfrac{3}{2}\right)\)

 d) Gọi G là trọng tâm tam giác ABC thì \(G=\left(\dfrac{x_A+x_B+x_C}{3};\dfrac{y_A+y_B+y_C}{3}\right)=\left(-\dfrac{2}{3};\dfrac{5}{3}\right)\)

 e) Gọi \(D\left(x_D;y_D\right)\) là điểm thỏa mãn ycbt.

Để ABCD là hình bình hành thì \(\overrightarrow{AB}=\overrightarrow{DC}\)

\(\Leftrightarrow\left(3;2\right)=\left(1-x_D;-1-y_D\right)\)

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

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

\(\Rightarrow D\left(-2;-3\right)\) 

f) Bạn xem lại đề nhé.