a. \(\left\{{}\begin{matrix}\overrightarrow{BA}=\left(3;2\right)\\\overrightarrow{BC}=\left(6;-\dfrac{5}{2}\right)\end{matrix}\right.\)
Do \(\dfrac{3}{6}\ne\dfrac{2}{-\dfrac{5}{2}}\Rightarrow\overrightarrow{BA}\) và \(\overrightarrow{BC}\) không cùng phương
\(\Rightarrow A;B;C\) lập thành 1 tam giác
Gọi G là trọng tâm tam giác \(\Rightarrow\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=4\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{23}{6}\end{matrix}\right.\)
\(\Rightarrow G\left(4;\dfrac{23}{6}\right)\)
b.
Gọi \(E\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AC}=\left(3;-\dfrac{9}{2}\right)\\\overrightarrow{BE}=\left(x-1;y-4\right)\end{matrix}\right.\)
ACEB là hbh khi: \(\overrightarrow{AC}=\overrightarrow{BE}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=3\\y-4=-\dfrac{9}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow E\left(4;-\dfrac{1}{2}\right)\)
c.
Gọi \(K\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{KA}=\left(4-x;6-y\right)\\\overrightarrow{KC}=\left(7-x;\dfrac{3}{2}-y\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{KA}-2\overrightarrow{BA}=\left(-2-x;2-y\right)\\5\overrightarrow{KC}+3\overrightarrow{BC}=\left(53-5x;-5y\right)\end{matrix}\right.\)
\(\overrightarrow{KA}-2\overrightarrow{BA}=5\overrightarrow{KC}+3\overrightarrow{BC}\Leftrightarrow\left\{{}\begin{matrix}-2-x=53-5x\\2-y=-5y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{55}{4}\\y=-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow K\left(\dfrac{55}{4};-\dfrac{1}{2}\right)\)
