a) Gọi \(K\left( {x;y} \right)\).
Ta có: \(\overrightarrow {AK} = \left( {x + 2;y - 1} \right),\overrightarrow {BC} = \left( {2;4} \right),\overrightarrow {BK} = \left( {x - 1;y + 2} \right)\)
\(\begin{array}{l}\overrightarrow {AK} - 2\overrightarrow {BC} + 3\overrightarrow {BK} = 0 \Leftrightarrow \left\{ \begin{array}{l}x + 2 - 2.2 + 3.\left( {x - 1} \right) = 0\\y - 1 - 2.4 + 3.\left( {y + 2} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4x - 5 = 0\\4y - 3 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{5}{4}\\y = \frac{3}{4}\end{array} \right. \Rightarrow K\left( {\frac{5}{4};\frac{3}{4}} \right)\end{array}\)
b) \(I\) là trung điểm \(AC\) nên \(\left\{ \begin{array}{l}{x_I} = \frac{{ - 2 + 3}}{2} = \frac{1}{2}\\{y_I} = \frac{{1 + 2}}{2} = \frac{3}{2}\end{array} \right.\) \( \Rightarrow I\left( {\frac{1}{2};\frac{3}{2}} \right)\)
Gọi \(M\left( {{x_M};{y_M}} \right)\).
\(I\) là trọng tâm của tam giác \(MBC\) \( \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{2} = \frac{{{x_M} + 1 + 3}}{3}\\\frac{3}{2} = \frac{{{y_M} + \left( { - 2} \right) + 2}}{3}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3 = 2{x_M} + 8\\9 = 2{y_M}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_M} = - \frac{5}{2}\\{y_M} = \frac{9}{2}\end{array} \right. \Rightarrow M\left( { - \frac{5}{2};\frac{9}{2}} \right)\)
