CHO tọa độ A(-1;0) B(4;5) A) XÁC ĐỊNH M THỎA MÃN :2 VECTO MA +3 VECTO MB =VECTO 0 CHỨNG TỎ M,A,B THẲNG HÀNG

$\begin{array}{l} A\left( { - 1;0} \right);\,\,B\left( {4;5} \right)\\ a)\,\,Goi\,\,M\left( {x;y} \right)\\ \Rightarrow \overrightarrow {MA} = \left( { - 1 - x; - y} \right);\,\,\overrightarrow {MB} = \left( {4 - x;5 - y} \right)\\ \,\,\,\,\,2\overrightarrow {MA} + 3\overrightarrow {MB} = \overrightarrow 0 \\ \Rightarrow 2\left( { - 1 - x; - y} \right) + 3\left( {4 - x;5 - y} \right) = \overrightarrow 0 \\ \Rightarrow \left( { - 2 - 2x; - 2y} \right) + \left( {12 - 3x;15 - 3y} \right) = \overrightarrow 0 \\ \Rightarrow \left( {10 - 5x;15 - 5y} \right) = \overrightarrow 0 \\ \Rightarrow \left\{ \begin{array}{l} 10 - 5x = 0\\ 15 - 5y = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 3 \end{array} \right. \Rightarrow M\left( {2;3} \right)\\ b)\,\,Ta\,\,co:\,\,\overrightarrow {MA} = \left( { - 3; - 3} \right);\,\,\overrightarrow {AB} = \left( {5;5} \right)\\ \Rightarrow \overrightarrow {AB} = - \frac{5}{3}\overrightarrow {MA} \\ \Rightarrow M,\,\,A,\,\,B\,\,thang\,\,hang. \end{array}$