Question

Trong mặt phẳng tọa độ Oxy , cho A(2;1) , B(-1; -2), C(-3;2) . Tìm tọa độ điểm M để thỏa vecto 3AM +vecto AB = vecto 0

Answer 1

\(A\left(2;1\right);B\left(-1;-2\right);M\left(x;y\right)\)

\(\overrightarrow{AM}=\left(x-2;y-1\right);\overrightarrow{AB}=\left(-3;-1\right)\)

\(3\cdot\overrightarrow{AM}+\overrightarrow{AB}=\overrightarrow{0}\)

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

=>\(\left\{{}\begin{matrix}3x-6-3=0\\3y-4=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=\dfrac{4}{3}\end{matrix}\right.\)

vậy: \(M\left(3;\dfrac{4}{3}\right)\)

Answer 2

Gọi tọa độ M có dạng \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-3;-3\right)\\\overrightarrow{AM}=\left(x-2;y-1\right)\end{matrix}\right.\)

\(\Rightarrow3\overrightarrow{AM}+\overrightarrow{AB}=\left(3x-9;3y-6\right)\)

\(3\overrightarrow{AM}+\overrightarrow{AB}=\overrightarrow{0}\Rightarrow\left\{{}\begin{matrix}3x-9=0\\3y-6=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow M\left(3;2\right)\)