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M(x;y); A(1;3); B(4;0); C(2;-5)
\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;0-y\right)=\left(4-x;-y\right)\) ; \(\overrightarrow{MC}=\left(2-x;-5-y\right)\)
\(\overrightarrow{MA}+\overrightarrow{MB}-3\cdot\overrightarrow{MC}=\overrightarrow{0}\)
=>\(\begin{cases}1-x+4-x-3\left(2-x\right)=0\\ 3-y-y-3\left(-5-y\right)=0\end{cases}\Rightarrow\begin{cases}-2x+5-6+3x=0\\ 3-2y+15+3y=0\end{cases}\)
=>\(\begin{cases}x-1=0\\ y+18=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=-18\end{cases}\)
=>M(1;-18)
Gọi tọa độ điểm \(M\) là \(M\left(x;y\right).\)
\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;-y\right);\overrightarrow{MC}=\left(2-x;-5-y\right).\)
Ta có: \(\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\overrightarrow{0}.\)
\(\left\{{}\begin{matrix}1-x+4-x-3\left(2-x\right)=0.\\3-y-y-3\left(-5-y\right)=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2x+5-6+3x=0.\\3-2y+15+3y=0.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0.\\y+18=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1.\\y=-18.\end{matrix}\right.\) \(\Rightarrow M\left(1;-18\right).\)
a: A(1;3); B(-2;5); C(-4;0)
\(\overrightarrow{AB}=\left(-2-1;5-3\right)=\left(-3;2\right)\)
\(\overrightarrow{AC}=\left(-4-1;0-3\right)=\left(-5;-3\right)\)
\(\overrightarrow{BC}=\left(-4+2;0-5\right)=\left(-2;-5\right)\)
\(\overrightarrow{CB}=\left(-2+4;5-0\right)=\left(2;5\right)\)
b: \(\overrightarrow{AB}\cdot\overrightarrow{CB}=-3\cdot2+2\cdot5=-6+10=4\)
\(\overrightarrow{AC}\cdot\overrightarrow{BC}=\left(-5\right)\cdot\left(-2\right)+\left(-3\right)\cdot\left(-5\right)=10+15=25\)
c: \(\overrightarrow{AB}=\left(-3;2\right)\)
=>\(AB=\sqrt{\left(-3\right)^2+2^2}=\sqrt{13}\)
\(\overrightarrow{BC}=\left(-2;-5\right)\)
=>\(BC=\sqrt{\left(-2\right)^2+\left(-5\right)^2}=\sqrt{4+25}=\sqrt{29}\)
e: \(\overrightarrow{AB}+2\cdot\overrightarrow{CB}\) =\(\left(-3+2\cdot2;2+2\cdot5\right)\)
=(-3+4;2+10)
=(1;12)
\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-2;-1\right)\\\overrightarrow{AC}=\left(-3;-2\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{AB}-\overrightarrow{AC}=\left(-2-\left(-3\right);-1-\left(-2\right)\right)=\left(1;1\right)\)
Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(1-x;3-y\right)\\\overrightarrow{MB}=\left(4-x;-y\right)\\\overrightarrow{MC}=\left(2-x;-5-y\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\left(x-1;y+18\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+18=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-18\end{matrix}\right.\)
\(\Rightarrow M\left(1;-18\right)\)