Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Gọi \(M\left(a;b\right)\)
\(\Rightarrow\overrightarrow{MB}=\left(2-a;3-b\right)\Rightarrow2\overrightarrow{MB}=\left(4-2a;6-2b\right)\)
\(\overrightarrow{MC}=\left(-1-a;-2-b\right)\Rightarrow3\overrightarrow{MC}=\left(-3-3a;-6-3b\right)\)
\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(1-5a;-5b\right)=\overrightarrow{0}\)
\(\Rightarrow\left\{{}\begin{matrix}1-5a=0\\-5b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{5}\\b=0\end{matrix}\right.\) \(\Rightarrow M\left(\frac{1}{5};0\right)\)
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: \(\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\overrightarrow{0}.\)
\(\left\{{}\begin{matrix}1-x+4-x-3\left(2-x\right)=0.\\3-y-y-3\left(-5-y\right)=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2x+5-6+3x=0.\\3-2y+15+3y=0.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0.\\y+18=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1.\\y=-18.\end{matrix}\right.\) \(\Rightarrow M\left(1;-18\right).\)
\(\text{Đặt }M\left(x;y\right)\\ \overrightarrow{MB}\left(-2-x,2-y\right);\overrightarrow{MC}\left(-x,1-y\right)\\ \left|\overrightarrow{MB}\right|=\left|2\overrightarrow{MC}\right|\Leftrightarrow\sqrt{\left(-2-x\right)^2+\left(2-y\right)^2}=2\sqrt{\left(-x\right)^2+\left(1-y\right)^2}\\ \Leftrightarrow x^2+4x+4+y^2-4y+4=2x^2+2y^2-4y+2\\ \Leftrightarrow x^2+y^2-4y-6=0\\ \text{Mà }M\in Ox\Leftrightarrow y=0\Leftrightarrow x^2-6=0\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}M\left(\sqrt{6};0\right)\\M\left(-\sqrt{6};0\right)\end{matrix}\right.\)
Bạn ghi lại đề, \(2\overrightarrow{MB}+3????\)