Do \(M\in\Delta\Rightarrow M\left(a;3a-5\right)\)
\(\overrightarrow{AB}=\left(-3;4\right)\Rightarrow AB=5\)
Phương trình AB: \(4\left(x-1\right)+3\left(y-0\right)=0\Leftrightarrow4x+3y-4=0\)
\(\overrightarrow{CD}=\left(4;1\right)\Rightarrow CD=\sqrt{17}\)
Phương trình CD: \(1\left(x+1\right)-4\left(y-4\right)=0\Leftrightarrow x-4y+17=0\)
\(S_{MAB}=S_{MCD}\Leftrightarrow d\left(M;AB\right).AB=d\left(M;CD\right).CD\)
\(\Leftrightarrow5.d\left(M;AB\right)=\sqrt{17}.d\left(M;CD\right)\)
\(\Leftrightarrow\frac{5.\left|4a+3\left(3a-5\right)-4\right|}{\sqrt{4^2+3^2}}=\frac{\sqrt{17}.\left|a-4\left(3a-5\right)+17\right|}{\sqrt{1+4^2}}\)
\(\Leftrightarrow\left|13a-19\right|=\left|-11a+27\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}13a-19=-11a+27\\13a-19=11a-27\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{23}{12}\\a=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}M\left(\frac{23}{12};\frac{3}{4}\right)\\M\left(-4;-17\right)\end{matrix}\right.\)
