ptts $d:\left\{ \begin{array}{l} x=t\\ y=3t-5 \end{array} \right.$
$M\in d\Rightarrow M(t;3t-5)$
Ta co:
$\overrightarrow{AB}=(-3;4)\Rightarrow AB=5$
$\overrightarrow{CD}=(4;1)\Rightarrow CD=\sqrt{17}$
ptdt $AB: 4(x-1)+3y=0$ hay $4x+3y-4=0$
ptdt$CD:(x+1)-4(y-4)=0$ hay $x-4y+17=0$
$d_{(M;AB)}=\frac{|4t+3(3t-5)-4|}{\sqrt{4^2+3^2}}=\frac{|13t-19|}{5}$
$d_{(M;CD)}=\frac{|t-4(3t-5)+17}{\sqrt{1+4^2}}=\frac{|27-11t|}{\sqrt{17}}$
$S_{(MAB)}=AB.d_{(M;AB)}/2=5.\frac{|13t-19|}{5}$
$S_{(MCD)}=CD.d_{(M:CD)}=\sqrt{17}.\frac{|27-11t|}{\sqrt{17}}$
theo bai ra ta co :
$S_{ABM}=S_{CDM}\Leftrightarrow |13t-19|=|27-11t|$
giai t ra nua la xong