Giải thích các bước giải:
Ta có :
$MN//AC\to \dfrac{S_{BMN}}{S_{BAC}}=(\dfrac{BM}{BA})^2=\dfrac{1}{1+8}=\dfrac19$
$\to \dfrac{BM}{BA}=\dfrac13\to BM=\dfrac13BA$
$\to \vec{BM}=\dfrac13\vec{BA}=\dfrac13(-2,6)=(-\dfrac23,2)$
$\to M(-\dfrac23,-1)$
Ta có : $AB//DC\to \dfrac{EA}{EC}=\dfrac{AB}{CD}=2$
$\to \vec{EC}=\dfrac12\vec{AE}=\dfrac12(4,-\dfrac{12}{5})=(2,-\dfrac{6}{5})$
$\to C(4, -\dfrac{3}{5})$
$\to \vec{BN}=\dfrac13\vec{BC}=\dfrac13(4,\dfrac{12}{5})=(\dfrac43,\dfrac45)$
$\to N(\dfrac43, -\dfrac{11}{5})$