3d) Gọi phương trình đường thẳng $d$ là $y=ax+b$
$M\in d$$\Rightarrow 2=a+b\Rightarrow b=2-a$
$d\cap Ox:y=0\Rightarrow 0=ax+b\Rightarrow x=\dfrac{-b}{a}\Rightarrow P(\dfrac{-b}{a};0)$
$d\cap Oy:x=0\Rightarrow y=b\Rightarrow Q(0;b)$
$S_{\Delta OPQ}=\dfrac{1}{2}|\dfrac{-b}{a}||b|=\dfrac{1}{2}|\dfrac{b^2}{a}|$
$=\dfrac{1}{2}|\dfrac{4-4a+a^2}{a}|=|\dfrac{4}{a}-4+a|$
Do $\dfrac{4}{a}+a\ge2\sqrt{\dfrac{4}{a}a}=4$
$\Rightarrow S\ge\dfrac{1}{2}|4-4|$
$S_{min}=0$ dấu bằng xảy ra $\Leftrightarrow \dfrac{4}{a}=a\Rightarrow a=2,b=0$
và $a=-2,b=4$
