Gọi $G(a;\ 1-2a) \in (d) \Rightarrow A(3a+1;\ -2-6a)$
Phương trình $(BC): x+y-2=0$
$BC= 5\sqrt 2$
$d(A;\ (BC))= \dfrac{|a +1-2a-2|}{\sqrt 2}=\dfrac{|-a-1|}{\sqrt 2}$
$S_{\Delta ABC}= \dfrac{1}{2}. d(A;\ (BC)) . BC = \dfrac{|a+1|}{2\sqrt 2} .5\sqrt 2 = \dfrac{5}{2}$
$\Rightarrow |a+1|=1 \Rightarrow a= 0;\ a=-2$
Có 2 điểm $A ....$