a,
Gọi đường thẳng cần tìm là $d$
$d//\Delta\to d: x+2y+c=0$ ($c\ne -3$)
Ta có: $d(A, d)=2$
$\to \dfrac{| 1+2.2+c|}{\sqrt{1+2^2}}=2$
$\to |c+5|=2\sqrt5$
$\to c=-5\pm2\sqrt5$ (TM)
Vậy $d: x+2y-5\pm 2\sqrt5$
b,
$\Delta: x=-2y+3$
$M\in \Delta\to M(-2t+3; t)$
$\to AM=\sqrt{(-2t+3-1)^2+(t-2)^2}=\sqrt{(2t-2)^2+(t-2)^2}=\sqrt{5t^2-12t+8}$
$AM=3\to AM^2=9$
$\to 5t^2-12t+8=9$
$\to t=\dfrac{6\pm\sqrt{41}}{5}$
Vậy $M\Big(\dfrac{6+\sqrt{41}}{5}; \dfrac{3-2\sqrt{41}}{5}\Big)$ hoặc $\Big(\dfrac{6-\sqrt{41}}{5};\dfrac{3+2\sqrt{41}}{5}\Big)$
c,
$AM\min\to AM^2\min$
Parabol $y(t)=5t^2-12t+8$ đạt $\min$ tại điểm $\Big(\dfrac{12}{2.5};\dfrac{4.5.8-12^2}{4.5}\Big)=\Big(\dfrac{6}{5};\dfrac{4}{5}\Big)$
Vậy $AM\min=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt5}$ khi $t=\dfrac{6}{5}$
$\to M\Big(\dfrac{3}{5};\dfrac{6}{5}\Big)$
