a) Gọi đường thẳng qua AB là $d:y=ax+b$

Ta có hệ:

$\left\{\begin{array}{l} a+b=-4\\ 2a+b=0\end{array} \right.<=>\left\{\begin{array}{l} a=4\\ b=-8\end{array} \right.\\ =>d:y=4x-8$

Xét điểm $C(5;-1)$ thấy $C \notin d=>A,B,C$ không thẳng hàng
$=>ABC$ tạo thành tam giác

$b)\left\{\begin{array}{l}x_G=\dfrac{x_A+x_B+x_C}{3}\\ y_G=\dfrac{y_A+y_B+y_C}{3}\end{array} \right.\\ \left\{\begin{array}{l}x_G=\dfrac{1+2+5}{3}\\ y_G=\dfrac{-4+0-1}{3}\end{array} \right.\\ <=>\left\{\begin{array}{l}x_G=\dfrac{8}{3}\\ y_G=\dfrac{-5}{3}\end{array} \right.\\ =>G:\left(\dfrac{8}{3};\dfrac{-5}{3}\right)$

c)Gọi $M,N,P$ lần lượt là trung điểm $AB,AC,BC$

$\left\{\begin{array}{l}x_M=\dfrac{x_A+x_B}{2}\\ y_M=\dfrac{y_A+y_B}{2}\end{array} \right.\\
<=>\left\{\begin{array}{l}x_M=\dfrac{3}{2}\\ y_M=-2\end{array} \right.\\
=>M\left(\dfrac{3}{2};-2\right)$

Tương tự $N\left(3;\dfrac{-5}{2}\right); P\left(\dfrac{7}{2};\dfrac{-1}{2}\right)$

$d) \left\{\begin{array}{l}x_A=\dfrac{x_D+x_C}{2}\\ y_A=\dfrac{y_D+y_C}{2}\end{array} \right.\\ \left\{\begin{array}{l}x_D=2x_A-x_C=-3\\ y_D=2y_A-y_C=-7\end{array} \right.\\=>DM\left(-3;-7\right)$