b)
Fie M(x; y), mijlocul laturii BC. unde B(-1; 1), C(4; 3).
[tex]\bf x=\dfrac{-1+4}{2}=\dfrac{3}{2}=1,5\\ \\ \\ y=\dfrac{1+3}{2}=2\\ \\ \\ Deci,\ \ avem\ \ M(1,5;\ 2),\ \ A(3;\ \ -2)\\ \\ \\ AM:\ \ \dfrac{x-x_A}{x_M-x_A}=\dfrac{y-y_A}{y_M-y_A}[/tex]
[tex]\it \dfrac{x-3}{1,5-3}=\dfrac{y-(-2)}{2-(-2)}\ \Rightarrow \dfrac{3-x}{1,5}=\dfrac{y+2}{4} \Rightarrow 1,5y+3=12-4x \Rightarrow\\ \\ \\ \Rightarrow AM:\ \ 4x+1,5y-9=0|_{\cdot2} \Rightarrow 8x+3y-18=0[/tex]