[tex]\it (BC):\ \dfrac{y-y_B}{y_C-y_B}=\dfrac{x-x_B}{x_C-x_B} \Rightarrow \dfrac{y+3}{-5+3}=\dfrac{x+1}{5+1} \Rightarrow \dfrac{y+3}{-2}=\dfrac{x+1}{6} \Rightarrow\\ \\ \\ \Rightarrow6y+18=-2x-2|_{-18} \Rightarrow 6y=-2x-20|_{:2} \Rightarrow 3x=-x-10|_{:3} \Rightarrow\\ \\ \\ x=-\dfrac{1}{3}x-\dfrac{10}{3} \ (ecua\c{\it t}ia\ dreptei BC)\\ \\ \\ m=-\dfrac{1}{3}\ \ (panta\ dreptei\ BC)[/tex]
[tex]\it \left.\begin{aligned} \it x_M=\dfrac{x_A+x_B}{2}=\dfrac{5-1}{2}=2\\ \\ \\ \it y_M=\dfrac{y_A+y_B}{2}=\dfrac{7-3}{2}=2 \end{aligned}\right\} \Rightarrow M(2,\ 2)\\ \\ \\ \left.\begin{aligned} \it x_P=\dfrac{x_A+x_C}{2}=\dfrac{5+5}{2}=5\\ \\ \\ \it y_P=\dfrac{y_A+y_C}{2}=\dfrac{7-5}{2}=1 \end{aligned}\right\} \Rightarrow P(5,\ 1)[/tex]
[tex]\it (MP): \dfrac{y-y_p}{y_M-y_P}=\dfrac{x-x_P}{x_M-x_P} \Rightarrow \dfrac{y-1}{2-1}=\dfrac{x-5}{2-5} \Rightarrow \dfrac{y-1}{1}=\dfrac{x-5}{-3} \Rightarrow\\ \\ \\ -3y+3=x-5|_{-3} \Rightarrow -3y=x-8|_{:(-3)} \Rightarrow y=-\dfrac{1}{3}x+\dfrac{8}{3}\ (ecua\c{\it t}ia\ lui\ MP)\\ \\ \\ m=-\dfrac{1}{3}\ (panta\ dreptei\ MP)[/tex]
Deoarece dreptele BC și MP au aceeași pantă ⇒ MP || BC.
