Vértices do triângulo :
A = (0,0) ; B = (-1,4) ; C = (5,2)
Ponto médio de BC :
[tex]\displaystyle \text{M(BC)}=(\ \frac{\text x_\text B+\text x_\text C}{2},\frac{\text Y_\text B+\text Y_\text C}{2}\ ) \\\\\\ \text{M(BC)}=(\ \frac{-1+5}{2},\frac{4+2}{2} \ ) \\\\\\ \text{M(BC)} = (2,3)[/tex]
Equação da reta que por A(0,0) e por M(BC) = (2,3) :
[tex]\displaystyle \text y-\text y_\text o =\text m(\text x-\text x_\text o) \\\\ \text m = \frac{\text y_2-\text y_1}{\text x_2-\text x_1} \to \text m =\frac{3-0}{2-0} \to \boxed{\text m =\frac{3}{2}} \\\\\\ \text y - \text 0 = \frac{3}{2}(\text x-0) \\\\ 2\text y = 3\text x \\\\\\ \underline{\text{Portanto a equa{\c c}{\~a}o da reta {\'e}}}: \\\\\\ \huge\boxed{3\text x-2\text y = 0 }\checkmark[/tex]
