[tex]\begin{gathered} \text{If we have a line} \\ y=m_1x+b, \\ \text{ Then the slope of a perpendicular line to the first must satisfy} \\ m_1m_2=-1 \\ \text{ That means that} \\ m_2=\frac{-1}{m_1} \end{gathered}[/tex][tex]\begin{gathered} \text{ In our case, the line ML has slope} \\ m_1=\frac{-3-6}{-2-(-5)}=\frac{-9}{3}=-3 \\ \text{ So, the slope of the perpendicular line to ML has slope} \\ m_2=\frac{-1}{m_1}=\frac{-1}{-3}=\frac{1}{3} \end{gathered}[/tex][tex]\begin{gathered} \text{ Then the line perpendicular to segment ML has the equation} \\ y=\frac{1}{3}x+b \\ \text{ And we have that the point (4,5) must be in the line, so} \\ 5=\frac{1}{3}(4)\text{ + b} \\ \text{ So } \\ b=5-\frac{4}{3} \\ b=\frac{15-4}{3}=\frac{11}{3} \\ \\ \text{and the equation of the line perpendicular do ML and that contains N is} \\ \\ y=\frac{1}{3}x+\frac{11}{3} \end{gathered}[/tex]
