First we need to find the slope of line m.
The slope of a line is:
[tex]\begin{gathered} P_1=(x_1,y_1),P_2=(x_2,y_2)_{} \\ m=\frac{(y_2-y_1)}{(x_2-x_1)} \end{gathered}[/tex]For line m we know P1=(-6,-5) and P2=(-4, 1), so the slope of m is:
[tex]\begin{gathered} m=\frac{(1-(-6))}{(-4-(-5))} \\ m=\frac{1+6}{-4+5}=\frac{7}{1}=7 \end{gathered}[/tex]Now, if two lines are perpendicular the slopes satisfy the following equation:
[tex]m_2=-\frac{1}{m_1}[/tex]The slope of line l is:
[tex]\begin{gathered} We\text{ call m2 the slope of line l and m1 the slope of line m:} \\ m_2=-\frac{1}{m_1}=-\frac{1}{7} \end{gathered}[/tex]Finally, the line l pass througth point (6, 4), so the equation is:
[tex]\begin{gathered} y=m_2x+b \\ 4=-\frac{1}{7}\cdot6+b_{} \\ b=4+\frac{6}{7}=\frac{4\cdot7+6}{7} \\ b=\frac{28+6}{7} \\ b=\frac{34}{7} \end{gathered}[/tex]So, teh equation of line l is:
[tex]y=-\frac{1}{7}x+\frac{34}{7}[/tex]
