The given line k having equation is:
[tex]y=\frac{2}{3}x-1[/tex]The slope is 2/3.
And the line I is perpendicular to line k, so the slope of line I is:
[tex]\begin{gathered} m_I=-\frac{1}{m_k} \\ m_I=-\frac{1}{\frac{2}{3}} \\ m_I=-\frac{3}{2} \end{gathered}[/tex]So slope of line I is -3/2
An equation for l in point-slope form is:
As it is passing through: (-1, 5/2)
[tex]\begin{gathered} (y-\frac{5}{2})=-\frac{3}{2}(x-(-1)) \\ (y-\frac{5}{2})=-\frac{3}{2}(x+1) \end{gathered}[/tex]An equation for l in slope-intercept form is:
[tex]\begin{gathered} (y-\frac{5}{2})=-\frac{3}{2}(x+1) \\ y-\frac{5}{2}=-\frac{3}{2}x-\frac{3}{2} \\ y=-\frac{3}{2}x+\frac{5}{2}-\frac{3}{2} \\ y=-\frac{3}{2}x+\frac{(5-3)}{2} \\ y=-\frac{3}{2}x+\frac{2}{2} \\ y=-\frac{3}{2}x+1 \end{gathered}[/tex]