Since the triangles are similiar, Let:
[tex]k\cdot FM=LK[/tex]where:
k = Constant of proportionality:
[tex]\begin{gathered} 8\cdot k=12 \\ solve_{\text{ }}for_{\text{ }}k\colon \\ k=\frac{12}{8} \\ k=\frac{3}{2} \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} k\cdot FJ=LA \\ so\colon \\ \frac{3}{2}\cdot6=LA \\ LA=9 \end{gathered}[/tex]