Answer:
[tex]\large\boxed{slope=-\dfrac{5}{3}}[/tex]
Step-by-step explanation:
[tex]\text{Let}\ k:y=m_1x+b_1,\ l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\==========================\\\\\text{We have}\ k:y=\dfrac{3}{5}x+1\to m_1=\dfrac{3}{5}.\\\\\text{Therefore}\ m_2=-\dfrac{1}{\frac{3}{5}}=-\dfrac{5}{3}[/tex]
