[tex]\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\-------------------------\\\\\text{We have}\ 5x+3y=8\qquad\text{subtract 5x from both sides}\\\\3y=-5x+8\qquad\text{divide both sides by 3}\\\\y=-\dfrac{5}{3}x+\dfrac{8}{3}\to m_1=-\dfrac{5}{3}\\\\\text{Therefore}\ m_2=-\dfrac{1}{-\frac{5}{3}}=\dfrac{3}{5}\\\\Answer:\ \boxed{b.\ \dfrac{3}{5}}[/tex]
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[tex]\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{We have}\ 3x-5y=10\qquad\text{subtract 3x from both sides}\\\\-5y=-3x+10\qquad\text{divide both sides by (-5)}\\\\y=\dfrac{3}{5}x-2\to m_1=\dfrac{3}{5}\\\\\text{Therefore we have}\ m_2=\dfrac{3}{5}.\\\\Answer:\ \boxed{D.\ y=\dfrac{3}{5}x}[/tex]
