Respuesta :

Neither. They can’t be parallel because both lines have different slopes. They aren’t perpendicular either because their slopes don’t multiply to equal -1.
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[tex]k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2[/tex]

[tex]k:5x+3y=7\ \ \ |-5x\\\\3y=-5x+7\ \ \ \ |:3\\\\y=-\dfrac{5}{3}x+\dfrac{7}{3}\to m_1=-\dfrac{5}{3}\\\\l:3x+5y=4\ \ \ \ |-3x\\\\5y=-3x+4\ \ \ \ |:5\\\\y=-\dfrac{3}{5}+\dfrac{4}{5}\to m_2=-\dfrac{3}{5}[/tex]

[tex]m_1\neq m_2\to\text{ not parallel}\\\\m_1\cdot m_2=\left(-\dfrac{5}{3}\right)\cdot\left(-\dfrac{3}{5}\right)=1\neq-1\to\text{ no perpendicular}[/tex]

Answer: NEITHER