Respuesta :

gmany

[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:y=-\dfrac{6}{5}x+2\to m_1=-\dfrac{6}{5}\\\\l:y=\dfrac{6}{5}x+2\to m_2=\dfrac{6}{5}\\\\m_1\neq m_2\to\text{no parallel}\\\\m_1\cdot m_2=-\dfrac{6}{5}\cdot\dfrac{6}{5}\neq-1\to\text{ no perpendicular}[/tex]

Answer: NEITHER

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