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]

We have:

[tex]4x-5y=2\ \ \ \ |-4x\\\\-5y=-4x+2\ \ \ \ |:(-5)\\\\y=\dfrac{4}{5}x-\dfrac{2}{5}\to m_1=\dfrac{4}{5}\\\\l:y=m_2x+b\\\\k\ \prep\ l\Rightarrow\dfrac{4}{5}m_2=-1\ \ \ \ |\cdot\dfrac{5}{4}\\\\m_2=-\dfrac{5}{4}[/tex]

[tex]l:y=-\dfrac{5}{4}x+b[/tex]

The line l is passing through point (0, 0). Substitute the coordinates of the point to the equation of a line l:

[tex]0=-\dfrac{5}{4}\cdot0+b\to b=0[/tex]

[tex]y=-\dfrac{5}{4}x\ \ \ \ |\cdot4\\\\4y=-5x\ \ \ \ |+5x\\\\5x+4y=0[/tex]

Answer: 5x + 4y = 0

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