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Answer:

[tex]\huge\boxed{y=\dfrac{4}{3}x-1}[/tex]

Step-by-step explanation:

Let

[tex]k:y=m_1x+b_1\\l:y=m_2x+b_2[/tex]

then

[tex]k\perp l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\k\ ||\ l\iff m_1=m_2[/tex]

==========================================

We have:

[tex]y=-\dfrac{3}{4}x-3\to m_1=-\dfrac{3}{4}[/tex]

Let

[tex]y=m_2x+b_2[/tex]

Therefore

[tex]m_2=-\dfrac{1}{-\frac{3}{4}}=\dfrac{4}{3}\Rightarrow y=\dfrac{4}{3}x+b_2[/tex]

Substitute

[tex](0,\ -1)\to x=0;\ y=-1[/tex]

[tex]-1=\dfrac{4}{3}\cdot0+b_2\\\\b_2=-1[/tex]

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