If they are parallel:
![m1=m2](https://img.qammunity.org/2023/formulas/mathematics/college/nfdvp7mtz4q3lr2oldef2azfj8vxpbv57i.png)
Since the slope for y=-5x+8 is m1=-5:
![m2=-5](https://img.qammunity.org/2023/formulas/mathematics/college/ewmwq4no8r4p4gguiiem7hg7eej8mkxleu.png)
Let:
(x1,y1)=(0,2)
Using the point-slope equation:
![\begin{gathered} y-y1=m(x-x1) \\ y-2=-5(x-0) \\ y-2=-5x \\ y=-5x+2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5pi60dn6kir74pyibu31d7gso53x5n9nq3.png)
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If is perpendicular to the line equation y=1/2x-3:
![\begin{gathered} m1* m2=-1 \\ \text{Where:} \\ m1=(1)/(2) \\ (1)/(2)* m2=-1 \\ m2=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ww6rvil1znhyir585bp4yur9yjunimvcuh.png)
Let:
(x1,y1)=(-5,2)
Using the point-slope equation:
![\begin{gathered} y-y1=m(x-x1) \\ y-2=-2(x-(-5)) \\ y-2=-2(x+5) \\ y-2=-2x-10 \\ y=-2x-8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/keqn68pvlppen8m8ghjzvcczlgac9mhxta.png)