hello
to solve this question, let's write the standard equation of a straight line
![\begin{gathered} y=mx+c \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5961lnvsggjgy5g8msxuydrg1harpxttib.png)
![y=(5)/(8)x+6](https://img.qammunity.org/2023/formulas/mathematics/college/cp1dnfqmezo6pgmwsoykwgtaxvgwwoevmc.png)
since, this equation passes the given gpoint and it's parallel to it, we will have to modify our equation
![y-y_1=m(x-x_1)+c](https://img.qammunity.org/2023/formulas/mathematics/college/o3tmqximl5sez72t422p0k1epx0v60qr8a.png)
the given points are (0, 0)
let's insert it and solve
![\begin{gathered} y-0=(5)/(8)(x-0)+6 \\ y=(5)/(8)x+6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zc62liew7b7seo1se35p3nentzla52mhhn.png)
from the calculation above, the equation of the line is y = 5/8x + 6