hello
to solve this question, let's write the standard equation of a straight line
[tex]\begin{gathered} y=mx+c \\ \end{gathered}[/tex][tex]y=\frac{5}{8}x+6[/tex]since, this equation passes the given gpoint and it's parallel to it, we will have to modify our equation
[tex]y-y_1=m(x-x_1)+c[/tex]the given points are (0, 0)
let's insert it and solve
[tex]\begin{gathered} y-0=\frac{5}{8}(x-0)+6 \\ y=\frac{5}{8}x+6 \end{gathered}[/tex]from the calculation above, the equation of the line is y = 5/8x + 6
