As indicated in the question, parallel lines have the same slope.
This means the slope of the fourth parallel line is also 2. A point on that line has been identified as;
[tex](3,15)[/tex]Using the point-lope form which is;
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{Where,} \\ x_1=3,y_1=15 \\ \text{The equation becomes;} \\ y-15=2(x-3) \end{gathered}[/tex]Further simplified, this now becomes;
[tex]\begin{gathered} y-15=2(x-3) \\ y-15=2x-6 \\ \text{Add 15 to both sides;} \\ y=2x+9 \end{gathered}[/tex]ANSWER:
It would be better to use the point-slope form to derive the equation of the 4th line because we already have the slope and one point on the equation.