Given;
[tex]3y-x=-12[/tex]Recall that the equation of a straight line is given as;
[tex]\begin{gathered} y=mx+c\text{ } \\ \text{Where m= slope of the line} \\ 3y=x-12 \\ y=\frac{1}{3}x-4 \\ \text{Thus, m=}\frac{1}{3} \end{gathered}[/tex]The slope of two parallel lines is equal.
Thus, the slope of the second line is also 1/3;
Then, the equation is;
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{Where x}_1=18_{} \\ \text{and y}_1=2 \\ y-2=\frac{1}{3}(x-18) \\ 3(y-2)=x-18 \\ 3y-6=x-18 \\ 3y-x=6-18 \\ 3y-x=-12 \end{gathered}[/tex]The equation of the line is also 3y - x = -12
