SOLUTION
[tex]\begin{gathered} For\text{ line FG, we have points:} \\ F(1,1)\text{ and G\lparen3,3\rparen} \\ \end{gathered}[/tex][tex]\begin{gathered} For\text{ line F'G', we have points:} \\ F^{\prime}(4,4)\text{ and G'\lparen12,12\rparen} \end{gathered}[/tex]That is the graph.
CONCLUSION
THE TWO LINES WILL HAVE THE SAME SLOPE.
Therefore, they could be considered as parallel since two parallel lines have the same slope.