If the equation y=16(x+12)y=\dfrac{1}{6} (x+12)y=61​(x+12) were graphed in the xy−planexy-planexy−plane, which of the following statements would be true of the graphed line? It would be perpendicular to the graph of y=16x+3y=\dfrac{1}{6}x+3y=61​x+3. It would be parallel to the graph of 12y=2x+312y=2x+312y=2x+3. It would have the same slope as the graph of x+6y=18x+6y=18x+6y=18. It would have the same yyy‑intercept as the graph of y=16x+12y=\dfrac{1}{6}x+12y=61​x+12. I would be guessing.

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