Point
\[P'\] is the image of
\[P(6,0)\] under the translation
\[(x,y)\to(x-6,y-1)\].
What are the coordinates of
\[P'\]?
1 answer
Applying the translation, the x-coordinate of $P'$ is found by subtracting 6 from the x-coordinate of $P$, and the y-coordinate of $P'$ is found by subtracting 1 from the y-coordinate of $P$. In this case, that gives $P' = \boxed{(0,-1)}$.