solution: If $P$ is equidistant from $A$ and $O$, it must lie on the perpendicular bisector of $AO$. Since $A$ has coordinates $(10,-10)$ and $O$ has coordinates $(0,0)$, $AO$ has slope $\frac{-10-0}{10-0}=-1$.  The perpendicular bisector of $AO$ must have slope $-\frac{1}{-1}=1$, and must also pass through the midpoint of $AO$, which is $(5,-5)$. Therefore, the perpendicular bisector has equation $y-(-5)=x-5$ or $y=x-10$.
  
  $P$ is the point of intersection of the lines $y=x-10$ and the line $y=-x+6$.  Setting these equations equal and solving for $x$ yields $-x+6=x-10 \Rightarrow x=8$.  It follows that $y=-8+6=-2$ and $P=(x,y)=\boxed{(8,-2)}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Determine the coordinates of the point $P$ on the line $y=-x+6$ such that $P$ is equidistant from the points $A(10,-10)$ and $O(0,0)$ (that is, so that $PA=PO$).  Express your answer as an ordered pair $(a,b)$.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: (8,-2)
extra_info.question: Determine the coordinates of the point $P$ on the line $y=-x+6$ such that $P$ is equidistant from the points $A(10,-10)$ and $O(0,0)$ (that is, so that $PA=PO$).  Express your answer as an ordered pair $(a,b)$.
extra_info.split: test