jmc

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 $P=(x,y)=\boxed{(8,-2)}$.