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 $(a,b)$.
I need to find a point $P$ on the line $y=-x+6$ that is the same distance from $A$ and $O$. One way to approach this problem is to use the distance formula. If I let $P=(x,y)$, then I can write $PA=PO$ as:

\begin{align*}
\sqrt{(x-10)^2+(y+10)^2}&=\sqrt{x^2+y^2}\\
\end{align*} Then I can square both sides to get rid of the square roots and simplify:

\begin{align*}
(x-10)^2+(y+10)^2&=x^2+y^2\\
x^2-20x+100+y^2+20y+100&=x^2+y^2\\
-20x+20y&=-200\\
x-y&=10\\
\end{align*} So I have a second equation that relates $x$ and $y$. I can use it along with the original equation of the line $y=-x+6$ to solve for $x$ and $y$. I can eliminate $y$ by adding the two equations:

\begin{align*}
x-y&=10\\
y&=-x+6\\
\hline
2x&=16\\
x&=8\\
\end{align*} Then I can plug in $x=8$ into either equation to find $y$:

\begin{align*}
y&=-x+6\\
y&=-8+6\\
y&=-2\\
\end{align*} So the point $P$ has the coordinates $(8,-2)$. 