<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 $(a,b)$.
</Question>

<think>
<Plan_1>
Okay, so I need to find a point P on the line y = -x + 6 such that the distance from P to point A(10, -10) is equal to the distance from P to the origin O(0, 0). Hmm, let me think about how to approach this.

First, since P lies on the line y = -x + 6, I can represent P as (x, -x + 6). That way, no matter what x is, the y-coordinate will satisfy the equation of the line.

Now, the key part is setting up the distances PA and PO and making them equal. The distance formula between two points (x1, y1) and (x2, y2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. But since we want PA = PO, maybe I can square both sides to eliminate the square roots and make the equation easier to handle. Let me try that.

Let’s denote point P as (x, -x + 6). Then, the distance PA squared would be (x - 10)^2 + [(-x + 6) - (-10)]^2. Similarly, the distance PO squared would be (x - 0)^2 + [(-x + 6) - 0]^2.

<prompt_1.1> Set up the equations for the squared distances PA² and PO² using the distance formula. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me compute each part step by step.

First, PA squared:

The x-coordinate difference is x - 10, and the y-coordinate difference is (-x + 6) - (-10) = -x + 6 + 10 = -x + 16. So PA² = (x - 10)^2 + (-x + 16)^2.

Now, PO squared:

The x-coordinate difference is x - 0 = x, and the y-coordinate difference is (-x + 6) - 0 = -x + 6. So PO² = x^2 + (-x + 6)^2.

Since PA = PO, their squares should be equal. Therefore:

(x - 10)^2 + (-x + 16)^2 = x^2 + (-x + 6)^2.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me expand both sides.

Starting with the left side:

<prompt_2.1> Expand and simplify both sides of the equation (x - 10)^2 + (-x + 16)^2 = x^2 + (-x + 6)^2. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
(x - 10)^2 = x^2 - 20x + 100

(-x + 16)^2 = x^2 - 32x + 256

So adding these together: x^2 - 20x + 100 + x^2 - 32x + 256 = 2x^2 - 52x + 356.

Now the right side:

x^2 + (-x + 6)^2

We already expanded (-x + 6)^2 earlier, which is x^2 - 12x + 36. So adding x^2 gives: x^2 + x^2 - 12x + 36 = 2x^2 - 12x + 36.

So putting it all together:

Left side: 2x^2 - 52x + 356

Right side: 2x^2 - 12x + 36

Subtract the right side from the left side to set the equation to zero:

(2x^2 - 52x + 356) - (2x^2 - 12x + 36) = 0

Simplify term by term:

2x^2 - 2x^2 = 0

-52x - (-12x) = -52x + 12x = -40x

356 - 36 = 320

So the equation becomes: -40x + 320 = 0

Solving for x:

-40x = -320

Divide both sides by -40:

x = (-320)/(-40) = 8

So x = 8.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, since the point P is on the line y = -x + 6, substituting x = 8 into that equation gives y = -8 + 6 = -2.

Wait, so the coordinates of P would be (8, -2).

Let me check if this is correct.

<prompt_3.1> Verify the solution by checking if the distances PA and PO are equal. </prompt_3.1>

But wait, let me double-check my calculations because sometimes when dealing with quadratics or absolute values, there might be another solution. Let me verify each step.

First, the setup: PA = PO, so squared distances equal.

<prompt_3.2> Re-verify the solution by re-evaluating the setup and calculations for distances PA² and PO². </prompt_3.2>

But let me visualize this. The line y = -x +6 is a straight line with slope -1 crossing the y-axis at 6.

<prompt_3.3> Visualize the problem to ensure the solution makes sense and consider the possibility of another intersection point. </prompt_3.3>

Alternatively, maybe there's another way to approach this problem. For instance, the set of all points equidistant from A and O is the perpendicular bisector of the segment AO.

<prompt_3.4> Use the concept of the perpendicular bisector to find the coordinates of point P. </prompt_3.4>

Therefore, the coordinates of point P are (8, -2).
**Final Answer**
The coordinates of point \( P \) are \boxed{(8, -2)}.
</Plan_3>

<Execution_3>
<execution_3.1>
First, let's compute PA. Point A is (10, -10), so the distance from (8, -2) to (10, -10):

PA = sqrt[(10 - 8)^2 + (-10 - (-2))^2] = sqrt[2^2 + (-8)^2] = sqrt[4 + 64] = sqrt[68].

Now, PO is the distance from (8, -2) to (0, 0):

PO = sqrt[(8 - 0)^2 + (-2 - 0)^2] = sqrt[64 + 4] = sqrt[68].
</execution_3.1>

<execution_3.2>
That seems right.

Then expanding PA²:

(x -10)^2 + (-x +16)^2. Let me check that again.

(x -10)^2 = x² -20x +100. Correct.

(-x +16)^2: (-x +16) is the same as (16 -x), squaring that is (16)^2 - 2*16*x +x² = 256 -32x +x². So that's correct. So adding to (x -10)^2 gives x² -20x +100 +x² -32x +256 = 2x² -52x +356. Correct.

PO²: x² + (-x +6)^2. (-x +6)^2 is x² -12x +36. So adding x² gives 2x² -12x +36. Correct.

Subtracting the two gives 2x² -52x +356 -2x² +12x -36 = (-40x +320) =0. So x=8. Then y= -8+6= -2. Correct.

But let's check the distances. Let me compute PA and PO.

PA: distance from (8, -2) to (10, -10).

Difference in x: 10 -8 =2, difference in y: -10 -(-2)= -8. So PA squared is 2² + (-8)^2=4 +64=68. So PA is sqrt(68).

PO: distance from (8, -2) to (0,0).

Difference in x:8-0=8, difference in y: -2-0=-2. So PO squared is 8² + (-2)^2=64 +4=68. So PO is sqrt(68). Therefore, PA=PO. Correct.

But wait, sqrt(68) is the same as 2*sqrt(17), but regardless, the distances are equal. So that checks out.
</execution_3.2>

<execution_3.3>
The point P is somewhere on this line. The condition PA = PO means that P is equidistant from A(10, -10) and O(0,0), so P lies on the perpendicular bisector of the segment OA.

Wait, is that correct? Wait, the set of points equidistant from A and O is indeed the perpendicular bisector of OA. So maybe another approach is to find the equation of the perpendicular bisector of OA and then find its intersection with the line y = -x +6.

Let me try that method to confirm.

First, find the midpoint of OA. O is (0,0) and A is (10, -10). The midpoint M is ((0 +10)/2, (0 + (-10))/2) = (5, -5).

The slope of OA is ( -10 -0 ) / (10 -0 ) = -10/10 = -1. Therefore, the perpendicular bisector will have a slope that is the negative reciprocal, which is 1.

So the equation of the perpendicular bisector is y - (-5) = 1*(x -5), which simplifies to y +5 = x -5, so y = x -10.

Now, the point P lies on both y = -x +6 and y = x -10. Let's solve these two equations simultaneously.

Set -x +6 = x -10

Adding x to both sides: 6 = 2x -10

Adding 10 to both sides: 16 = 2x

Therefore, x =8. Then y = -8 +6 = -2, so y = -2. Therefore, P is (8, -2). Same result as before. So this confirms that the coordinates are indeed (8, -2).
</execution_3.3>

<execution_3.4>
So if I find the equation of the perpendicular bisector of AO, then the intersection of this bisector with the line y = -x +6 should give the point P.

Let me try this method to confirm.

First, find the midpoint of AO. Point A is (10, -10) and O is (0,0). The midpoint M is ((10+0)/2, (-10+0)/2) = (5, -5).

The slope of AO is ( -10 - 0 ) / (10 - 0 ) = -10/10 = -1. Therefore, the perpendicular bisector will have a slope that is the negative reciprocal, which is 1.

So the equation of the perpendicular bisector is y - (-5) = 1*(x -5), which simplifies to y +5 = x -5, so y = x -10.

Now, find the intersection of y = x -10 and y = -x +6.

Set x -10 = -x +6

Adding x to both sides: 2x -10 =6

Adding 10: 2x=16

x=8

Then y=8 -10= -2. So y=-2.

Therefore, the intersection point is (8, -2), which matches the previous result.
</execution_3.4>
</Execution_3>

<Plan_4>
Based on execution_3.4:
**Final Answer**
The coordinates of point \( P \) are \boxed{(8, -2)}.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the coordinates of the point $P$ on the line $y=-x+6$ that is equidistant from points $A$ and $O$ are $\boxed{(8,-2)}$.
</Final_answer>
</think>