Okay, let's see. I need to find the point P on the line y = -x + 6 such that the distance from P to A(10, -10) is equal to the distance from P to O(0, 0). Hmm. So, P has to be equidistant from A and O, and it's on the given line. 

First, let me recall that the set of all points equidistant from two points A and O is the perpendicular bisector of the segment AO. So maybe I can find the equation of the perpendicular bisector of AO and then find its intersection with the line y = -x + 6. That intersection point should be P. Let me check if that approach works.

Wait, but if I do that, I need to make sure that the point P is on both lines. So first, find the midpoint of AO. The midpoint M would be ((10 + 0)/2, (-10 + 0)/2) = (5, -5). Then, the slope of AO is ( -10 - 0 ) / (10 - 0 ) = -10 / 10 = -1. Therefore, the perpendicular bisector would have a slope that's the negative reciprocal of -1, 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 must lie on both y = -x + 6 and y = x - 10. So let's solve these two equations simultaneously. 

Set them equal: -x + 6 = x - 10. Bring variables to one side: -x - x = -10 - 6 → -2x = -16 → x = 8. Then substitute back into one of the equations, say y = x - 10: y = 8 - 10 = -2. So P is (8, -2). 

Wait, let me verify if this is correct. Let's compute PA and PO. 

PA is the distance from (8, -2) to A(10, -10). Using the distance formula: sqrt[(10-8)^2 + (-10 - (-2))^2] = sqrt[4 + (-8)^2] = sqrt[4 + 64] = sqrt[68]. 

PO is the distance from (8, -2) to O(0,0): sqrt[(8 - 0)^2 + (-2 - 0)^2] = sqrt[64 + 4] = sqrt[68]. 

Yes, both distances are equal. So that seems correct. 

Alternatively, maybe there's another approach using coordinates. Let me try setting up equations. Let P be (x, y). Since it's on the line y = -x + 6, so y = -x + 6. Then, the distance from P to A(10, -10) is sqrt[(x - 10)^2 + (y + 10)^2], and the distance from P to O(0,0) is sqrt[x^2 + y^2]. Set them equal:

sqrt[(x - 10)^2 + (y + 10)^2] = sqrt[x^2 + y^2]

Square both sides to eliminate the square roots:

(x - 10)^2 + (y + 10)^2 = x^2 + y^2

Expand the left side:

x^2 - 20x + 100 + y^2 + 20y + 100 = x^2 + y^2

Simplify:

x^2 - 20x + 100 + y^2 + 20y + 100 = x^2 + y^2

Subtract x^2 + y^2 from both sides:

-20x + 100 + 20y + 100 = 0

Combine like terms:

-20x + 20y + 200 = 0

Divide both sides by 20:

-x + y + 10 = 0

So y = x - 10. Wait, that's the same equation as before. So the perpendicular bisector is y = x - 10. Then solving with y = -x + 6 gives x = 8, y = -2. So same result. 

Therefore, the coordinates of P are (8, -2). 

But let me check again if there's any mistake here. Wait, when I set up the equation, after expanding and simplifying, I get y = x - 10. Then solving with y = -x + 6 gives x = 8, y = -2. That seems correct. 

Alternatively, maybe there's a geometric interpretation. The point P is equidistant from A and O, so it lies on the perpendicular bisector of AO, which we found as y = x - 10. Intersecting that with the given line y = -x + 6 gives P(8, -2). 

Alternatively, if I use parametric equations for the line y = -x + 6, let me parameterize it. Let x = t, then y = -t + 6. So any point on the line can be written as (t, -t + 6). Then the distance from (t, -t + 6) to A(10, -10) must equal the distance to O(0,0). 

So set up the equation:

sqrt[(t - 10)^2 + (-t + 6 + 10)^2] = sqrt[t^2 + (-t + 6)^2]

Square both sides:

(t - 10)^2 + (-t + 16)^2 = t^2 + (-t + 6)^2

Expand both sides:

Left side: (t^2 - 20t + 100) + (t^2 - 32t + 256) = 2t^2 - 52t + 356

Right side: t^2 + (t^2 - 12t + 36) = 2t^2 - 12t + 36

Set equal:

2t^2 - 52t + 356 = 2t^2 - 12t + 36

Subtract 2t^2 from both sides:

-52t + 356 = -12t + 36

Bring variables to left and constants to right:

-52t + 12t = 36 - 356

-40t = -320

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

So t = 8, then y = -8 + 6 = -2. So P is (8, -2). Same answer. 

Therefore, regardless of the method, P is (8, -2). 

I think that's thorough. So the answer should be (8, -2).
Therefore, the coordinates of the point $P$ that is equidistant from points $A$ and $O$ and lies on the line $y = -x + 6$ are $\boxed{(8, -2)}$.
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