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Determine the coordinates of the point $P$ on the line $y=-x+6$ such that $P$ is equidistant from the points $A(10,-12)$ and $O(2,8)$ (that is, so that $PA=PO$).  Express your answer as an ordered pair $(a,b)$.

 Jan 14, 2024
 #1
avatar+128732 
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Let P  = (x , 6-x)

 

We can equate distances (squared)

 

(10 - x)^2  + ( -12 - (6-x)^2   =  ( 2 -x)^2  + (8 - (6-x) )^2

 

(x -10)^2  + ( x - 18)^2  = (x - 2)^2 +  ( x + 2)^2

 

x^2 -20x + 100  + x^2 -36x + 324  =  x^2 - 4x + 4  + x^2 + 4x + 4

 

2x^2  - 56x + 424 =  2x^2 + 8

 

416  =  56x

 

x = 416 / 56   =   52  /  7

 

y = 6 - (52/7)  =  -10/7

 

(x, y)  = (52/7 , -10/7)

 

 

cool cool cool

 Jan 14, 2024

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