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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 $(a,b)$.

 Jul 23, 2024
 #1
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Let P  have the coordinates  ( x , -x + 6)

 

PA^2  = ( x - 10)^2 + (- x + 6 - -12)^2  =  (x -10)^2 + ( -x + 18)^2

PO^2 = ( x - 2)^2 + ( -x + 6 - 8)^2 = (x -2^2 + ( -x - 2)^2

 

We want   PA^2 = PO^2....so....

 

(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

 

-56x + 424  =  8

 

-56x = -416

 

x = 416/56 = 52/7

 

y = - 52/7 + 6 =  -10/7 

 

P = (52/7, -10/7)

 

 

cool cool cool

 Jul 23, 2024

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