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Points $A,$ $B,$ and $C$ are given in the coordinate plane. There exists a point $Q$ and a constant $k$ such that for any point $P$,
\[PA^2 + PB^2 + PC^2 = 3PQ^2 + k.\]
If $A = (7,-11),$ $B = (10,13),$ and $C = (18,-22)$, then find the constant $k$.

 Mar 24, 2024
 #1
avatar+128406 
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Let P= (x,y)

 

PA^2  + PB^2  + PC^2   =  

 

(x - 7)^2  + (y + 11)^2   + ( x-10)^2 + (y -13)^2  + ( x-18)^2  + (y + 22)^2   =

 

3x^2  - 70x + 3y^2 + 40y +  1247          

 

3 [ x^2 - (70/3x + y^2  +(40/3)y + 1247/3) ]   complete the square on x,y

 

3 [ (x^2  - (70/3)x + 1225/9 + y^2 + (40/3)y + 400/9  +  1247/3 - 1225/9 - 400/9)

 

3 [ (x - 35/3)^2  + (y + 20/3)^2 ] + 3 [ 1247/3 -1225/9 - 400/9 ] 

 

3 [ (x - 35/3)^2  + (y + 20/3)^2 ] + 2216/3

 

Q = ( 35/3 , -20/3)

    

k = 2216 / 3

 

 

cool cool cool

 Mar 24, 2024

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