P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:

P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+

1.	P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:
Answer» `\|(m-1)/(1+m)\|=\|(1+7)/(1-7)\|` `\|(m-1)/(1+m)\|=4/3` `(m-1)/(1+m)=4/3,-4/3` `3m-3=4+4m,3m-3=-4=4m` `m=-7,-1/7` `(y-1/2)=-1/7(x+1/2)` `7y+x-3=0` `x^2+y^2=9` `48y^2-42y+9+y^2=9` `50y^2-42y=0` `y=0,21/25` `x=3,-72/25` `(3,0),(-72/25,21/25)`.

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P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:

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