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P is a point on the circle x^2+y^2=9 Q i...

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:

Text Solution

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`|(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`
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