The point A (2,7) lies on the perpendicular bisector of the line segment joining the points P (5,-3) and Q(0,-4).

Explanation

If A (2,7) lies on perpendicular bisector of P(6,5) and Q(0,-4) , then AP = AQ
`:.` AP `=sqrt((6-2)^(2)+(5-7)^(2))`
`=sqrt((4)^(2)+(-2)^(2))`
`=sqrt(16+4)=sqrt(20)`
and A `=sqrt((0-2)^(2)+(-4-7)^(2))`
`=sqrt((-2)^(2)+(-11)^(2))`
`=sqrt(4+121)=sqrt(125)`
So, A does not lies on the perpendicular bisector of PQ.
Alternate Method
If the point A(2,7) lies on the perpendicular bisector of the line segment , then the point A satisfy the equation of perpendicular bisector.
Now , we find the equation of equation of perpendicular bisector . For this , we find the slope perpendicular bisector.
`:.` Slope of perpendicular `=(-1)/("Slop of line segment joining the points (5,-3) and (0,-4)")`
[`:' "slope"=(y_(2)-y_(1))/(x_(2)-x_(1))]`
`=(-1)/((-4-(-3))/(0-5))=5`
[since, perpendicular bisector is perpendicular to the line segment , so its slopes have the condition , `m_(1)*m_(2)=-1]`
Since , the perpendicular bisector passes through the mid- point of the line segment joining the points (5,-3) and (0,4).
`:.` Mid - point of PQ `=((5+0)/(2),(-3-4)/(2))=((5)/(2),(-7)/(2))`
So , the equation of perpendicular bisector having slop `(1)/(3)` and passes through the mid - point `((5)/(2),(-7)/(2))` is
`(y+(7)/(2))=(x-(5)/(2))`
[`:'`equation of line is `(y-y_(1))=m(x-x_(1))]`
`rArr 2y+7=10x-25`
`rArr10x-2y=32=0`
`rArr10x-2y=32`
`rArr5x-y=16` ...(i)
Now , check whether the point A(2,7) lie on the Eq. (i) or not.
`5xx2-7=10-7=3ne16`
Hence, the point A(2,7) does not lie on the perpendicular bisector of the line segment.