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Explanation

The equation of the line `AB` is
`(x)/(7)+(y)/(-5)=1`……`(i)`
`implies5x-7y=35`
Equation of line perpendicular to `AB` is
`7x+5y=lambda` ……..`(ii)`
It meets `X`-axis at `P(lambda//7,0)` and `Y`-axis at `Q(0,lambda//5)`.
The equations of lines `AQ` and `BP` are `(x)/(7)+(5y)/(lambda)=1` and `(7x)/(lambda)-(y)/(5)=1`, respectively.
Let `R(h,k)` be their point of intersection of lines `AQ` and `BP`.
Then, `(h)/(7)+(5k)/(lambda)=1`
and `(7h)/(lambda)-(k)/(5)=1`
`implies(1)/(5k)(1-(h)/(7))=(1)/(7h)(1+(k)/(5))` [on eliminating `lambda`]
`impliesh(7-h)=k(5+k)`
`impliesh^(2)+k^(2)-7h+5k=0`
Hence, the locus of a point is
`x^(2)+y^(2)-7x+5y=0`.