Find the locus of a point which is equidistant from (1,3) and x-axis.

$\text{Let p(h,k) be any point on the locus and let A(1,3) and B(h,0).Then,}PA=PB\Rightarrow P{A}^2=P{B}^2\Rightarrow (1-h{)}^2+(3-k{)}^2=(h-h{)}^2+(0-k{)}^2\Rightarrow 1+h^2-2h+9+k^2-6k=0+k^2\Rightarrow h^2-2h-6k+10=0Hence,locusof(h,k)is{x}^2-2x-6y+10=0$