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

Explanation

Let `p(h, k)` be the point that is equidistant from `A(1,3)` and the x-axis.

Now, the distance of the point `p(h,k)` from x-axis is `k`.

` AP=k=> AP^2=k^2`

`(h-1)^2+(k-3)^2=k^2`

`h^2-2h-6k+10=0`

Hence, locus of `(h, k)` is `x^2-2x-6y+10=0`