Find the point equidistant from the points P(1, 3) and Q(-5, -5), and which lies on the line y= -1.

Let the point be A. Since it lies on y = -1, its y coordinate will be -1. Thus, the coordinates of point A will be (x, -1).

$\text{Distance AP = Distance AQ}\Rightarrow {AP}^2={AQ}^2$
$\Rightarrow (x-1{)}^2+(-1-3{)}^2=(x-(-5){)}^2+(-1-(-5){)}^2\Rightarrow (x-1{)}^2+16=(x+5{)}^2+16\Rightarrow x^2-2x+1+16=x^2+10x+25+16\Rightarrow -2x+17=10x+41\Rightarrow x=-2$

Thus, the coordinates of point A will be (-2, -1).