Hello, nerdzor!
Find the equation of the locus of the point P(x,y) which moves so that
it is always equidistant from the point $\displaystyle S(1,2)$ and the line $\displaystyle L\!:\:y= 4$ First, make a sketch . . . Code:

 *S(1,2)
 \
 \
+\
 *P(x,y)
 :
 :
 :
4+    *    L
 Q
The distance from $\displaystyle P\text{ to }S$ is: .$\displaystyle \sqrt{(x1)^2 + (y2)^2}$
The distance from $\displaystyle P$ to line $\displaystyle L$ is: .$\displaystyle PQ \:=\:y(4) \:=\:y+4$
Since $\displaystyle PS = PQ$, we have: .$\displaystyle \sqrt{(x1)^2 + (y2)^2} \:=\:y+4$
That is the equation of the locus, so technically, we're done.
. . But, of course, we are expected to simplify our equation.
. . Square both sides: .$\displaystyle (x1)^2 + (y2)^2 \:=\:(y+4)^2$
. . Expand: .$\displaystyle x^2  2x + 1 + y^2  4y + 4 \:=\:y^2 + 8y + 16$
. . And we have: .$\displaystyle x^2  2x \:=\:12y + 11$
. . Add 1 to both sides: .$\displaystyle x^2  2x + 1 \:=\:12y + 12$
. . Factor: .$\displaystyle \boxed{(x1)^2 \:=\:12(y+1)}$
The locus is a parabola.
. . It opens upward, and its vertex is at $\displaystyle (1,1)$