Find the locus that is equidistant from the point (3, -3) and y=x-2?

Question

Please don't use parabola general formula.

Answer 1

Writing distances
(x-3)^2+(y+3)^2 = (x-y-2)^2/2
Remember that the distance of a point (p,q) to a line
ax+by+c=0 is
abs(ap+bq+c)/sqrt(a^2+b^2)
so
(x^2-6x+9+y^2+6y+9)*2 = x^2+y^2-2xy-4x+4y
x^2+y^2+2xy-8x+8y+36=0

Answer 2

Okay... If you plot the line and point on a graph, and find the midpoints of the lines created by connecting the point to various points of the line, you create another line. This is your locus:
y = x - 4

Answer 3

(2.75,-2.75)