Find the locus of a point which is equidistant from the origin and the point (-2,5)

Answer 1

the locus is the perpendicular bisector of the line from (0,0) to (-2,5).

It's easy to find the midpoint, and the slope of the line.

Now just find the line with perpendicular slope, through the midpoint.

Answer 2

Let a point on the locus be P (x,y)
then
√(x^2 + y^2) = √((x+2)^2 + (y-5)^2)
square both sides:
x^2 + y^2 = (x+2)^2 + (y-5)^2
x^2 + y^2 = x^2 + 4x + 4 + y^2 - 10y + 25
4x - 10y = -29

which is the perpendicular bisector of the line joining (0,0) and (-2,5) , as analysed by Steve above.

Answer 3

draw the grid