Can I ask for your help regarding a problem in Algebra?

Let the moving point P be represented by (x,y)

Distance from P to (-1,0) = distance from P to (0,-2)

[ ( x +1)^2 +( y-0)^2 ]^1/2 = [ (x-0)^2 + (y +2)^2 ]^1/2

square both sides

(x+1)^2 + y^2 = x^2 + (y+2)^2

x^2 + 2x + 1 + y^2 = x^2 + y^2 + 4y +4

2x + 1 = 4y + 4

2x -3 = 4y

y = 2/4 x - 3/4

y = 1/2 x - 3/4

A:(-1,0) and B:(0,-2)

The gradient of the line that passes through A and B is of the form:

m=(y1-y2)/(x1-x2)=(-2-0)/(0--1)=-2

So the gradient that passes through A and B is minus 2.

The locus of points that are equidistant from A and B will be on a line with the gradient equal to the normal of the gradient we just calculated. As they are perpendicular.

The new gradient is

n=-1/m=1/2

This line will pass through the midpoint of A and B ( as at that point they are equidistant).

M;(-1/2 , -1)

The equation will have the form:

y-y1=n(x-x1)

Substituting values gives

y+1 = (1/2)(x+1/2)

Solving for y and expanding gives:

y=x/2 -3/4

The point P forms a line which is the perpendicular bisector of the two points point 1 and 2. Try to imagine this.

Slope of the locus of P = -1/slope of 12 = -1/(-2/1) = 1/2

y = x/2 + c in the general form

now since perpendicular bisector will pass through mid of 12, mid of 12 = (-1/2,-1)

using this, c = -3/4

y = x/2 - 3/4

or 2x - y - 3 = 0

The point P is always equidistant from A(-1,0) and B(0,-2)

Let (x,y) be P.

PA=PB

PA^2=PB^2

(x+1)^2+y^2=x^2+(y+2)^2 ...........(Distance formula)

On simplification,we obtain,

2x+1=4y+4

2x-4y=3...........is the Locus of the point P.

Geometrically speaking,this line will be the perpendicular bisector of the line segment joining the 2 points