Question

Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, -1).

Answer 1

Let the point P (x, y, z) which is equidistance from A(1, 2, 3) and B(3, 2, -1),

so AP = BP

$(AP{)}^2=(BP{)}^2$

$(x-1{)}^2+(y-2{)}^2+(z-3{)}^2=(x-3{)}^2+(y-2{)}^2+(z+1{)}^2$

$x^2+1-2x+y^2+4-4y+z^2+9-6z$

=$x^2+9-6x+y^2+4-4y+z^2+1+2z$

4x-8z=14-14

4x-8z = 0

x-2z = 0