Question

Find the locus of the point which is equidistant from the points A(0,2,3) and (2,-2,1).

Answer 1

Let P(x,y,z) be any point which is equidistant from A(0,2,3) and B(2,-2,1). Then,

PA=PB

$\Rightarrow P{A}^2=P{B}^2$

$\Rightarrow \sqrt{(x-0{)}^2+(y-2{)}^2+(z-3{)}^2}=\sqrt{(x-2{)}^2+(y+2{)}^2+(z-1{)}^2}$

$\Rightarrow 4x-8y-4z+4=0$ or $x-2y-z+1=0$

Hence, the required locus is $x-2y-z+1=0$.