Question

The locus of a point which is equidistant from the points $(1,2,3)$ and $(3,2,-1)$ is

• A. $x+2y-3z=3$
• B. $x-2z=0$
• C. $x-y+3z=3$
• D. $x+y+2z=6$

Answer 1

The correct option is B $x-2z=0$

Let $P(x,y,z)$ be the point that is equidistant from points $A(1,2,3)$ and $B(3,2,-1)$
i.e. $PA=PB$
$P{A}^2=P{B}^2$
Using Distance formula,
$\Rightarrow (x-1{)}^2+(y-2{)}^2+(z-3{)}^2=(x-3{)}^2+(y-2{)}^2+(z+1{)}^2$
$\Rightarrow x^2+1-2x+y^2+4-4y+z^2+9-6z=x^2+9-6x+y^2+4-4y+z^2+1+2z$
$\Rightarrow -2x-6z+14=-6x+2z+14$
$\Rightarrow -2x-6z+6x-2z=0$
$\Rightarrow 4x-8z=0$
$\Rightarrow x-2z=0$