Question

The point in YZ-plane which is equidistant from three points $A(2,0,3),B(0,3,2)$ and $C(0,0,1)$ is

• A. (0,3,1)
• B. (0,1,3)
• C. (1,3,0)
• D. (3,1,0)

Answer 1

Answer: (B)

Since x-coordinate of every point in YZ-plane is zero.
Let $P(0,y,z)$ be a point on the YZ-plane such that
$PA=PB=PC.\text{ Now, }PA=PB$
$\Rightarrow (0-2{)}^2+(y-0{)}^2+(z-3{)}^2$
$=(0-0{)}^2+(y-3{)}^2+(z-2{)}^2$
i.e., $z-3y=0\text{ and }PB=PC$
$\Rightarrow y^2+9-6y+z^2+4-4z=y^2+z^2+1-2z,$
i.e., $3y+z=6$
On simplifying the two equations, we get $y=1$ and $z=3$. Here, the coordinate of the point $P$ are $(0,1,3)$.