Question

The point in the xy-plane which is equidistant from $(2,0,3),(0,3,2)$ and $(0,0,1)$ is

• A. (1, 2, 3)
• B. (-3, 2, 0)
• C. (3, -2, 0)
• D. (3, 2, 0)
• E. (3, 2, 1)

Answer 1

Answer: (D)

Let the points are $A(2,0,3),B(0,3,2)$ and $D(0,0,1).$
We know that $Z$ -coordinate of every point an $xy$ -plane is zero so let $p(x,y,0)$ be a point on $xy$ -plane such that $PA=PB=PC$.
Now, $PA=PB$
$\Rightarrow P{A}^2=P{B}^2$
$\Rightarrow (x-2{)}^2+(y-0{)}^2+(0-3{)}^2=(x-0{)}^2+(y-3{)}^2+(0-2)$
$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0$ ...(i)
and, $PB=PC$
$\Rightarrow P{B}^2=P{C}^2$
$\Rightarrow (x-0{)}^2+(y-3{)}^2+(0-2{)}^2=(x-0{)}^2+(y-0{)}^2$
$+(0-1{)}^2$
$\Rightarrow -6y+12=0$
$\Rightarrow y=2$ ...(ii)
Putting $y=2$ in Eq. (i), we get $x=3$ Hence, the required point is $(3,2,0)$.