Question
The coordinates of point in $$xy$$ -plane which is equidistant from three points $$A\left( {2,\,0,\,3} \right),\,B\left( {0,\,3,\,2} \right)$$ and $$C\left( {0,\,0,\,1} \right)$$ are :
A.
$$\left( {3,\,2,\,0} \right)$$
B.
$$\left( {3,\,4,\,0} \right)$$
C.
$$\left( {0,\,0,\,3} \right)$$
D.
$$\left( {2,\,3,\,0} \right)$$
Answer :
$$\left( {3,\,2,\,0} \right)$$
Solution :
We know that $$z$$-co-ordinate of every point on $$xy$$ -plane is zero. So, let $$P\left( {x,\,y,\,0} \right)$$ be a point in $$xy$$ -plane such that
$$\eqalign{
& PA = PB = PC \cr
& {\text{Now}},\,PA = PB \Rightarrow P{A^2} = P{B^2} \cr
& \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {0 - 3} \right)^2} = {\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {0 - 2} \right)^2} \cr
& \Rightarrow 4x - 6y = 0 \cr
& \Rightarrow 2x - 3y = 0......\left( 1 \right) \cr
& PB = PC \Rightarrow P{B^2} = P{C^2} \cr
& \Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {0 - 2} \right)^2} = {\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {0 - 1} \right)^2} \cr
& \Rightarrow - 6y + 12 = 0 \cr
& \Rightarrow y = 2......\left( 2 \right) \cr
& {\text{Putting }}y = 2{\text{ in }}\left( 1 \right){\text{, we obtain }}x = 3 \cr
& {\text{Hence, the required point is }}\left( {3,\,2,\,0} \right) \cr} $$