Question

Determine the point in xy-plane which is equidistant from three points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).

Answer 1

We know that z-coordinate of every point on xy-plane is zero. So, let P(x, y, 0) be a point on xy-plane such that PA=PB=PC.

Now, PA=PB

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

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

$\Rightarrow 4x-6y=0\Rightarrow 2x-3y=0$

and, PB=PC

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

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

$\Rightarrow -6y+12=0$

$\Rightarrow y=2$

Putting y=2 in (i), we obtain x=3.

Hence, the required point is (3, 2, 0).