Question

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

Answer 1

We know that, x-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, $PA=PB\Rightarrow P{A}^2=P{B}^2$

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

$\Rightarrow x^2-4x+4+y^2+9=x^2+y^2-6y+9+4$

$\Rightarrow 2x-3y=0$

and $PB=PC\Rightarrow P{B}^2=P{C}^2$

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

$\Rightarrow x^2+y^2-6y+9+4=x^2+y^2+1$

$\Rightarrow -6y+12=0\Rightarrow y=2$

Substitute the value of y=2 in Eq. )i) we get x=3.

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