Find the point in xy-plane which is equidistant from the points

Solution:

The general point on xy plane is D(x, y, 0). Consider this point is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).

∴ AD = BD

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

Squaring both sides

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

$x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}-6 y+9+4$

$-4 x=-6 y \ldots(1)$

Also, AD = CD

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

Squaring both sides,

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

$x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}+1$

$-4 x=-12$ ..........(2)

Simultaneously solving equation (1) and (2) we get

X = 3, y = 2.

The point which is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1) is (3, 2, 0).