Question

Find the coordinates of the point which is equidistant from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).

Answer 1

Let the required point be P(x, y, z)

Here, 0(0, 0, 0), A(2, 0, 0), B(0, 3, 0), C(0, 0, 8)

Since, $(OP{)}^2$=$(AP{)}^2$

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

$x^2+y+z^2=x^2-4x+4+y^2+z^2$

4x=4

x=1

$(OP{)}^2=(BP{)}^2$

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

$x^2+y^2+z^2=x^2+y^2-6y+9+z^2$

6y=9

y=$\frac{3}{2}$

$(OP{)}^2=(CP{)}^2$

$(x-0{)}^2+(y-0{)}^2+(z-0{)}^2=(x-0{)}^2+(y-0{)}^2+(z-8{)}^2$

$x^2+y^2+z^2=x^2+y^2+z^2-16z+64$

16z=64

z = 4

The required point = $(1,\frac{3}{2},4)$