Question

Find the coordinates of the point which is equidistant from the points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).

Answer 1

Consider, D(x,y,z) point equidistant from points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0). 

∴ AD = 0D

\(\sqrt{(x - a)^2 +(y-0)^2+(z-0)^2}\) = \(\sqrt{(x - 0)^2 +(y-0)^2+(z-0)^2}\)

Squaring both sides, 

(x - a)² + (y - 0)² + (z - 0)² = (x - 0)² + (y - 0)² + (z - 0)²

x² +2ax + a² + y² + z² = x² + y² + z² 

a(2x-a) = 0 

as a ≠ 0. 

X = a/2 

∴ BD = 0D

\(\sqrt{(x - a)^2 +(y-0)^2+(z-0)^2}\) = \(\sqrt{(x - 0)^2 +(y-0)^2+(z-0)^2}\)

Squaring both sides,

(x - 0)² + (y - b)² + (z - 0)² = (x - 0)² + (y - 0)² + (z - 0)²

x² + y² + 2by + b² + z² = x² + y² + z² 

b(2y-b) = 0 

as b ≠ 0. 

y= b/2 

∴ CD = 0D

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

Squaring both sides,

(x - 0)² + (y - 0)² + (z - c)² = (x - 0)² + (y - 0)² + (z - 0)²

x² + y² + z² + 2cz + c² = x² + y² + z² 

c(2z-c) = 0 

as c ≠ 0. 

z = c/2 

Therefore, the pint D(a/2,b/2,c/2) is equidistant to points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).