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Home » 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).
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).
CBSE | Class 11 | Exercise 26B | Maths | RS Aggarwal | Three Dimensional Geometry
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:

Let us take,

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 = OD

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

Squaring on 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  ;a ≠ 0

x = a/2

∴ BD = OD

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

Squaring on both sides,

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

b(2y-b) = 0    ;b ≠ 0

y= b/2

∴ CD = OD

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

Squaring on both sides,

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

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

c(2z-c) = 0      ;c ≠ 0

z= c/2

Therefore, the point 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).

 

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