# 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).

Given: Points are O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8)

To find: the coordinates of point which is equidistant from the points

Let required point P(x, y, z)

According to question:

PA = PB = PC = PO

PA2 = PB2 = PC2 = PO2

Formula used:

Distance between any two points (a, b, c) and (m, n, o) is given by, Therefore,

The distance between P(x, y, z) and O(0, 0, 0) is PO,  Distance between P(x, y, z) and A(2, 0, 0) is PA,  Distance between P(x, y, z) and B(0, 3, 0) is PB,  Distance between P(x, y, z) and C(0, 0, 8) is PC,  As PO2 = PA2

x2+ y2 + z2 = (x – 2)2 + y2 + z2

x2= x2+ 4 – 4x

4x = 4

x = 1

As PO2 = PB2

x2+ y2 + z2 = x2+ (y – 3)2 + z2

y2= y2+ 9 – 6y

6y = 9  As PO2 = PC2

x2+ y2 + z2 = x2 + y2 + (z – 8)2

z2= z2+ 64 – 16x

16z = 64

z = 4

Hence point is equidistant from given points

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