Q. 155.0( 2 Votes )

# 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 :

**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

⇒ PA^{2} = PB^{2} = PC^{2} = PO^{2}

**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 PO^{2} = PA^{2}

x^{2}+ y^{2} + z^{2} = (x – 2)^{2} + y^{2} + z^{2}

⇒ x^{2}= x^{2}+ 4 – 4x

⇒ 4x = 4

⇒ x = 1

As PO^{2} = PB^{2}

x^{2}+ y^{2} + z^{2} = x^{2}+ (y – 3)^{2} + z^{2}

⇒ y^{2}= y^{2}+ 9 – 6y

⇒ 6y = 9

As PO^{2} = PC^{2}

x^{2}+ y^{2} + z^{2} = x^{2} + y^{2} + (z – 8)^{2}

⇒ z^{2}= z^{2}+ 64 – 16x

⇒ 16z = 64

⇒ z = 4

Hence **point** is equidistant from given points

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