Q. 265.0( 2 Votes )

# Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer :

Let A (1, 2, 3) & B (3, 2, – 1)

Let point P be (x, y, z)

Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1)

i.e. PA = PB

__Calculating PA__

P ≡ (x, y, z) and A ≡ (1, 2, 3)

Distance PA

Here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 1, y_{2} = 2, z_{2} = 3

Distance PA

__Calculating PB__

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

Distance PB

Here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 3, y_{2} = 2, z_{2} = – 1

Since PA = PB

Squaring both sides, we get –

PA^{2} = PB^{2}

⇒ (1 – x)^{2} + (2 – y)^{2} + (3 – z)^{2} = (3 – x)^{2} + (2 – y)^{2} + (– 1 – z)^{2}

⇒ (1 + x^{2} – 2x) + (4 + y^{2} – 4y) + (9 + z^{2} – 6z)

= (9 + x^{2} – 6x) + (4 + y^{2} - 4y) + (1 + z^{2} + 2z)

⇒ – 2x – 4y – 6z + 14 = – 6x - 4y + 2z + 14

⇒ 4x – 8z = 0

⇒ x – 2z = 0

This is the required equation.

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