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

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,

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Distance PA

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

Distance PB

Here,

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = – 1

Since PA = PB

Squaring both sides, we get –

PA2 = PB2

(1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2

(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)

= (9 + x2 – 6x) + (4 + y2 - 4y) + (1 + z2 + 2z)

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

4x – 8z = 0

x – 2z = 0

This is the required equation.

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