# Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, -1)

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Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, -1).

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Given: Points are A(1, 2, 3) and B(3, 2, -1)

To find: the locus of points which are equidistant from the given points

Let the required point P(x, y, z)

According to the question:

PA = PB

⇒ PA2  = PB2

Formula used:

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

$\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}$

Therefore, The distance between P(x, y, z) and A(1, 2, 3) is PA,

$\sqrt{(x-1)^2+(y−2)^2+(z-3)^2}$

The distance between P(x, y, z) and B(3, 2, -1) is PB,

=$\sqrt{(x-3)^2+(y−2)^2+(z-(-1))^2}$

$\sqrt{(x-3)^2+(y−2)^2+(z+1))^2}$

As PA2 = PB

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

⇒ x2+ 1 – 2x + y2 + 4 – 4y + z2 + 9 – 6z = x2+ 9 – 6x + y2 + 4 – 4y + z2 + 1 + 2z

⇒ x2+ 1 – 2x + y2 + 4 – 4y + z2 + 9 – 6z – x2– 9 + 6x – y2 – 4 + 4y – z2 – 1 – 2z = 0

⇒ 4x – 8z = 0

⇒ 4(x – 2z) = 0

⇒ x – 2z = 0

Hence locus of point P is x – 2z = 0