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

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,



Therefore,


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



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




As PA2 = PB2


(x – 1)2+ (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


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