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 = 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
