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Question
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Solution
Let P (x, y, z) be any point that is equidistant from the points A (1, 2, 3) and B (3, 2, −1).
Then, we have:
PA = PB
\[\Rightarrow \sqrt{\left( x - 1 \right)^2 + \left( y - 2 \right)^2 + \left( z - 3 \right)^2} = \sqrt{\left( x - 3 \right)^2 + \left( y - 2 \right)^2 + \left( z + 1 \right)^2}\]
\[ \Rightarrow x^2 - 2x - 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1\]
\[ \Rightarrow - 2x - 4y - 6z + 14 = - 6x - 4y + 2z + 14\]
\[ \Rightarrow - 2x + 6x - 4y + 4y - 6z - 2z = 14 - 14\]
\[ \Rightarrow 4x - 8z = 0\]
\[ \Rightarrow x - 2z = 0\]
Hence, the locus is x\[-\]2z = 0
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