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Find the Locus of the Points Which Are Equidistant from the Points (1, 2, 3) and (3, 2, –1). - Mathematics

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प्रश्न

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

उत्तर

Let P (xyz) 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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अध्याय 28: Introduction to three dimensional coordinate geometry - Exercise 28.2 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 21 | पृष्ठ १०

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