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

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Payal Gupta | Contributor-Level 10

8 months ago

8. Let P(x, y, z) be the point equidistant from the given points (1, 2, 3) say A and (3, 2, –1) say B.

So, PA = PB

=> ${(1 - x)}^{2} + {(2 - y)}^{2} + {(3 - z)}^{2}$ = ${(3 - x)}^{2} + {(2 - y)}^{2} + {(- 1 - z)}^{2}$

Squaring both sides,

=> ${(1 - x)}^{2} + {(2 - y)}^{2} + {(3 - z)}^{2}$ = ${(3 - x)}^{2} + {(2 - y)}^{2} + {(- 1 - z)}^{2}$

=> ${(1 - x)}^{2} + {(3 - z)}^{2} = {(3 - x)}^{2} + [{(-)}^{2} {(1 + z)}^{2}$

=> $(1^{2} + x^{2} - 2 x) + (3^{2} + z^{2} - 2.3 . z) = (3^{2} + x^{2} - 2.3 . x) + (1^{2} + z^{2} + 2 z)$

=> $1 + x^{2} - 2 x + 9 + z^{2} - 6 z = 9 + x^{2} - 6 x + 1 + z^{2} + 2 z$

=> $6 x - 2 x - 6 z - 2 z = 0$

=> $4 x - 8 z = 0$

=> $x - 2 z = 0$

Therefore, the required equation of point is $x - 2 z = 0$

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