Find the equation of the set of points which | vedclass

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Question

(A) Let $P(x, y, z)$ be the point that is equidistant from points $A(1, 2, 3)$ and $B(3, 2, -1)$.Accordingly,$PA = PB$.$\Rightarrow PA^2 = PB^2$.$\Rig

Vedclass · Accepted Answer

$x - 2z = 0$

Vedclass · Answer

(A) Let $P(x, y, z)$ be the point that is equidistant from points $A(1, 2, 3)$ and $B(3, 2, -1)$.Accordingly,$PA = PB$.$\Rightarrow PA^2 = PB^2$.$\Rightarrow (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$.Expanding both sides:$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1$.Simplifying the equation:$-2x - 6z + 14 = -6x + 2z + 14$.$-2x + 6x - 6z - 2z = 0$.$4x - 8z = 0$.Dividing by $4$,we get $x - 2z = 0$.Thus,the required equation is $x - 2z = 0$.

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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