Find the equation of the set of points which | vedclass

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Question

(A) Let $P(x, y, z)$ be any point such that $PA = PB$. Using the distance formula,we have: $\sqrt{(x-3)^2 + (y-4)^2 + (z+5)^2} = \sqrt{(x+2)^2 + (y-1)

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$10x + 6y - 18z - 29 = 0$

Vedclass · Answer

(A) Let $P(x, y, z)$ be any point such that $PA = PB$. Using the distance formula,we have: $\sqrt{(x-3)^2 + (y-4)^2 + (z+5)^2} = \sqrt{(x+2)^2 + (y-1)^2 + (z-4)^2}$ Squaring both sides: $(x-3)^2 + (y-4)^2 + (z+5)^2 = (x+2)^2 + (y-1)^2 + (z-4)^2$ Expanding the squares: $(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25) = (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 8z + 16)$ Simplifying the equation: $-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$ Rearranging the terms: $10x + 6y - 18z - 29 = 0$

Find the equation of the set of points which are equidistant from the points $A(3, 4, -5)$ and $B(-2, 1, 4)$.

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