Find the point in the XY-plane which is equid | vedclass

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Question

(A) Since the point $P$ lies in the $XY$-plane,its $z$-coordinate is $0$. Let the point be $P(x, y, 0)$.Given that $P$ is equidistant from $A(2, 0, 3)

Vedclass · Accepted Answer

$(3, 2, 0)$

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(A) Since the point $P$ lies in the $XY$-plane,its $z$-coordinate is $0$. Let the point be $P(x, y, 0)$.Given that $P$ is equidistant from $A(2, 0, 3)$,$B(0, 3, 2)$,and $C(0, 0, 1)$,we have $PA = PB = PC$,which implies $PA^2 = PB^2$ and $PB^2 = PC^2$.First,$PA^2 = PB^2$:$(x - 2)^2 + (y - 0)^2 + (0 - 3)^2 = (x - 0)^2 + (y - 3)^2 + (0 - 2)^2$$x^2 - 4x + 4 + y^2 + 9 = x^2 + y^2 - 6y + 9 + 4$$-4x + 13 = -6y + 13$$4x = 6y \implies 2x = 3y \quad (i)$Next,$PB^2 = PC^2$:$(x - 0)^2 + (y - 3)^2 + (0 - 2)^2 = (x - 0)^2 + (y - 0)^2 + (0 - 1)^2$$x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 + 1$$-6y + 13 = 1$$6y = 12 \implies y = 2$Substituting $y = 2$ into equation $(i)$:$2x = 3(2) \implies 2x = 6 \implies x = 3$Thus,the required point is $(3, 2, 0)$.

Find the point in the $XY$-plane which is equidistant from the three points $A(2, 0, 3)$,$B(0, 3, 2)$,and $C(0, 0, 1)$.

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