If the vertices of Delta ABC are A(0, 0),B(4, | vedclass

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(D) Let the coordinates of point $P$ be $(x, y)$.Since $P$ is equidistant from $A(0, 0)$,$B(4, 0)$,and $C(0, 6)$,we have $PA = PB = PC$,which implies

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$(2, 3)$

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(D) Let the coordinates of point $P$ be $(x, y)$.Since $P$ is equidistant from $A(0, 0)$,$B(4, 0)$,and $C(0, 6)$,we have $PA = PB = PC$,which implies $PA^2 = PB^2 = PC^2$.From $PA^2 = PB^2$,we get $x^2 + y^2 = (x - 4)^2 + (y - 0)^2$.$x^2 + y^2 = x^2 - 8x + 16 + y^2$.$8x = 16 \implies x = 2$.From $PA^2 = PC^2$,we get $x^2 + y^2 = (x - 0)^2 + (y - 6)^2$.$x^2 + y^2 = x^2 + y^2 - 12y + 36$.$12y = 36 \implies y = 3$.Thus,the coordinates of point $P$ are $(2, 3)$.

If the vertices of $\Delta ABC$ are $A(0, 0)$,$B(4, 0)$ and $C(0, 6)$,then the coordinates of the point $P$ which is equidistant from $A, B$ and $C$ and in the plane of $\Delta ABC$ is $\ldots \ldots \ldots$

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