The point P is equidistant from A(1, 3),B(-3, | vedclass

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(D) Let the coordinates of $P$ be $(x, y)$.Since $P$ is equidistant from $A, B, C$,we have $PA^2 = PB^2$ and $PB^2 = PC^2$.From $PA^2 = PB^2$:$(x-1)^2

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$5 \sqrt{10}$

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(D) Let the coordinates of $P$ be $(x, y)$.Since $P$ is equidistant from $A, B, C$,we have $PA^2 = PB^2$ and $PB^2 = PC^2$.From $PA^2 = PB^2$:$(x-1)^2 + (y-3)^2 = (x+3)^2 + (y-5)^2$$x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 6x + 9 + y^2 - 10y + 25$$-2x - 6y + 10 = 6x - 10y + 34$$8x - 4y + 24 = 0 \Rightarrow 2x - y + 6 = 0$ ... $(i)$From $PB^2 = PC^2$:$(x+3)^2 + (y-5)^2 = (x-5)^2 + (y+1)^2$$x^2 + 6x + 9 + y^2 - 10y + 25 = x^2 - 10x + 25 + y^2 + 2y + 1$$16x - 12y + 8 = 0 \Rightarrow 4x - 3y + 2 = 0$ ... (ii)Solving $(i)$ and (ii):Multiply $(i)$ by $2$: $4x - 2y + 12 = 0$ ... (iii)Subtract (ii) from (iii): $(-2y - (-3y)) + (12 - 2) = 0$ $\Rightarrow y + 10 = 0$ $\Rightarrow y = -10$Substitute $y = -10$ into $(i)$: $2x - (-10) + 6 = 0$ $\Rightarrow 2x + 16 = 0$ $\Rightarrow x = -8$Thus,$P$ is $(-8, -10)$.$PA = \sqrt{(-8-1)^2 + (-10-3)^2} = \sqrt{(-9)^2 + (-13)^2} = \sqrt{81 + 169} = \sqrt{250} = 5 \sqrt{10}$.

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

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