A point P(x, y) moves such that the sum of th | vedclass

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(B) Let $P = (x, y)$. The sum of the squares of the distances from $(1, 2)$ and $(-2, 1)$ is $14$: $(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$ Expand

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$\frac{3 \sqrt{17}}{2}$

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(B) Let $P = (x, y)$. The sum of the squares of the distances from $(1, 2)$ and $(-2, 1)$ is $14$: $(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$ Expanding the terms: $(x^2 - 2x + 1) + (y^2 - 4y + 4) + (x^2 + 4x + 4) + (y^2 - 2y + 1) = 14$ $2x^2 + 2y^2 + 2x - 6y + 10 = 14$ $2x^2 + 2y^2 + 2x - 6y - 4 = 0$ $x^2 + y^2 + x - 3y - 2 = 0$ For $x$-intercepts,set $y = 0$: $x^2 + x - 2 = 0$ $\Rightarrow (x+2)(x-1) = 0$ $\Rightarrow x = -2, 1$. So $A = (-2, 0)$ and $B = (1, 0)$. For $y$-intercepts,set $x = 0$: $y^2 - 3y - 2 = 0$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $y = \frac{3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{17}}{2}$. So $C = (0, \frac{3+\sqrt{17}}{2})$ and $D = (0, \frac{3-\sqrt{17}}{2})$. The quadrilateral $ACBD$ is a rhombus-like shape with diagonals along the axes. The length of the horizontal diagonal $AB = |1 - (-2)| = 3$. The length of the vertical diagonal $CD = |\frac{3+\sqrt{17}}{2} - \frac{3-\sqrt{17}}{2}| = \sqrt{17}$. The area of the quadrilateral is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 3 	imes \sqrt{17} = \frac{3\sqrt{17}}{2}$.

$A$ point $P(x, y)$ moves such that the sum of the squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is $14$. Let $f(x, y) = 0$ be the locus of $P$,which intersects the $x$-axis at points $A, B$ and the $y$-axis at points $C, D$. Then the area of the quadrilateral $ACBD$ is equal to:

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