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Step-by-Step Solution
Step 1: Express the Locus Condition Mathematically
Let the moving point be $P(h, k)$. According to the problem, the sum of the squares of its distances from the points $(1,2)$ and $(-2,1)$ is 14. Hence:
$$(h - 1)^2 + (k - 2)^2 + (h + 2)^2 + (k - 1)^2 = 14.$$
Step 2: Expand and Simplify
Expand each term:
$$(h - 1)^2 = h^2 - 2h + 1,$$
$$(k - 2)^2 = k^2 - 4k + 4,$$
$$(h + 2)^2 = h^2 + 4h + 4,$$
$$(k - 1)^2 = k^2 - 2k + 1.$$
Summing them up gives:
$$h^2 - 2h + 1 + k^2 - 4k + 4 + h^2 + 4h + 4 + k^2 - 2k + 1 = 14.$$
Combine like terms:
$$2h^2 + 2k^2 + (-2h + 4h) + (-4k - 2k) + (1 + 4 + 4 + 1) = 14.$$
$$2h^2 + 2k^2 + 2h - 6k + 10 = 14.$$
$$2h^2 + 2k^2 + 2h - 6k - 4 = 0.$$
Divide throughout by 2 to simplify:
$$h^2 + k^2 + h - 3k - 2 = 0.$$
Thus, the locus $f(x, y) = 0$ is:
$$x^2 + y^2 + x - 3y - 2 = 0.$$
Step 3: Find Intersections with the Axes
3.1 Intersection with the x-axis
On the $x$-axis, $y = 0$. Substitute $y = 0$ into the locus:
$$x^2 + (0)^2 + x - 3(0) - 2 = 0,$$
which simplifies to
$$x^2 + x - 2 = 0.$$
Solve for $x$:
$$x^2 + x - 2 = 0.$$
We can factor or use the quadratic formula. By factoring:
$$(x + 2)(x - 1) = 0.$$
Hence,
$$x = -2 \quad\text{or}\quad x = 1.$$
Thus, the points of intersection on the x-axis are $A(-2, 0)$ and $B(1, 0)$.
3.2 Intersection with the y-axis
On the $y$-axis, $x = 0$. Substitute $x = 0$ into the locus:
$$(0)^2 + y^2 + (0) - 3y - 2 = 0,$$
which simplifies to
$$y^2 - 3y - 2 = 0.$$
Solve for $y$:
$$y^2 - 3y - 2 = 0.$$
Apply the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -3$, and $c = -2$:
$$y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}.$$
Thus, the points of intersection on the y-axis are:
$$C\left(0, \frac{3 + \sqrt{17}}{2}\right) \quad \text{and} \quad D\left(0, \frac{3 - \sqrt{17}}{2}\right).$$
Step 4: Determine the Required Area
The quadrilateral $ACBD$ is formed by the points of intersection $A(-2, 0)$, $B(1, 0)$, $C\left(0, \frac{3 + \sqrt{17}}{2}\right)$, and $D\left(0, \frac{3 - \sqrt{17}}{2}\right)$.
One convenient way to find its area is to notice that $ACBD$ can be viewed as a rectangle in terms of absolute lengths along the axes. Specifically, the length along the $x$-axis side is $|x_A| + |x_B| = |-2| + |1| = 2 + 1 = 3,$ and the length along the $y$-axis side is $|y_C| + |y_D| = \left|\frac{3 + \sqrt{17}}{2}\right| + \left|\frac{3 - \sqrt{17}}{2}\right|.$
Notice $\frac{3 + \sqrt{17}}{2}$ is positive and $\frac{3 - \sqrt{17}}{2}$ is also positive if $3 > \sqrt{17} \approx 4.123$, but here $3 - \sqrt{17}$ is negative. Thus the absolute value sum is:
$$\left|\frac{3 + \sqrt{17}}{2}\right| + \left|\frac{3 - \sqrt{17}}{2}\right| = \frac{3 + \sqrt{17}}{2} + \left|\frac{3 - \sqrt{17}}{2}\right|.$$
Since $\frac{3 - \sqrt{17}}{2}$ is negative, its absolute value is $\frac{\sqrt{17} - 3}{2}$. Hence,
$$\text{Sum} = \frac{3 + \sqrt{17}}{2} + \frac{\sqrt{17} - 3}{2} = \frac{(3 + \sqrt{17}) + (\sqrt{17} - 3)}{2} = \frac{2\sqrt{17}}{2} = \sqrt{17}.$$
Thus, the total length along the $y$ direction is $\sqrt{17}$. As a result, the area of the quadrilateral, viewed as a product of these distances (since it's effectively formed by bounding lines on both axes), is:
$$\text{Area} = \frac{1}{2} \times \bigl(\text{distance on x-axis}\bigr)\times\bigl(\text{distance on y-axis}\bigr).$$
However, this is a known result for the area of a rectangle formed by intercepts but effectively halved when joined as a quadrilateral around the origin. In direct usage:
$$\text{Area} = \frac{1}{2} ( (|-2| + |1|) \times (|y_C| + |y_D| ) ) = \frac{1}{2} (3 \times \sqrt{17}) = \frac{3\sqrt{17}}{2}.$$
Step 5: Final Answer
Therefore, the area of the quadrilateral $ACBD$ is
$$\displaystyle \frac{3\sqrt{17}}{2}.$$
