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

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9/2

$\frac{3 \sqrt{17}}{4}$

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

answer is B.

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Detailed Solution

$\begin{array}{l} (x - 1)^{2} + (y - 2)^{2} + (x + 2)^{2} + (y - 1)^{2} = 14 \\ \Rightarrow x^{2} + y^{2} + x - 3 y - 2 = 0 \\ Put x = 0 \\ \Rightarrow y^{2} - 3 y - 2 = 0 \\ \Rightarrow y = \frac{3 \pm \sqrt{17}}{2} \\ Put y = 0 \\ \Rightarrow x^{2} + x - 2 = 0 \\ (x + 2) (x - 1) = 0 \\ ∴ A (- 2, 0), B (1, 0), C (0, \frac{3 + \sqrt{17}}{2}), D (0, \frac{3 - \sqrt{17}}{2}) \\ Area = \frac{1}{2} \cdot 3 \cdot \sqrt{17} = \frac{3 \sqrt{17}}{2} \end{array}$