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A point P moves so that the sum of squares of its distances from the points $\mathrm{(1,2)}$ and $\mathrm{(-2,1)}$ is $\mathrm{14}$ . Let $\mathrm{f(x, y)=0}$ be the locus of $\mathrm{p,}$ which intersects the x-axis at the points $\mathrm{A,B}$ and the y-axis at the points $\mathrm{C,D}$ . Then the area of the quadrilateral $\mathrm{\mathrm{ACBD}}$ is equal to:
Option: 1
$\frac{9}{2}$

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

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

Option: 4
$9$

Answers (1)

best_answer

$\mathrm{\Rightarrow (x-1)^{2}+(y-2)^{2}+(x+2)^{2}+(y-1)^{2}=14} \\$

$\mathrm{\Rightarrow x^{2}+y^{2}+x-3 y-2=0} \\$

$\mathrm{ \text { Put } x=0 }\\$

$\mathrm{\Rightarrow y^{2}-3 y-2=0} \\$

$\mathrm{\Rightarrow y=\frac{3 \pm \sqrt{17}}{2}} \\$

$\mathrm{\Rightarrow x^{2}+x-2=0 \Rightarrow(x+2)(x-1)=0 }\\$

$\mathrm{\therefore A(-2,0), B(1,0), C\left(0, \frac{3+\sqrt{17}}{2}\right), D\left(0, \frac{3-\sqrt{17}}{2}\right)} \\$

$\mathrm{\text { Area }=\frac{1}{2} \times 3 \times \sqrt{17}=\frac{3 \sqrt{17}}{2} }$

Hence correct option is 2

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SANGALDEEP SINGH

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