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Let a, b, $x, y$ be real numbers such that $a^{2}+b^{2}=81, x^{2}+y^{2}=121$ and $a x+b y=99$. Then the set of all possible values of ay $-\mathrm{bx}$ is -
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Verified Answer
The correct answer is:
$\{0\}$
$a^{2}+b^{2}=81$
$x^{2}+y^{2}=121$
$a x+b y=99$
$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)=(81)(121)$ ...(1)
and $(a x+b y)^{2}=(99)^{2}$ ...(2)
$(1)-(2)$
(ay $-b x)^{2}=0$
ay $-b x=0$
$x^{2}+y^{2}=121$
$a x+b y=99$
$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)=(81)(121)$ ...(1)
and $(a x+b y)^{2}=(99)^{2}$ ...(2)
$(1)-(2)$
(ay $-b x)^{2}=0$
ay $-b x=0$
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