Let a, b, x, y be real numbers such that $a^{2}+b^{2}=25, x^{2}+y^{2}=169,$ and $a x+b y=65$. If $k=a y-b x$ , then

  1. $\mathrm{k}=0$
  2. $0<\mathrm{k} \leq \frac{5}{13}$
  3. $\mathrm{k}=\frac{5}{13}$
  4. $\mathrm{k}>\frac{5}{13}$
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Correct Answer: Option: 1


We can take a=5, b=0, x=13 and y=0 as values which satisfies all three equations.

Hence, $k=ay-bx=5\times 0-0\times 13=0$

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