AP EAMCET · Maths · Pair of Lines
The product of the perpendicular distances drawn from the origin to the pair of straight lines \(6 x^2-5 x y-6 y^2+x+5 y-1=0\) is
- A \(1\)
- B \(\frac{1}{12}\)
- C \(\frac{1}{13}\)
- D \(13\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{13}\)
Step-by-step Solution
Detailed explanation
\(P = \frac{|c|}{\sqrt{(a-b)^2 + 4h^2}}\) \(a=6, b=-6, h=-\frac{5}{2}, c=-1\) \(P = \frac{|-1|}{\sqrt{(6 - (-6))^2 + 4(-\frac{5}{2})^2}}\) \(P = \frac{1}{\sqrt{(12)^2 + 4(\frac{25}{4})}}\) \(P = \frac{1}{\sqrt{144 + 25}}\) \(P = \frac{1}{\sqrt{169}}\) \(P = \frac{1}{13}\)
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