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AP EAMCET · Maths · Pair of Lines

If \(\mathrm{P}\) is the set of all real numbers \(\alpha\) such that the product of the lengths of perpendiculars from \((\alpha, 1)\) to the pair of straight lines \(3 x^2+7 x y+2 y^2=0\) is \(\frac{\sqrt{2}}{5}\) then the sum of the elements in \(\mathrm{P}\) is

  1. A \(-\frac{11}{3}\)
  2. B \(-\frac{14}{3}\)
  3. C \(\frac{11}{3}\)
  4. D \(\frac{14}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-\frac{14}{3}\)

Step-by-step Solution

Detailed explanation

\(3x^2+7xy+2y^2=0 \implies (3x+y)(x+2y)=0\) \(L_1: 3x+y=0\), \(L_2: x+2y=0\) \(d_1 = \frac{|3(\alpha)+1|}{\sqrt{3^2+1^2}} = \frac{|3\alpha+1|}{\sqrt{10}}\) \(d_2 = \frac{|\alpha+2(1)|}{\sqrt{1^2+2^2}} = \frac{|\alpha+2|}{\sqrt{5}}\)…