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JEE Mains · Maths · STD 11 - 9. straight line

बिंदु \((1,3)\) से होकर जाने वाली तथा सरल रेखा \(y +1=3 \sqrt{2} x\) से \(\tan ^{-1}(\sqrt{2})\) का कोण बनाने वाली रेखाओं में से एक का समीकरण है

  1. A \(4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0\)
  2. B \(5 \sqrt{2} x +4 y -(15+4 \sqrt{2})=0\)
  3. C \(4 \sqrt{2} x+5 y-4 \sqrt{2}=0\)
  4. D \(4 \sqrt{2} x-5 y-(5+4 \sqrt{2})=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0\)

Step-by-step Solution

Detailed explanation

\(y=m x+c\) \(3= m + c\) \(\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|\) \(=6 m+\sqrt{2}=m-3 \sqrt{2}\) \(=\sin =-4 \sqrt{2} \rightarrow m =\frac{-4 \sqrt{2}}{5}\) \(=6 m-\sqrt{2}=m-3 \sqrt{2}\) \(=7 m -2 \sqrt{2} \rightarrow m =\frac{2 \sqrt{2}}{7}\) According to…
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