MHT CET · Maths · Pair of Lines
If the acute angle between the lines \(x^{2}-4 x y+y^{2}=0\) is \(\tan ^{-1}(k)\), then \(\mathrm{k}=\)
- A \(\frac{1}{\sqrt{3}}\)
- B \(\sqrt{3}\)
- C \(\frac{1}{6}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Acute angle between the lines is given by
\(\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
\(\begin{array}{l}
\text { Here } a=1,2 h=-4 \Rightarrow h=-2, b=1 \\
\therefore \tan \theta=\left|\frac{2 \sqrt{4-1}}{1+1}\right| \Rightarrow \tan \theta=\sqrt{3} \\
\theta=\tan ^{-1}(\sqrt{3}) \Rightarrow k=\sqrt{3}
\end{array}\)
\(\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
\(\begin{array}{l}
\text { Here } a=1,2 h=-4 \Rightarrow h=-2, b=1 \\
\therefore \tan \theta=\left|\frac{2 \sqrt{4-1}}{1+1}\right| \Rightarrow \tan \theta=\sqrt{3} \\
\theta=\tan ^{-1}(\sqrt{3}) \Rightarrow k=\sqrt{3}
\end{array}\)
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