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If the pair of straight lines \(6 x^2-5 x y+y^2=0\) makes angles \(\alpha\) and \(\beta\) with the \(X\)-axis, then \(\tan (\alpha-\beta)=\)
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Verified Answer
The correct answer is:
\(\frac{1}{7}\)
\(6 x^2-5 x y+y^2=0\)
\(\begin{aligned}
& \Rightarrow \quad\left(\frac{y}{x}\right)^2-5\left(\frac{y}{x}\right)+6=0 \\
& \Rightarrow \quad\left(\frac{y}{x}-3\right)\left(\frac{y}{x}-2\right)=0 \\
\end{aligned}\)
\(\Rightarrow \quad y=3 x\) and \(y=2 x\) are straight lines \(\tan \alpha=3, \tan \beta=2\)
\(\tan (\alpha-\beta)=\frac{3-2}{1+6}=\frac{1}{7}\)
\(\begin{aligned}
& \Rightarrow \quad\left(\frac{y}{x}\right)^2-5\left(\frac{y}{x}\right)+6=0 \\
& \Rightarrow \quad\left(\frac{y}{x}-3\right)\left(\frac{y}{x}-2\right)=0 \\
\end{aligned}\)
\(\Rightarrow \quad y=3 x\) and \(y=2 x\) are straight lines \(\tan \alpha=3, \tan \beta=2\)
\(\tan (\alpha-\beta)=\frac{3-2}{1+6}=\frac{1}{7}\)
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