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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Aicharya · Accepted Answer

$6 x^2-5 x y+y^2=0$ $\begin{aligned} & \Rightarrow \quad\left(\frac{y}{x}\right)^2-5\left(\frac{y}{x}\right)+6=0 \ & \Rightarrow \quad\left(\frac{y}{x}-3\right)\left(\frac{y}{x}-2\right)=0 \ \end{aligned}$ $\Rightarrow \quad y=3 x$ and $y=2 x$ are straight lines $\tan \alpha=3, \tan \beta=2$ $\tan (\alpha-\beta)=\frac{3-2}{1+6}=\frac{1}{7}$

If the pair of straight lines \(6 x^2-5 x y+y^2=0\) makes angles \(\alpha\) and \(\beta\) with the…

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