If the pair of straight lines 6x^2 - 5xy + y^ | vedclass

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(B) Given the equation of the pair of straight lines is $6x^2 - 5xy + y^2 = 0$. Dividing by $x^2$ (assuming $x 
eq 0$),we get: $\left(\frac{y}{x}ig

Vedclass · Accepted Answer

$\frac{1}{7}$

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(B) Given the equation of the pair of straight lines is $6x^2 - 5xy + y^2 = 0$. Dividing by $x^2$ (assuming $x 
eq 0$),we get: $\left(\frac{y}{x}ight)^2 - 5\left(\frac{y}{x}ight) + 6 = 0$. Let $m = \frac{y}{x}$,then $m^2 - 5m + 6 = 0$. Factoring the quadratic equation: $(m - 3)(m - 2) = 0$. Thus,the slopes of the lines are $m_1 = 	an \alpha = 3$ and $m_2 = 	an \beta = 2$. Using the formula for the tangent of the difference of two angles: $	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$. Substituting the values: $	an(\alpha - \beta) = \frac{3 - 2}{1 + (3)(2)} = \frac{1}{1 + 6} = \frac{1}{7}$.

If the pair of straight lines $6x^2 - 5xy + y^2 = 0$ makes angles $\alpha$ and $\beta$ with the $X$-axis,then $\tan(\alpha - \beta) = $

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