If the angle between the two straight lines represented by 2x^​2 + 5xy + 3y² + 7y - 98 = 0 is tan^-1m then m equals to

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$$\begin{gathered} \mathrm{If}{\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0\mathrm{represents}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{straight}\mathrm{lines} \\ \mathrm{then}: \\ \mathrm{Angle}\mathrm{between}\mathrm{lines}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \mathrm{tan\theta }=\frac{2\sqrt{{\mathrm{h}}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}} \\ \mathrm{Given}, \\ 2{\mathrm{x}}^2+5\mathrm{xy}+3{\mathrm{y}}^2+7\mathrm{y}-98=0 \\ 2\mathrm{h}=5 \\ \mathrm{h}=\frac{5}{2} \\ \mathrm{a}=2 \\ \mathrm{b}=3 \\ =2\frac{\sqrt{{\left(\frac{5}{2}\right)}^2-2\left(3\right)}}{5} \\ =\frac{2\sqrt{{\displaystyle \frac{25}{4}}-6}}{5} \\ =\frac{2\sqrt{\frac{1}{4}}}{5}=\frac{1}{5} \\ =\frac{1}{5} \\ \Rightarrow \mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{1}}{\mathbf{5}}\right) \\ \mathbf{hence}\mathbf{,}\mathbf{ }\mathbf{m}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}} \end{gathered}$$

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