Mathematics - Straight Lines Question with Solution | TestHub

The required line is at a distance of 3 units from the given line and parallel to it. Hence it is $x+2y-4-3\sqrt{5}=0$, cuts x-axis at $C(4+3\sqrt{5},0)$ with slope $\tan \theta =\frac{-1}{2}$. After rotation about C by $30^{\circ }$, slope becomes $m=\tan \left(\theta -30^{\circ }\right)=\frac{-(2+\sqrt{3})}{2\sqrt{3}-1}=\frac{-(4+3\sqrt{3})}{11}\Rightarrow [m]=-1$.