• # question_answer The equation of the pair of straight lines, each of which makes an angle $\alpha$with the line $y=x$, is            [MP PET 1990] A)            ${{x}^{2}}+2xy\sec 2\alpha +{{y}^{2}}=0$ B)  ${{x}^{2}}+2xy\,\text{cosec}\,2\alpha +{{y}^{2}}=0$      C)            ${{x}^{2}}-2xy\,\text{cosec}\,2\alpha +{{y}^{2}}=0$          D)            ${{x}^{2}}-2xy\sec 2\alpha +{{y}^{2}}=0$

Any line through the origin is$y=mx$. If it makes an angle $\alpha$with the line$y=x$, then we should have            $\tan \alpha =\pm \left\{ \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right\}=\pm \frac{(m-1)}{1+m}$            or ${{(1+m)}^{2}}{{\tan }^{2}}\alpha ={{(m-1)}^{2}}$            $\Rightarrow {{m}^{2}}-2m\left\{ \frac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha } \right\}+1=0$            $\Rightarrow {{m}^{2}}-2m\sec 2\alpha +1=0$,      $\left\{ \because \frac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }=\sec 2\alpha \right\}$            But $m=\frac{y}{x},$hence on eliminating m, we get the required equation ${{y}^{2}}-2xy\sec 2\alpha +{{x}^{2}}=0$