JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • 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)            \[{{x}^{2}}-2xy\sec 2\alpha +{{y}^{2}}=0\]

    Correct Answer: D

    Solution :

               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 \[{{y}^{2}}-2xy\sec 2\alpha +{{x}^{2}}=0\]


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