JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer Slope of a line passing through \[P(2,3)\]and intersecting the line, \[x+y=7\]at a distance of 4 units from P, is [JEE Main 9-4-2019 Morning]

    A) \[\frac{\sqrt{5}-1}{\sqrt{5}+1}\]                  

    B) \[\frac{1-\sqrt{5}}{1+\sqrt{5}}\]

    C) \[\frac{1-\sqrt{7}}{1+\sqrt{7}}\]

    D) \[\frac{\sqrt{7}-1}{\sqrt{7}+1}\]

    Correct Answer: C

    Solution :

    \[x=2+r\cos \theta \]           \[y=3+r\sin \theta \]           \[\Rightarrow 2+r\cos \theta +3+r\sin \theta =7\]           \[\Rightarrow r(\cos \theta +\sin \theta )=2\]           \[\Rightarrow \sin \theta +\cos \theta =\frac{2}{r}=\frac{2}{\pm 4}=\pm \frac{1}{2}\]           \[\Rightarrow 1+\sin 2\theta =\frac{1}{4}\]\[\Rightarrow \sin 2\theta =-\frac{3}{4}\]           \[\Rightarrow \frac{2m}{1+{{m}^{2}}}=-\frac{3}{4}\]\[\Rightarrow 3{{m}^{2}}+8m+3=0\]           \[\Rightarrow m=\frac{-4\pm \sqrt{7}}{1-7}\]           \[\frac{1-\sqrt{7}}{1+\sqrt{7}}=\frac{{{\left( 1-\sqrt{7} \right)}^{2}}}{1-7}=\frac{8-2\sqrt{7}}{-6}=\frac{-4+\sqrt{7}}{3}\]            

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