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  • Angles made with the -axis by the two lines through the point (1,2) and cutting the line <img alt="\ma

Angles made with the \mathrm{x}-axis by the two lines through the point (1,2) and cutting the line \mathrm{x+y=4} at a distance \mathrm{{\frac{1}{3}} \sqrt{6}} from the point (1,2) are

Option: 1

\mathrm{\frac{\pi}{6}\; and\: \frac{\pi}{3}}


Option: 2

\mathrm{\frac{\pi}{8}\; and\: \frac{3\pi}{8}}


Option: 3

\mathrm{\frac{\pi}{12}\; and\: \frac{5\pi}{12}}


Option: 4

none of these


Answers (1)

best_answer

Any line through (1,2) can be written as \mathrm{\frac{x-1}{\cos \theta}=\frac{y-2}{\sin \theta}=r}

where \mathrm{\theta} is the angle which this line makes with positive direction of \mathrm{x}-axis. Any point on this line is
\mathrm{(r \cos \theta+1, r \cos \theta+2)\: when \: |r|=\frac{1}{3} \sqrt{6}}, this point lies on the line \mathrm{x+y=4}.

\mathrm{i.e. \, r \cos \theta+1+r \sin \theta+2=4},

\mathrm{ |r|=\frac{1}{3} \sqrt{6} \Rightarrow r(\cos \theta+\sin \theta)=1,|r|=\frac{1}{3} \sqrt{6} }

\mathrm{ \Rightarrow r^{2}(1+2 \sin \theta \cos \theta)=1, r^{2}=\frac{6}{9} \Rightarrow 1+\sin 2 \theta=\frac{1}{r^{2}}=\frac{9}{6} \Rightarrow \mathrm{ \sin 2 \theta=\frac{1}{2}}}

\mathrm{ \Rightarrow 2 \theta=\frac{\pi}{6} \text { or } \frac{5 \pi}{6} \Rightarrow \theta=\frac{\pi}{12} \text { or } \frac{5 \pi}{12}}

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