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Line \mathrm{ a x+b y+p=0} makes angle \mathrm{\frac{\pi}{4}} with \mathrm{x \cos a+y \sin \alpha=p, p \in R+}. If these lines and the line \mathrm{x\: \sin \alpha-y \cos \alpha=0} are concurrent then -
 

Option: 1

\mathrm{a_2+b_2=1}
 


Option: 2

\mathrm{\mathrm{a}_2+\mathrm{b} 2=2}
 


Option: 3

\mathrm{2\left(\mathrm{a}_2+\mathrm{b} 2\right)=1}

 


Option: 4

\text{None of these}


Answers (1)

best_answer

Lines \mathrm{\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\: \: and \: \: \mathrm{x} \sin \alpha-\mathrm{y} \cos \alpha=0}are mutually perpendicular. Thus \mathrm{\mathrm{ax}+\mathrm{by}+\mathrm{p}=0} will be equally inclined to these line and would be the angle bisector of these lines. Now equations of angle bisectors is,

\mathrm{x \sin \alpha-y \cos \alpha= \pm(x \cos \alpha+y \sin \alpha-p) }

\mathrm{ \Rightarrow x(\cos \alpha-\sin \alpha)+y(\sin \alpha+\cos \alpha)=p }

\mathrm{ \text { or } x(\sin \alpha+\cos \alpha)-y(\cos \alpha-\sin \alpha)=p }

Comparing these lines with  \mathrm{ a x+b y+p=0 }, we get

\mathrm{ \frac{a}{\cos \alpha-\sin \alpha}=\frac{b}{\sin \alpha+\cos \alpha}=1 }

\mathrm{ \Rightarrow a_2+b_2=2 }

\mathrm{ \text { or } \frac{a}{\sin \alpha+\cos \alpha}=\frac{b}{\sin \alpha-\cos \alpha}=1 }

\mathrm{ \Rightarrow a_2+b_2=2 }

Hence option 2 is correct.





 

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Nehul

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