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  • Line  makes angle <img alt="\frac{\pi}{4} \text { with } \mathrm{x}

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

Option: 1

a^2+b^2=1


Option: 2

a^2+b^2=2


Option: 3

2\left(a^2+b^2\right)=1


Option: 4

none of these


Answers (1)

best_answer

Lines \mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}  and x \sin \alpha-y \cos \alpha=0 are mutually perpendicular. Thus a x+b y+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-\mathrm{y} \cos \alpha= \pm(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})
\Rightarrow \quad x(\cos \alpha-\sin \alpha)+y(\sin \alpha+\cos \alpha)=p
\text { or } \quad \mathrm{x}(\sin \alpha+\cos \alpha)-\mathrm{y}(\cos \alpha-\sin \alpha)=\mathrm{p}
Comparing these lines with ax + by + p = 0, we get
\frac{a}{\cos \alpha-\sin \alpha}=\frac{b}{\sin \alpha+\cos \alpha}=1
\Rightarrow \quad a^2+b^2=2
\text { or } \quad \frac{a}{\sin \alpha+\cos \alpha}=\frac{b}{\sin \alpha-\cos \alpha}=1
\Rightarrow \quad a^2+b^2=2

Posted by

Ritika Jonwal

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