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Question Number 211180 by Durganand last updated on 30/Aug/24

Answered by mr W last updated on 30/Aug/24

$$(ax+by+c)(\frac{x}{a}+\frac{y}{b}+d)$$$$x^2+y^2+(\frac{a}{b}+\frac{b}{a})xy+(ad+\frac{c}{a})x+(bd+\frac{c}{b})+cd$$$$=x^2+y^2-\frac{2}{\sin \theta }xy$$$$\Rightarrow cd=0$$$$\Rightarrow ad+\frac{c}{a}=0$$$$\Rightarrow bd+\frac{c}{b}=0$$$$\Rightarrow \frac{a}{b}+\frac{b}{a}=-\frac{2}{\sin \theta }$$$$\Rightarrow c=d=0$$$$say-\frac{a}{b}=k=\tan {\alpha }_1$$$$\Rightarrow k+\frac{1}{k}=\frac{2}{\sin \theta }$$$$\Rightarrow k^2-\frac{2k}{\sin \theta }+1=0$$$$\Rightarrow k=\frac{1\pm \cos \theta }{\sin \theta }=\tan \frac{\theta }{2}or\tan (\frac{\pi }{2}-\frac{\theta }{2})$$$$\Rightarrow {\alpha }_1=\frac{\theta }{2},{\alpha }_2=\frac{\pi }{2}-\frac{\theta }{2}$$$$\Rightarrow {\alpha }_1=\frac{\pi }{2}-\frac{\theta }{2},{\alpha }_2=\frac{\theta }{2}$$$$line1:ax+by=0\Rightarrow y=-\frac{a}{b}x=kx=\left(\tan \frac{\theta }{2}\right)x$$$$line2:\frac{x}{a}+\frac{y}{b}=0\Rightarrow y=-\frac{b}{a}x=\frac{x}{k}=\frac{x}{\tan \frac{\theta }{2}}$$$$anglebetweenthem=\frac{\pi }{2}\pm \theta ✓$$

Commented by mr W last updated on 30/Aug/24

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