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Question-211180




Question Number 211180 by Durganand last updated on 30/Aug/24
Answered by mr W last updated on 30/Aug/24
(ax+by+c)((x/a)+(y/b)+d)  x^2 +y^2 +((a/b)+(b/a))xy+(ad+(c/a))x+(bd+(c/b))+cd  =x^2 +y^2 −(2/(sin θ))xy  ⇒cd=0  ⇒ad+(c/a)=0  ⇒bd+(c/b)=0  ⇒(a/b)+(b/a)=−(2/(sin θ))  ⇒c=d=0  say −(a/b)=k=tan α_1   ⇒k+(1/k)=(2/(sin θ))  ⇒k^2 −((2k)/(sin θ))+1=0  ⇒k=((1±cos θ)/(sin θ))=tan (θ/2) or tan ((π/2)−(θ/2))  ⇒α_1 =(θ/2), α_2 =(π/2)−(θ/2)  ⇒α_1 =(π/2)−(θ/2), α_2 =(θ/2)  line 1: ax+by=0 ⇒y=−(a/b)x=kx=(tan (θ/2))x  line 2: (x/a)+(y/b)=0 ⇒y=−(b/a)x=(x/k)=(x/(tan (θ/2)))  angle between them =(π/2)±θ ✓
$$\left({ax}+{by}+{c}\right)\left(\frac{{x}}{{a}}+\frac{{y}}{{b}}+{d}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right){xy}+\left({ad}+\frac{{c}}{{a}}\right){x}+\left({bd}+\frac{{c}}{{b}}\right)+{cd} \\ $$$$={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{sin}\:\theta}{xy} \\ $$$$\Rightarrow{cd}=\mathrm{0} \\ $$$$\Rightarrow{ad}+\frac{{c}}{{a}}=\mathrm{0} \\ $$$$\Rightarrow{bd}+\frac{{c}}{{b}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}}{{b}}+\frac{{b}}{{a}}=−\frac{\mathrm{2}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{c}={d}=\mathrm{0} \\ $$$${say}\:−\frac{{a}}{{b}}={k}=\mathrm{tan}\:\alpha_{\mathrm{1}} \\ $$$$\Rightarrow{k}+\frac{\mathrm{1}}{{k}}=\frac{\mathrm{2}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{k}^{\mathrm{2}} −\frac{\mathrm{2}{k}}{\mathrm{sin}\:\theta}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}\pm\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:{or}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\alpha_{\mathrm{1}} =\frac{\theta}{\mathrm{2}},\:\alpha_{\mathrm{2}} =\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\alpha_{\mathrm{1}} =\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}},\:\alpha_{\mathrm{2}} =\frac{\theta}{\mathrm{2}} \\ $$$${line}\:\mathrm{1}:\:{ax}+{by}=\mathrm{0}\:\Rightarrow{y}=−\frac{{a}}{{b}}{x}={kx}=\left(\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right){x} \\ $$$${line}\:\mathrm{2}:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{0}\:\Rightarrow{y}=−\frac{{b}}{{a}}{x}=\frac{{x}}{{k}}=\frac{{x}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${angle}\:{between}\:{them}\:=\frac{\pi}{\mathrm{2}}\pm\theta\:\checkmark \\ $$
Commented by mr W last updated on 30/Aug/24

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