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Question Number 21341 by gopikrishnan005@gmail.com last updated on 21/Sep/17
the angle between the straight lines x^2 +4xy+3y^2 =0 is
$${the}\:{angle}\:{between}\:{the}\:{straight}\:{lines}\:{x}^{\mathrm{2}} +\mathrm{4}{xy}+\mathrm{3}{y}^{\mathrm{2}} =\mathrm{0}\:{is} \\ $$
Answered by mrW1 last updated on 21/Sep/17
x^2 +4xy+3y^2 =0  ⇒(x+3y)(x+y)=0  ⇒x+3y=0 or x+y=0  line 1: y=−(1/3)x  θ_1 =−tan^(−1) (1/3)  line 2: y=−x  θ_2 =−tan^(−1) 1=−(π/4)  ⇒Δθ=θ_1 −θ_2 =(π/4)−tan^(−1) (1/3) (≈26.6°)
$$\mathrm{x}^{\mathrm{2}} +\mathrm{4xy}+\mathrm{3y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{3y}\right)\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}+\mathrm{3y}=\mathrm{0}\:\mathrm{or}\:\mathrm{x}+\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{line}\:\mathrm{1}:\:\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x} \\ $$$$\theta_{\mathrm{1}} =−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{line}\:\mathrm{2}:\:\mathrm{y}=−\mathrm{x} \\ $$$$\theta_{\mathrm{2}} =−\mathrm{tan}^{−\mathrm{1}} \mathrm{1}=−\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\Delta\theta=\theta_{\mathrm{1}} −\theta_{\mathrm{2}} =\frac{\pi}{\mathrm{4}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:\left(\approx\mathrm{26}.\mathrm{6}°\right) \\ $$
Answered by $@ty@m last updated on 22/Sep/17

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