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(C) The given line is $x+y=0$,which has a slope $m_{1} = -1$. Let the slope of the required lines be $m$. Since the angle between the lines is $30^{\c

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$x^{2}+4xy+y^{2}=0$

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(C) The given line is $x+y=0$,which has a slope $m_{1} = -1$. Let the slope of the required lines be $m$. Since the angle between the lines is $30^{\circ}$,we use the formula $	an 	heta = \left| \frac{m - m_{1}}{1 + m m_{1}} ight|$. Substituting the values: $	an 30^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} ight| = \left| \frac{m+1}{1-m} ight|$. $\frac{1}{\sqrt{3}} = \left| \frac{m+1}{1-m} ight|$. Squaring both sides: $\frac{1}{3} = \frac{(m+1)^{2}}{(1-m)^{2}}$. $(1-m)^{2} = 3(m+1)^{2} \Rightarrow 1 - 2m + m^{2} = 3(m^{2} + 2m + 1)$. $1 - 2m + m^{2} = 3m^{2} + 6m + 3$. $2m^{2} + 8m + 2 = 0 \Rightarrow m^{2} + 4m + 1 = 0$. Substituting $m = \frac{y}{x}$: $\left( \frac{y}{x} ight)^{2} + 4\left( \frac{y}{x} ight) + 1 = 0$. Multiplying by $x^{2}$: $y^{2} + 4xy + x^{2} = 0$ or $x^{2} + 4xy + y^{2} = 0$.

The joint equation of two lines passing through the origin,each of which makes an angle of $30^{\circ}$ with the line $x+y=0$,is

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