If P is the set of all real numbers alpha suc | vedclass

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(B) The given pair of straight lines is $3x^2+7xy+2y^2=0$. Factorizing the equation: $3x^2+6xy+xy+2y^2 = 3x(x+2y)+y(x+2y) = (3x+y)(x+2y) = 0$. The two

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$-\frac{14}{3}$

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(B) The given pair of straight lines is $3x^2+7xy+2y^2=0$. Factorizing the equation: $3x^2+6xy+xy+2y^2 = 3x(x+2y)+y(x+2y) = (3x+y)(x+2y) = 0$. The two lines are $L_1: 3x+y=0$ and $L_2: x+2y=0$. The lengths of the perpendiculars from $(\alpha, 1)$ to $L_1$ and $L_2$ are $p_1 = \frac{|3\alpha+1|}{\sqrt{3^2+1^2}} = \frac{|3\alpha+1|}{\sqrt{10}}$ and $p_2 = \frac{|\alpha+2|}{\sqrt{1^2+2^2}} = \frac{|\alpha+2|}{\sqrt{5}}$. The product of the lengths is $p_1 p_2 = \frac{|3\alpha+1| |\alpha+2|}{\sqrt{50}} = \frac{|3\alpha^2+7\alpha+2|}{5\sqrt{2}}$. Given $p_1 p_2 = \frac{\sqrt{2}}{5}$,we have $\frac{|3\alpha^2+7\alpha+2|}{5\sqrt{2}} = \frac{\sqrt{2}}{5} \implies |3\alpha^2+7\alpha+2| = 2$. Case $1$: $3\alpha^2+7\alpha+2 = 2 \implies 3\alpha^2+7\alpha = 0 \implies \alpha(3\alpha+7) = 0$. So,$\alpha = 0$ or $\alpha = -\frac{7}{3}$. Case $2$: $3\alpha^2+7\alpha+2 = -2 \implies 3\alpha^2+7\alpha+4 = 0 \implies (3\alpha+4)(\alpha+1) = 0$. So,$\alpha = -\frac{4}{3}$ or $\alpha = -1$. The set $P = \{0, -\frac{7}{3}, -\frac{4}{3}, -1\}$. The sum of the elements in $P$ is $0 - \frac{7}{3} - \frac{4}{3} - 1 = -\frac{11}{3} - 1 = -\frac{14}{3}$.

If $P$ is the set of all real numbers $\alpha$ such that the product of the lengths of perpendiculars from $(\alpha, 1)$ to the pair of straight lines $3x^2+7xy+2y^2=0$ is $\frac{\sqrt{2}}{5}$,then the sum of the elements in $P$ is

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