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(A) The given pair of lines is $2x^2+5xy+3y^2=0$. Factoring this,we get $(x+y)(2x+3y)=0$. Thus,the two lines are $x+y=0$ and $2x+3y=0$. The third line

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$\pm 1$

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(A) The given pair of lines is $2x^2+5xy+3y^2=0$. Factoring this,we get $(x+y)(2x+3y)=0$. Thus,the two lines are $x+y=0$ and $2x+3y=0$. The third line is $y=x+c$,or $x-y+c=0$. The vertices of the triangle are the intersection points of these lines: $1$. Intersection of $x+y=0$ and $2x+3y=0$ is the origin $(0,0)$. $2$. Intersection of $x+y=0$ and $x-y+c=0$: Adding the equations,$2x+c=0 \Rightarrow x=-\frac{c}{2}$. Then $y=\frac{c}{2}$. So,$B = (-\frac{c}{2}, \frac{c}{2})$. $3$. Intersection of $2x+3y=0$ and $x-y+c=0$: From the second,$x=y-c$. Substituting into the first,$2(y-c)+3y=0$ $\Rightarrow 5y=2c$ $\Rightarrow y=\frac{2c}{5}$. Then $x=\frac{2c}{5}-c=-\frac{3c}{5}$. So,$C = (-\frac{3c}{5}, \frac{2c}{5})$. The area of the triangle with vertices $(0,0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is $\frac{1}{2} |x_1 y_2 - x_2 y_1|$. Area $= \frac{1}{2} |(-\frac{c}{2})(\frac{2c}{5}) - (-\frac{3c}{5})(\frac{c}{2})| = \frac{1}{2} |-\frac{c^2}{5} + \frac{3c^2}{10}| = \frac{1}{2} |\frac{c^2}{10}| = \frac{c^2}{20}$. Given area $= \frac{1}{20}$,so $\frac{c^2}{20} = \frac{1}{20}$ $\Rightarrow c^2=1$ $\Rightarrow c = \pm 1$.

If the area of the triangle formed by the lines $y=x+c$ and $2x^2+5xy+3y^2=0$ is $\frac{1}{20}$ sq. units,then $c=$

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