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(B) Let the two lines passing through the origin be $L_1$ and $L_2$.Line $L_1$ is perpendicular to $x+2y+7=0$. The slope of $x+2y+7=0$ is $m = -1/2$.

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$6x^2+5xy-4y^2=0$

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(B) Let the two lines passing through the origin be $L_1$ and $L_2$.Line $L_1$ is perpendicular to $x+2y+7=0$. The slope of $x+2y+7=0$ is $m = -1/2$. Thus,the slope of $L_1$ is $m_1 = 2$. The equation of $L_1$ is $y = 2x$,or $2x-y=0$.Line $L_2$ is parallel to $3x+4y+5=0$. The slope of $3x+4y+5=0$ is $m = -3/4$. Thus,the slope of $L_2$ is $m_2 = -3/4$. The equation of $L_2$ is $y = -3/4x$,or $3x+4y=0$.The joint equation of the pair of lines is $(2x-y)(3x+4y) = 0$.Expanding this,we get $6x^2 + 8xy - 3xy - 4y^2 = 0$,which simplifies to $6x^2 + 5xy - 4y^2 = 0$.

The equation of a pair of lines is given by a second-degree homogeneous equation. If one of them is perpendicular to the line $x+2y+7=0$ and the other is parallel to the line $3x+4y+5=0$,then the equation of that pair of lines is:

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