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(B) The slope of the line $x+2y+7=0$ is $m_1 = -\frac{1}{2}$. The line perpendicular to this has a slope $m = -\frac{1}{m_1} = 2$. Thus,the equation o

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

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(B) The slope of the line $x+2y+7=0$ is $m_1 = -\frac{1}{2}$. The line perpendicular to this has a slope $m = -\frac{1}{m_1} = 2$. Thus,the equation of the first line is $y = 2x$,or $2x-y=0$.The slope of the line $3x+4y+5=0$ is $m_2 = -\frac{3}{4}$. The line parallel to this has the same slope $m = -\frac{3}{4}$. Thus,the equation of the second line is $y = -\frac{3}{4}x$,or $3x+4y=0$.The combined 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 line is parallel to the line $3x+4y+5=0$,then the equation of that pair of lines is:

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