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(D) Let the slope of the line be $m$. The equation of the line passing through $(-5, 4)$ is $y-4=m(x+5)$,which simplifies to $mx-y+5m+4=0$. The distan

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$2x-y+14=0$

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(D) Let the slope of the line be $m$. The equation of the line passing through $(-5, 4)$ is $y-4=m(x+5)$,which simplifies to $mx-y+5m+4=0$. The distance between the parallel lines $x+2y+1=0$ and $x+2y-1=0$ is $d = \frac{|1 - (-1)|}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{5}}$. Since the intercept made by the line between these two parallel lines is equal to the distance between them,the line must be perpendicular to the given parallel lines. The slope of the lines $x+2y+1=0$ and $x+2y-1=0$ is $m_1 = -\frac{1}{2}$. Therefore,the slope of the required line is $m = -\frac{1}{m_1} = 2$. Substituting $m=2$ into the equation $y-4=m(x+5)$,we get $y-4=2(x+5)$,which simplifies to $2x-y+14=0$.

If a straight line passes through the point $(-5, 4)$ and makes an intercept of length $\frac{2}{\sqrt{5}}$ between the lines $x+2y+1=0$ and $x+2y-1=0$,then the equation of that line is

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