If a line L is perpendicular to the line 5x - | vedclass

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(B) The equation of a line perpendicular to $5x - y = 1$ is of the form $x + 5y = c$.The intercepts of this line on the coordinate axes are found by s

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$\frac{5}{\sqrt{13}}$

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(B) The equation of a line perpendicular to $5x - y = 1$ is of the form $x + 5y = c$.The intercepts of this line on the coordinate axes are found by setting $y=0$ and $x=0$:For $y=0$,$x=c$. So,the $x$-intercept is $(c, 0)$.For $x=0$,$5y=c$,so $y=c/5$. The $y$-intercept is $(0, c/5)$.The area of the triangle formed by this line and the coordinate axes is given by $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}| = 5$.$\frac{1}{2} 	imes |c| 	imes |\frac{c}{5}| = 5$$\frac{c^2}{10} = 5$ $\Rightarrow c^2 = 50$ $\Rightarrow c = \pm 5\sqrt{2}$.Thus,the equation of line $L$ is $x + 5y = \pm 5\sqrt{2}$.The distance $d$ between the parallel lines $x + 5y = c$ and $x + 5y = 0$ is given by $d = \frac{|c - 0|}{\sqrt{1^2 + 5^2}} = \frac{|c|}{\sqrt{26}}$.Substituting $c = \pm 5\sqrt{2}$:$d = \frac{5\sqrt{2}}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{2} 	imes \sqrt{13}} = \frac{5}{\sqrt{13}}$.

If a line $L$ is perpendicular to the line $5x - y = 1$,and the area of the triangle formed by the line $L$ and the coordinate axes is $5$,then the distance of line $L$ from the line $x + 5y = 0$ is

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