Question

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

• A. $\frac{7}{\sqrt{5}}$
• B. $\frac{5}{\sqrt{13}}$
• C. $\frac{7}{\sqrt{13}}$
• D. $\frac{5}{\sqrt{7}}$

Answer 1

Answer: (B)

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

$y=5x-1$ is perpendicular to $L.$
Hence the slope of the required line will be equal to $\frac{-1}{5}$.
Therefore the equation of the required line will be of the type
$x+5y=c$ ...(i)
$\frac{x}{c}+\frac{y}{\frac{c}{5}}=1$.
Hence the corresponding intercepts are
$(c,\frac{c}{5})$.
Therefore the area of the triangle thus formed will be
$\frac{1}{2}(base\times height)$
$=\frac{c\times \frac{c}{5}}{2}$
$=\frac{c^2}{10}$
$=5$ ...(given)
Hence
$c^2=50$
$c=5\sqrt{2}$.
Hence the equation of the line is
$x+5y=5\sqrt{2}$.
Now this is parallel to the line $x+5y=0$.
Hence applying the formula of distance between two parallel lines, we get
$=\frac{5\sqrt{2}}{\sqrt{26}}$
$=\frac{5}{\sqrt{13}}$.