If a line L is perpendicular to the line 5x - y = 1, and the area...

The equation of a line perpendicular to the line $ax + by + c = 0$ is $bx - ay + k = 0, k \in R .$

Given, line $L$ is perpendicular to $5x - y = 1,$ hence the equation of $L$ is $x +5 y + k =0$

For finding $x$-Intercept, put $y = 0$

$⇒ x + k = 0, ⇒ x = - k$

$∴$ $x$-intercept$= - k$

Similarly, for finding $y$- Intercept, put $x = 0$

$⇒ 5y + k = 0, ⇒ y = - \dfrac{k}{5}$

$∴$ $y$-intercept$= - \dfrac{k}{5}$

Given, the area formed by the line $L$ with the coordinate axes is $5$ sq units.

Hence, the area of $\text{\Delta =}\dfrac{1}{2} \times \left(- k\right)\left(- \dfrac{k}{5}\right) = 5$

$⇒ {k}^{2} = 50$

$⇒ k = \pm \text{5}\sqrt{2}$

Hence, the equation of the line is $L: x +5 y \pm 5 \sqrt{2} = 0$

Now, we know that the distance between the parallel lines $ax + by + c = 0$ and $ax + by + {c}_{1} = 0$ is $\dfrac{\left|c - {c}_{1}\right|}{\sqrt{{a}^{2} + {b}^{2}}}$

Thus, the distance between the lines $x +5 y =0$ and $x +5 y\text{ \pm 5}\sqrt{2} = 0$ is

$d = \left|\dfrac{\pm 5\sqrt{2} - 0}{\sqrt{{1}^{2} + {5}^{2}}}\right|$

$⇒ d = \dfrac{5\sqrt{2}}{\sqrt{26}}$

$⇒ d = \dfrac{5\sqrt{2}}{\sqrt{13 \times 2}} = \dfrac{5}{\sqrt{13}}$ units.