JEE Main 2014 — Straight Lines Question with Solution

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 $x+5y+k=0$

For finding $x$-Intercept, put $y=0$

$\Rightarrow x+k=0, \Rightarrow x=-k$

$\therefore$ $x$-intercept$=-k$

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

$\Rightarrow 5y+k=0, \Rightarrow y=-\frac{k}{5}$

$\therefore$ $y$-intercept$=-\frac{k}{5}$

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

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

$\Rightarrow k^2=50$

$\Rightarrow k=\pm \text{5}\sqrt{2}$

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

Now, we know that the distance between the parallel lines $ax+by+c=0$ and $ax+by+c_1=0$ is $\frac{\left|c-c_1\right|}{\sqrt{a^2+b^2}}$

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

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

$\Rightarrow d=\frac{5\sqrt{2}}{\sqrt{26}}$

$\Rightarrow d=\frac{5\sqrt{2}}{\sqrt{13\times 2}}=\frac{5}{\sqrt{13}}$ units.