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 sq units, then the…

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

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

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

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

$\therefore$∴ $x$x-intercept$=-k$=−k

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

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

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

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

Hence, the area of $\text{Δ=}\frac{1}{2}\times (-k)(-\frac{k}{5})=5$Δ=21​×(−k)(−5k​)=5

$$\Rightarrow {k}^{2}=50$$⇒k2=50

$$\Rightarrow k=\pm \text{5}\sqrt{2}$$⇒k=±52​

Hence, the equation of the line is $L:x⁡+5y⁡\pm 5\sqrt{2}=0$L:x⁡+5y⁡±52​=0

Now, we know that the distance between the parallel lines $ax+by+c=0$ax+by+c=0 and $ax+by+{c}_{1}=0$ax+by+c1​=0 is $\frac{|c-{c}_{1}|}{\sqrt{{a}^{2}+{b}^{2}}}$a2+b2​∣c−c1​∣​

Thus, the distance between the lines $x+5y=0$x+5y=0 and $x+5y\text{⁡±5}\sqrt{2}=0$x+5y⁡±52​=0 is

$$d=|\frac{\pm 5\sqrt{2}-0}{\sqrt{{1}^{2}+{5}^{2}}}|$$d=∣12+52​±52​−0​∣

$$\Rightarrow d=\frac{5\sqrt{2}}{\sqrt{26}}$$⇒d=26​52​​

$\Rightarrow d=\frac{5\sqrt{2}}{\sqrt{13\times 2}}=\frac{5}{\sqrt{13}}$⇒d=13×2​52​​=13​5​ units.