A straight line L is perpendicular to the line $5x-y=1.$ The area of the triangle formed by the line L and coordinate axes is 5. The equation of the line L is 

Let $L = 3x – 4y – 8$. Then the value of L at $(3, 4)$ is $3\times 3-4\times 4-8=-15<0$. Hence, for the point P(x, y), we should have $L > 0$ or $3x – 4y – 8 > 0$ or $3x – 4(–3x) – 8 > 0$             [$\therefore$ $P(x, y)$, lies on$y = –3x$] $\text{ or }x>\frac{18}{15}\text{ and }-y-4y-8>0\text{ or }y<-\frac{8}{5}$