IIT-JEE 1980Question 1
A straight line $L$ is perpendicular to the line $5x-y=1$. The area of the triangle formed by the line $L$ and the coordinate axes is $5$. Find the equation of the line $L$.
Points and Straight Lines
Answer$x+5y-5\sqrt2=0$ or $x+5y+5\sqrt2=0$
Solution
$L$: $x+5y=c$ has intercepts $c$ and $\frac c5$; area $\frac{c^2}{10}=5$ gives $c=\pm5\sqrt2$, so $x+5y=\pm5\sqrt2$.