Question

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

• A. $x+5y=5$
• B. $x+5y=\pm 5\sqrt{2}$
• C. $x-5y=5$
• D. $x-5y=5\sqrt{2}$

Answer 1

The correct option is B $x+5y=\pm 5\sqrt{2}$

A line perpendicular to the line $5x-y=1$ is given by $x+5y-\lambda =0=L$, (given)
In intercept form $\frac{x}{\lambda }+\frac{y}{\frac{\lambda }{5}}=1$
So, area of triangle is $\frac{1}{2}\times$ (Multiplication of intercepts)
$\Rightarrow \frac{1}{2}\left(\lambda \right)\times \left(\frac{\lambda }{5}\right)=5\Rightarrow \lambda =\pm 5\sqrt{2}$
Hence the equation of required straight line is .
$x+5y=\pm 5\sqrt{2}$.