Question

If a line is perpendicular to the line $5x-y=0$ and forms a triangle of area 5 square units with co-ordinate axes, then its equation is

• A. $x+5y\pm 5\sqrt{2}=0$
• B. $x-5y\pm 5\sqrt{2}=0$
• C. $5x+y\pm 5\sqrt{2}=0$
• D. $5x-y\pm 5\sqrt{2}=0$

Answer 1

The correct option is A $x+5y\pm 5\sqrt{2}=0$

Solution:

Equation of the line perpendicular to the line $5x-y=0$

is $x+5y=\lambda$

Now, area of the $△AOB=5$ sq. units

or, $\frac{1}{2}\times \lambda \times \frac{\lambda }{5}=5$

or, ${\lambda }^2=50$

or, $\lambda =\pm 5\sqrt{2}$

Equation of the line are $x+5y=\pm 5\sqrt{2}$

$x+5y\pm 5\sqrt{2}=0$

Hence, A is the correct option.