Straight Line Question 284

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[IIT 1980; RPET 1997]

Options:

A) $ x+5y=5 $

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

C) $ x-5y=5 $

D) $ x-5y=5\sqrt{2} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • 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}{\lambda /5}=1 $ So, area of triangle is $ \frac{1}{2} $ $ \times $ (Multiplication of intercepts)
    Þ $ \frac{1}{2}(\lambda )\times ( \frac{\lambda }{5} )=5\Rightarrow \lambda =\pm 5\sqrt{2} $ Hence the equation of required straight line is $ x+5y=\pm 5\sqrt{2} $ .