Question

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. Find the equation of the line:

• A. $x+5y=\pm 5\sqrt{2}$
• B. $x-3y=0$
• C. $2x+y=0$
• D. $x+4y=5\sqrt{2}$

Answer 1

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

Equation of any line L perpendiculuar to 5x - y = 1 is x + 5y = k......(1)
Where k is an arbitrary constant. If this line cuts x - axis at A and y - axis at B, then for A, y = 0
from (1) y = $\frac{k}{5}$ i.e., B is the point $(0,\frac{k}{5})$
$\therefore$ Area of the given $\Delta OAB=\frac{1}{2}(x_1,y_2-x_2,y_1)=\frac{1}{2}(\frac{k^2}{5}-0)=\frac{k^2}{10}$
But the given condition $\frac{k^2}{10}=5or{k}^2=50\therefore k=\pm 5\sqrt{2}$
Hence, from (1), the required equation of the line is
$x+5y=5\sqrt{2}orx+5y=-5\sqrt{2}$