Question

If a line L is perpendicular to the line 5x - y = 1, and the area of the triangle formed by the line L and the coordinates axes is 5, then the distance of line L from the line x + 5y = 0 is

• A. $\frac{7}{\sqrt{5}}$
• B. $\frac{5}{\sqrt{1}3}$
• C. $\frac{7}{\sqrt{1}3}$
• D. $\frac{5}{\sqrt{7}}$

Answer 1

The correct option is B $\frac{5}{\sqrt{1}3}$

Slope of line $5x-y=1,m_1=5$

$\therefore$slope of line $L,m_2=-\frac{1}{m_1}=\frac{-1}{5}$

$\therefore$ Equation of line $L:y=\frac{-1}{5}x+C$

Y-intercept$=C$, then x-intercept$=5C$

$\therefore$ Area=$\frac{1}{2}\left(5C\right)\left(C\right)=5$

$⟹C^2=2$

$C=\pm \sqrt{2}$

$\therefore$ Equation of line $L$ is $y=-\frac{x}{5}\pm \sqrt{2}$

$⟹x+5y=\pm 5\sqrt{2}$

Distance of line $L$ from $x+5y=\pm 5\sqrt{2}$

$d=\frac{\pm 5\sqrt{2}}{\sqrt{(1{)}^2+(5{)}^2}}$

$d=\frac{5\sqrt{2}}{\sqrt{26}}=\frac{5}{\sqrt{13}}$