Question

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

• A. $x+5y=5$
• B. $x+5y=\pm 5\surd 2$
• C. $x–5y=5$
• D. $x–5y=\pm 5\surd 2$

Answer 1

The correct option is B

$x+5y=\pm 5\surd 2$

Explanation for the correct option:

Step 1. Finding the equation of line $L$:

Given, $5x–y=1$ …..(I)

The slope of line (i) is

$m_1=5$

The slope of perpendicular line $L=-\frac{1}{5}$

$\therefore$The equation of line $L$ is

$\begin{array}{rcl}y& =& \left(-\frac{1}{5}\right)x+C\end{array}$

$\begin{array}{rcl}x+5y& =& 5C\end{array}$ …..(ii)

When

$\begin{array}{rcl}x& =& 0,y=C\\ y& =& 0,x=5C\end{array}$

So the coordinates are $(0,C),(0,0)$ and $(5C,0)$

Step 2. Put the value of $OB$ in area of triangle, we get

$\begin{array}{rcl}\frac{1}{2}\times OB\times OA& =& 5\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{1}{2}\times 5C\times C& =& 5\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{C}^2& =& 2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}C& =& \pm \sqrt{2}\end{array}$

Step 3. Substitute the value of $C$ in equation (ii), we get

$\begin{array}{rcl}x+5y& =& 5\times \left(\pm \sqrt{2}\right)\end{array}$

$\therefore$the required equation of the line is $\begin{array}{rcl}x+5y& =& \pm 5\sqrt{2}\end{array}$

Hence, option B is the answer.