Question

A line is drawn perpendicular to line $y=5x$, meeting the coordinate axes at $A$ and $B.$ If the area of triangle $OAB$ is $10sq$. units where $O$ is the origin, then the equation of drawn line is

• A. $3x-y-9$
• B. $x+5y=10$
• C. $x+4y=10$
• D. $x-4y=10$

Answer 1

The correct option is B $x+5y=10$

Using the concept: line perpendicular to
$ax+by+c=0$ is $bx-ay+c^{\prime }=0.$

Let the required line be $x+5y=c.$

$x+5y=c\Rightarrow \frac{x}{c}+\frac{y}{\frac{c}{5}}=1$

x-intercept$=c$,y-intercept$=\frac{c}{5}$

Now area of triangle $=\frac{1}{2}\times base\times height$ $⟹area=\frac{1}{2}\times c\times \frac{c}{5}=\frac{c^2}{10}=10$

$⟹c^2=100$ $⟹c=\pm 10$

$\therefore$ the required line is $x+5y=10$ or $x+5y+10=0$