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 $$10\ sq$$. units where $$O$$ is the origin, then the equation of drawn line is

A
3xy9
B
x+5y=10
C
x+4y=10
D
x4y=10

Solution

## The correct option is B $$x+5y=10$$Using the concept: line perpendicular to $$ax+by+c=0$$ is $$bx-ay+c'=0.$$Let the required line be $$x+5y=c.$$$$x+5y=c\Rightarrow \dfrac xc+\dfrac {y}{\dfrac c5}=1$$x-intercept$$=c$$,y-intercept$$=\dfrac c5$$Now area of triangle  $$= \dfrac{1}{2} \times base \times height$$ $$\implies area=\dfrac{1}{2} \times c \times \dfrac{c}{5}=\dfrac{c^2}{10} =10$$$$\implies c^2=100$$ $$\implies c=\pm 10$$$$\therefore$$ the required line is $$x+5y=10$$ or $$x+5y+10=0$$Maths

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