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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
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B
x+5y=10
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C
x+4y=10
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D
x4y=10
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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 $$x+5y+10=0$$


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