question_answer

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)  \[3x-y=9\]

B)  \[x+5y=-10\]

C)  \[x+4y=10\]

D)  \[x-4y=10\]

Correct Answer: B

Solution :

Let the equation of line be x/a +y/ b= 1

AB is perpendicular to y = 5x

Hence,

\[-b/a\times 5=-1\]\[\Rightarrow \]\[5b=a\]

Area of \[\Delta OAB=\frac{1}{2}\left| ab \right|\]

\[\Rightarrow \]   \[10=\frac{1}{2}\left| 5{{b}^{2}} \right|\]

\[\Rightarrow \]   \[{{b}^{2}}=4\]

\[\Rightarrow \]   \[b=\pm 2,a\]

\[=\pm 10\]

The line can be \[x/10+y/2=1\]or

\[x/10+y/2=-1\]