question_answer

If a line L is perpendicular to the line \[5x-y=1\], and the area of the triangle formed by the line L and the coordinate axes is 5, then the distance of line L from the line \[x+5y=0\] is:     JEE Main Online Paper (Held On 19 April 2016)

A) \[\frac{7}{\sqrt{5}}\]                                     

B) \[\frac{5}{\sqrt{13}}\]

C) \[\frac{7}{\sqrt{13}}\]                   

D) \[\frac{5}{\sqrt{7}}\]

Correct Answer: B

Solution :

Let equation of line L, perpendicular to  \[5xy=1\]be \[x+5y=c\] Given that area of \[\Delta AOB\] is 5. We know \[\left\{ area,A=\frac{1}{2}\left[ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right] \right\}\] \[\Rightarrow \] \[5=\frac{1}{2}\left[ c\left( \frac{c}{5} \right) \right]\] \[\left( \begin{align}   & \because ({{x}_{1}},{{y}_{1}})(10,0),({{x}_{3}},{{y}_{3}})=\left( 0,\frac{c}{5} \right) \\  & ({{x}_{2}},{{y}_{2}})=(c,0) \\ \end{align} \right)\] \[\Rightarrow \] \[c=\pm \sqrt{50}\] \[\therefore \] Equation of line L is \[x+5y=\pm \sqrt{50}\] Distance between L and line \[x+5y=0\]is \[d=\left| \frac{\pm \sqrt{50}-0}{\sqrt{{{1}^{2}}+{{5}^{2}}}} \right|=\frac{\sqrt{50}}{26}=\frac{5}{\sqrt{13}}\]