A) \[x+5y=5\]
B) \[x+5y=\pm 5\sqrt{2}\]
C) \[x-5y=5\]
D) \[x-5y=5\sqrt{2}\]
Correct Answer: B
Solution :
A line perpendicular to the line \[5x-y=1\]is given by\[x+5y-\lambda =0=L\], (given) In intercept form \[\frac{x}{\lambda }+\frac{y}{\lambda /5}=1\] So, area of triangle is \[\frac{1}{2}\]\[\times \](Multiplication of intercepts) Þ \[\frac{1}{2}(\lambda )\times \left( \frac{\lambda }{5} \right)=5\Rightarrow \lambda =\pm 5\sqrt{2}\] Hence the equation of required straight line is\[x+5y=\pm 5\sqrt{2}\] .
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