lf a line L is perpendicular to the line 5x - | vedclass

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Question

Let equation of line $L$, perpendicular to $5x-y=1$ be $x+5y=c$

Given that are of $\Delta AOB$  is $5$.

We know

$\left\{ {are,A = \

Vedclass · Accepted Answer

$\frac{5}{{\sqrt {13} }}$

Vedclass · Answer

Let equation of line $L$, perpendicular to $5x-y=1$ be $x+5y=c$

Given that are of $\Delta AOB$  is $5$.

We know

$\left\{ {are,A = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_2}} ight) + {x_2}\left( {{y_3} - {y_1}} ight) + {x_3}\left( {{y_1} - {y_2}} ight)} ight]} ight\}$

$ \Rightarrow 5 = \frac{1}{2}\left[ {c\left( {\frac{c}{5}} ight)} ight]$

$\because \left ( \begin{array}{l}
\left( {{x_1},{y_1}} ight) = \left( {10,0} ight)\left( {{x_3},{y_3}} ight) = \left( {0,\frac{c}{5}} ight)\
\left( {{x_2},{y_2}} ight) = \left( {c,0} ight)
\end{array} ight)$

$ \Rightarrow c =  \pm \sqrt {50} $

$	herefore $ Equation of line $L$ is $x + 5y =  \pm \sqrt {50} $

Distance betwween $L$ and line $x+5y=0$ is 

$d = \left. {\frac{{ \pm \sqrt {50}  - 0}}{{\sqrt {{1^2} + {5^2}} }}} ight| = \frac{{\sqrt {50} }}{{\sqrt {26} }} = \frac{5}{{\sqrt {13} }}$

lf 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

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