If a line L is perpendicular to the line 5x-y=1 and the area of triangle formed by the line L and the co-ordinate axis is 5 sq. units, then the distance of the line L from the line x+5y = 0 is
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If a line L is perpendicular to the line 5x-y=1 and the area of triangle formed by the line L and the co-ordinate axis is 5 sq. units, then the distance of the line L from the line x+5y = 0 is
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Step-by-step explanation:
5x−y=1
y=5x−1
m=5
m′=
5
−1
△AOB=
2
1
×c×5c
=
2
5c
2
We know △AOB=5sq.unite
2
1
5c
2
=5
c=±
2
y=−
5
1
x±
2
5y=−x±5
2
x+5y=±5
2
SOLUTION
GIVEN
A line L is perpendicular to the line 5x - y = 1 and the area of triangle formed by the line L and the co-ordinate axis is 5 sq. units
TO DETERMINE
The distance of the line L from the line
x + 5y = 0
EVALUATION
Here the given equation of the line is
[tex] \sf{5x - y = 1} \: \: \: \: \: .....(1)[/tex]
Now the line L is perpendicular to the line given by Equation (1)
Therefore the equation of the line L can be taken as
[tex] \sf{x + 5y = k} \: \: \: \: ......(2)[/tex]
Now we rewrite the equation in Intercept form as below
[tex] \displaystyle \sf{ \frac{x}{k} + \frac{5y}{k} = 1}[/tex]
[tex] \implies \displaystyle \sf{ \frac{x}{k} + \frac{y}{ \frac{k}{5} } = 1}[/tex]
So the line given by the equation (2) intersects x axis at (k, 0) & y axis at
[tex] \displaystyle \sf{ \bigg( 0 \: , \: \frac{k}{5} \bigg)}[/tex]
So the area of triangle formed by the line L and the co-ordinate axis is
[tex] \displaystyle \sf{ = \frac{1}{2} \times k \: \times \: \frac{k}{5} } \: \: \: \: sq \: unit[/tex]
[tex] \displaystyle \sf{ = \frac{ {k}^{2} }{10}} \: \: \: \: sq \: unit[/tex]
So by the given condition
[tex] \displaystyle \sf{ \frac{ {k}^{2} }{10} = 5 }[/tex]
[tex] \implies \displaystyle \sf{ {k}^{2} =5 0 }[/tex]
[tex] \implies \displaystyle \sf{ k = \pm \: 5 \sqrt{2} }[/tex]
∴ The equation of the line L is
[tex] \sf{x + 5y = \pm \: 5 \sqrt{2} } \: \: \: .....(3)[/tex]
Again the equation of the given line is
[tex] \sf{x + 5y = 0} \: \: \: \: .....(4)[/tex]
Hence the required distance between the lines given by (3) & (4) is
[tex] = \displaystyle \sf{ \bigg| \frac{ \pm \: 5 \sqrt{2} - 0 }{ \sqrt{ {1}^{2} + {5}^{2} } } \bigg| } \: \: unit[/tex]
[tex] = \displaystyle \sf{ \bigg| \frac{ \pm \: 5 \sqrt{2} }{ \sqrt{1 + 25 } } \bigg| } \: \: unit[/tex]
[tex] = \displaystyle \sf{ \frac{5 \sqrt{2} }{ \sqrt{26} } } \: \: unit[/tex]
[tex] = \displaystyle \sf{ \frac{5 }{ \sqrt{13} } } \: \: unit[/tex]
FINAL ANSWER
The required distance
[tex] = \displaystyle \sf{ \frac{5 }{ \sqrt{13} } } \: \: unit[/tex]
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