Question

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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Answer 1

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

Answer 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

$\sf{5x - y = 1} \: \: \: \: \: .....(1)$

Now the line L is perpendicular to the line given by Equation (1)

Therefore the equation of the line L can be taken as

$\sf{x + 5y = k} \: \: \: \: ......(2)$

Now we rewrite the equation in Intercept form as below

$\displaystyle \sf{ \frac{x}{k} + \frac{5y}{k} = 1}$

$\implies \displaystyle \sf{ \frac{x}{k} + \frac{y}{ \frac{k}{5} } = 1}$

So the line given by the equation (2) intersects x axis at (k, 0) & y axis at

$\displaystyle \sf{ \bigg( 0 \: , \: \frac{k}{5} \bigg)}$

So the area of triangle formed by the line L and the co-ordinate axis is

$\displaystyle \sf{ = \frac{1}{2} \times k \: \times \: \frac{k}{5} } \: \: \: \: sq \: unit$

$\displaystyle \sf{ = \frac{ {k}^{2} }{10}} \: \: \: \: sq \: unit$

So by the given condition

$\displaystyle \sf{ \frac{ {k}^{2} }{10} = 5 }$

$\implies \displaystyle \sf{ {k}^{2} =5 0 }$

$\implies \displaystyle \sf{ k = \pm \: 5 \sqrt{2} }$

∴ The equation of the line L is

$\sf{x + 5y = \pm \: 5 \sqrt{2} } \: \: \: .....(3)$

Again the equation of the given line is

$\sf{x + 5y = 0} \: \: \: \: .....(4)$

Hence the required distance between the lines given by (3) & (4) is

$= \displaystyle \sf{ \bigg| \frac{ \pm \: 5 \sqrt{2} - 0 }{ \sqrt{ {1}^{2} + {5}^{2} } } \bigg| } \: \: unit$

$= \displaystyle \sf{ \bigg| \frac{ \pm \: 5 \sqrt{2} }{ \sqrt{1 + 25 } } \bigg| } \: \: unit$

$= \displaystyle \sf{ \frac{5 \sqrt{2} }{ \sqrt{26} } } \: \: unit$

$= \displaystyle \sf{ \frac{5 }{ \sqrt{13} } } \: \: unit$

FINAL ANSWER

The required distance

$= \displaystyle \sf{ \frac{5 }{ \sqrt{13} } } \: \: unit$

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