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A line L is a drawn from P(4,3) to meet ...

A line `L` is a drawn from `P(4,3)` to meet the lines `L_1a n dL_2` given by `3x+4y+5=0` and `3x+4y+15=0` at points A and B , respectively. From `A` , a line perpendicular to `L` is drawn meeting the line `L_2` at `A_1` Similarly, from point `B`,a line perpendicular to `L` is drawn meeting the line `L_1` at `B_1` Thus, a parallelogram `A A_1B B_1` is formed. Then the equation of `L` so that the area of the parallelogram `A A_1B B_1` is formed. Then the equation of L so that the area of the parallelogram AA 1 ​ BB 1 ​ is least is

A

(a) `x-7y+17=0`

B

(b) `7x+y+31=0`

C

(c) `x-7y-17=0`

D

(d) `x+7y-31=0`

Text Solution

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The correct Answer is:
`7x+y-31=0`
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ARIHANT MATHS-THE STRAIGHT LINES-Exercise For Session 2
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  3. Let the algebraic sum of the perpendicular distance from the points (2...

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  4. If the quadrilateral formed by the lines ax+by+c=0. a'x+b'y+c=0, ax+by...

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  5. Prove that the area of the parallelogram formed by the lines 3x-4y+a=0...

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