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Let L be the line belonging to the famil...

Let L be the line belonging to the family of straight lines (a+2b) x+(a-3b)y+a-8b =0, a, b `in `R, which is the farthest from the point (2, 2).
Area enclosed by the line L and the coordinate axes is

A

x+4y+7=0

B

2x+3y+4=0

C

4x-y-6=0

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A

Given lines (x+y+1) +b(2x-3y-8) = 0 are concurrent at the point of intersection of the line x+y+1=0 and 2x-3y-8=0, which is (1, -2). Now, the line through A(1, -2), which is farthest from the point B(2, 2), is perpendicular to AB. Now, the slope of AB is 4. Then the required line is y+2 =-(1/4)(x-1) or x+4y=0.
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Let l be the line belonging to the family of straight lines (a + 2b)x+ (a - 3b)y +a-8b = 0, a, b in R , which is farthest from the point (2, 2), then area enclosed by the line L and the coordinate axes is

Let L be the line belonging to the family of straight lines (a+2b) x+(a-3b)y+a-8b =0, a, b in R, which is the farthest from the point (2, 2). If L is concurrent with the lines x-2y+1=0 and 3x-4y+ lambda=0 , then the value of lambda is

Knowledge Check

  • The mid-point of the line joining (-a, 2b) and (-3a, -4b) is

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    (-2a, -b)
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  • In the equation of a striaght line ax + by + c = 0 , if a,b,c are in arithmetic progression then the point on the straight line is

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    (1,2)
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    (1, -2)
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