A family of lines is given by (1+2λ)x+(1−λ)y+λ=0,λ being the parameter. The line belonging to this family at the maximum distance from the point (1,4) is
A
33x+12y−7=0
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B
33x+12y+7=0
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C
12x+33y−7=0
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D
12x+33y+7=0
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Solution
The correct option is C12x+33y−7=0 Given : (1+2λ)x+(1−λ)y+λ=0 ⇒x+y+λ(2x−y+1)=0 Clearly, it represents a family of lines passing through the intersection of the lines x+y=0 and 2x−y+1=0 i.e., the point (−13,13)
The required line passes through (−13,13) and is perpendicular to the line joining (1,4) and (−13,13) So, its equation is y−13=−1+134−13(x+13)⇒3y−13=−411(x+13)∴12x+33y−7=0