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Q.

A) Lines x−2y−6=0, 3x+y−4=0 and λx+4y+λ2=0 are concurrent. The value of λ isB) The points (λ+1,1), (2λ+1,3) and (2λ+2,2λ) are collinear, then the value of λ isC) If line x+y−1−λ=0, passing through the intersections of x - y + 1 = 0 and 3x+y−5=0 is perpendicular to one of them then the value of λ isD) If the line y−x−1+λ=0 is equally inclined to axes and equidistant from the points (1, -2) and (3, 4) then λ isp) 2q) -4r) -12s) 1

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a

A-p,q, B-p,r, C-p, D-p

b

A-p,q, B-p,r, C-q, D-r

c

A-p, B-p,r, C-p, D-p

d

A-p,q, B-r, C-p, D-p

answer is A.

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Detailed Solution

A) 1−2−631−4λ4λ2=0⇒1(λ2+16)+2(3λ2+4λ)−6(12−λ)=0⇒7λ2+14λ−56=0⇒λ2+2λ−8=0⇒λ=2,−4B) Collinear ⇒2λ=2λ−3⇒λ=−12,2C) Point of intersection of x-y+1=0 and 3x+y-5=0 is (1,2). (1,2) lies on-x+y-1-λ=0⇒λ=2 D) Perpendicular distance form (1,-2) and (3,4) to y-x-1 +λ=0 are equal⇒|λ−4|2=|λ|2⇒λ−4=−λ⇒λ=2
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