If the lines x+y+1=0 ; 4 x+3 y+4=0 and x+α y+β=0, where α2+β2=2, are concurrent then

Question

If the lines x+y+1=0 ; 4 x+3 y+4=0 and x+α y+β=0, where α2+β2=2, are concurrent then (A) α=1, β=-1 (B) α=1, β= ± 1 (C) α=-1, β= ± 1 (D) α= ± 1, β=1

Tardigrade Admin · Accepted Answer

Lines are x+y+1=0 ; 4 x+3 y+4=0 and x+α y+β=0, where α2+β2=2 |1 1 1 4 3 4 1 α β|=0 1(3 β-4 α)-1(4 β-4)+1(4 α-3) =3 β-4 α-4 β+4+4 α-3 =-β+1=0 ⇒ β=1 ∴ α= ± 1

Q. If the lines $x + y + 1 = 0; 4 x + 3 y + 4 = 0$ and $x + α y + β = 0$ , where $α^{2} + β^{2} = 2$ , are concurrent then

$α = 1, β = - 1$

$α = 1, β = \pm 1$

$α = - 1, β = \pm 1$

$α = \pm 1, β = 1$

Lines are $x + y + 1 = 0; 4 x + 3 y + 4 = 0$ and $x + α y + β = 0$ , where $α^{2} + β^{2} = 2$
$∣ ∣ 141 13 α 14 β ∣ ∣ = 0$
$1 (3 β - 4 α) - 1 (4 β - 4) + 1 (4 α - 3) = 3 β - 4 α - 4 β + 4 + 4 α - 3$
$= - β + 1 = 0 \Rightarrow β = 1$
$∴ α = \pm 1$

Q. If the lines x+y+1=0;4x+3y+4=0 and x+αy+β=0, where α2+β2=2, are concurrent then

α=1,β=−1

α=1,β=±1

α=−1,β=±1

α=±1,β=1