Question

If the lines $x+y+1=0 ; 4 x+3 y+4=0$ and $x+\alpha y+\beta=0$, where $\alpha^2+\beta^2=2$, are concurrent then

• A. $\alpha=1, \beta=-1$
• B. $\alpha=1, \beta= \pm 1$
• C. $\alpha=-1, \beta= \pm 1$
• D. $\alpha= \pm 1, \beta=1$

Answer 1

Answer: (D)

Lines are $ x+y+1=0 ; 4 x+3 y+4=0$ and $x+\alpha y+\beta=0$, where $\alpha^2+\beta^2=2$
$\begin{vmatrix}1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta\end{vmatrix}=0$
$1(3 \beta-4 \alpha)-1(4 \beta-4)+1(4 \alpha-3) =3 \beta-4 \alpha-4 \beta+4+4 \alpha-3 $
$=-\beta+1=0 \Rightarrow \beta=1$
$\therefore \alpha= \pm 1$