Question

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

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

Answer 1

The correct option is B $\beta =1,\alpha =\pm 1$

Since lines are concurrent,

$\left|\begin{array}{ccc}1& 1& 1\\ 4& 3& 4\\ 1& \alpha & \beta \end{array}\right|=0$ $\therefore 1(3\beta -4\alpha )-1(4\beta -4)+1(4\alpha -3)=0$

$\therefore 3\beta -4\alpha -4\beta +4+4\alpha -3=0$

$\therefore -\beta +1=0$

$\therefore \beta =1$

$\therefore {\alpha }^2+{\beta }^2=2$

$\therefore {\alpha }^2=2-1=1$

$\therefore \alpha =\pm 1$

$\therefore \beta =1$