Question

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

Answer 1

Answer: (A)

$\alpha =1,\beta =-1$

Consider the given lines are

$x+y+1=0......\left(1\right)$

$4x+3y+4=0.......\left(2\right)$

$x+\alpha y+\beta =0.......\left(3\right)$

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

Then solve by matrix form,(1), (2) and (3) to, we get

$1$ $1$ $1$

$4$ $3$ $4$

$1$ $\alpha$ $\beta$

$1(3\beta -4\alpha )-1(4\beta -4)+1(4\alpha -3)=0$

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

$\Rightarrow -\beta +1=0$

$\Rightarrow \beta =1$

Put the value of $\beta$ in given equation.

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

${\alpha }^2+1^2=2$

${\alpha }^2=1$

$\alpha =\pm 1$

Hence, this is the amswer.