Question

If $c<1$ and the straight lines $x+y-1=0,$ $2x-y-c=0$ and $-bx+3by-c=0$ are concurrent, then the possible range of $b$ is

• A. $b\in (-3,\frac{3}{4})$
• B. $b\in (-\frac{3}{4},3)-\left\{0\right\}$
• C. $b\in (-\frac{3}{2},4)$
• D. No real value exists

Answer 1

The correct option is B $b\in (-\frac{3}{4},3)-\left\{0\right\}$

The given lines are concurrent.
$\therefore △=\left|\begin{array}{ccc}1& 1& -1\\ 2& -1& -c\\ -b& 3b& -c\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}0& 0& -1\\ 3& -1-c& -c\\ -4b& 3b-c& -c\end{array}\right|=0(C_1\to C_1-C_2,C_2\to C_2+C_3)$

Expanding along $R_1,$ we get
$-1(9b-3c-4b-4bc)=0\Rightarrow 3c+4bc=5b$
$\Rightarrow c=\frac{5b}{4b+3}$
Now, $c<1$
$\Rightarrow \frac{5b}{4b+3}<1$
$\Rightarrow \frac{b-3}{4b+3}<0$
$\therefore b\in (-\frac{3}{4},3)$
Also $b\ne 0$ because at $b=0,$ $-bx+3by-c=0$ does not represent a straight line.