Question

Find the value of $\lambda$ such that lines $x+2y=3,3x-y=1$ and $\lambda x+y=2$ can not form a triangle.

Answer 1

$x+2y-3=0\to \left(i\right)$

$3x-y-1=0\to \left(ii\right)$

$\lambda x+y-2=0\to \left(iii\right)$

$areaoftriangle=0$

$\Rightarrow \left|\begin{array}{ccc}1& 2& -3\\ 3& -1& -1\\ \lambda & 1& -2\end{array}\right|=0\Rightarrow 1\left(2+1\right)-2\left(-6+\lambda \right)-3\left(3+\lambda \right)=0$

$\Rightarrow 3+12-2\lambda -9-3\lambda =0$

$\Rightarrow \lambda =\frac{6}{5}$