Question

The set of real values of `K' for which the lines x+3y+1=0,Kx+2y-2=0 and 2x-y+3=0 form a triangle is

• A. $R-\{-4,\frac{2}{3}\}$
• B. $R-\{-4,\frac{2}{3},\frac{-6}{5}\}$
• C. $R-\{\frac{2}{3},4\}$
• D. $R$

Answer 1

The correct option is B $R-\{-4,\frac{2}{3},\frac{-6}{5}\}$

Lines form a triangle. Therefore
$x+3y+1=0$ is not parallel to $Kx+2y-2=0\Rightarrow K\ne \frac{2}{3}$
$2x-y+3=0$ is not parallel to $Kx+2y-2=0\Rightarrow K\ne -4$
and lines must not be concurrent,
i.e.$\left|\begin{array}{ccc}1& 3& 1\\ K& 2& -2\\ 2& -1& 3\end{array}\right|\ne 0\Rightarrow K\ne \frac{-6}{5}$