Question

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

Answer 1

Answer: (B)

$R-\{\frac{2}{3},-4,\frac{-6}{5}\}$

$L_1:x+3y+1=0$

$L_2:kx+2y-2=0$

$L_3:2x-y+3=0$

$\Delta$ will not be formed if $L_1||L_2$

i.e., $\frac{1}{k}=\frac{3}{2}\Rightarrow k=\frac{2}{3}$

So, $k\ne \frac{2}{3}$

If $L_2||L_3\frac{k}{2}=-2$

$k=-4$

So, $k\ne -4$

$\Delta$ will not form if $L_1,L_2$ and $L_3$ become concurrent, i.e.,

$\left|\begin{array}{ccc}1& 3& 1\\ k& 2& -2\\ 2& -1& 3\end{array}\right|=0$

$\Rightarrow -10k-12=0$

$k=-\frac{6}{5}$

So, $k\ne -\frac{6}{5}$

Hence , set of values of k is $R-\{\frac{2}{3},-4,\frac{-6}{5}\}$