Mathematics - Determinant Question with Solution | TestHub

(A) Solving equations $L_1$ and $L_3$, $$\begin{array}{rl}\frac{x}{-36+10}& =\frac{y}{+12-25}=\frac{1}{2-15}\\ x& =2,y=1\end{array}$$ $L_1,L_2,L_3$ are concurrent, if point $(2,1)$ lies on $L_2$. $$\therefore 6-k-1=0\Rightarrow k=5$$ (B) Either $L_1$ is parallel to $L_2$, or $L_3$ is parallel to $L_2$, then $$\begin{array}{rl}& \frac{1}{3}=\frac{3}{-k}\text{ or }\frac{3}{5}=\frac{-k}{2}\\ & k=-9\text{ or }k=-\frac{6}{5}\end{array}$$ $$\Rightarrow k=-9\text{ or }k=-\frac{6}{5}$$ (C) $L_1,L_2,L_3$ form a triangle, if they are not concurrent, or not parallel. $$\therefore k\ne 5,-9,-\frac{6}{5}\Rightarrow k=\frac{5}{6}$$ (D) $L_1,L_2,L_3$ do not form a triangle, if $$k=5,-9,-\frac{6}{5}$$