JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
If the system of linear equations $\lambda x+y-z=-1$, $x-y-3z=2$, $-x+y+z=\mu$ has infinitely many solutions, then the equation of the line passing through the points $(\lambda+2\mu,2\lambda+\mu)$ and $(1,\lambda\mu)$ is
Determinant
Solution
Coefficient determinant $\begin{vmatrix}\lambda&1&-1\\1&-1&-3\\-1&1&1\end{vmatrix}=0\Rightarrow\lambda=-1$. Then the rows give $-x+y-z=-1$, $x-y-3z=2$, $-x+y+z=\mu$; adding first two: $-4z=1$, so $z=-\tfrac14$; row 3 minus row 1: $2z=\mu+1\Rightarrow\mu=-\tfrac32$. Points: $(\lambda+2\mu,2\lambda+\mu)=(-4,-\tfrac72)$ and $(1,\lambda\mu)=(1,\tfrac32)$. Slope $=\dfrac{5}{5}=1$. $y-\tfrac32=x-1\Rightarrow 2x-2y=-1$. Option (C).