Matrices and Determinants 4 Question 5
5. If the system of linear equations $x-2 y+k z=1, \quad 2 x+y+z=2,3 x-y-k z=3$ has a solution $(x, y, z), z \neq 0$, then $(x, y)$ lies on the straight line whose equation is
(2019 Main, 8 April II)
(a) $3 x-4 y-4=0$
(b) $3 x-4 y-1=0$
(c) $4 x-3 y-4=0$
(d) $4 x-3 y-1=0$
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Answer:
Correct Answer: 5. (c)
Solution:
- Given system of linear equations
$ \begin{array}{r} x-2 y+k z=1 \\ 2 x+y+z=2 \\ \text { and } \quad 3 x-y-k z=3 \end{array} $
has a solution $(x, y, z), z \neq 0$.
On adding Eqs. (i) and (iii), we get
$ \begin{array}{rlrl} & & x-2 y+k z+3 x-y-k z & =1+3 \\ & & 4 x-3 y & =4 \\ & & 4 x-3 y-4 & =0 \end{array} $
This is the required equation of the straight line in which point $(x, y)$ lies.
