If the system of linear equations $\mathrm{x}-2 \mathrm{y}+\mathrm{kz}=1$ $2 \mathrm{x}+\mathrm{y}+\mathrm{z}=2$ $3 x-y-k z=3$ has a solution $(x, y, z), z \neq 0$, then $(x, y)$ lies on the straight line whose equation is:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Matrices

If the system of linear equations

x - 2 y + kz = 1

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:

Answer: (B)

There is a quick way in: add the first and third equations. The

k z

term appears as

+ k z

in the first equation and

- k z

in the third, so it cancels, and

(x - 2 y + k z) + (3 x - y - k z) = 1 + 3

collapses to

4 x - 3 y = 4

. This holds for whatever solution the system has, independent of

k

and

z

, so the point

(x, y)

always lies on the straight line

4 x - 3 y - 4 = 0

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option