Question

If the system of linear equations
x-2y + kz = 1, 2x + y+ z = 2, 3x-y-kz = 3 has a solution (x, y, z), `z ne 0`, then (x, y) lies on the straight line whose equation is
A. 3x-4y-4 =0
B. 3x-4y-1=0
C. 4x-3y-4 =0
D. 4x-3y-1=0

Answer 1

Given system of linear equations
`x-2y + kz = 1 " "….(i)`
`2x + y + z = 2 " "…(ii)`
`"and " 3x-y-kz = 3 " "...(iii)`
has a solution `(x, y, z), z ne 0`.
On adding Eqs. (i) and (iii), we get
`x-2y + kz + 3x-y-kz = 1+3`
4x-3y = 4
`rArr 4x-3y-4 = 0`
This is the required equation of the straight line in which point (x, y) lies.