Question -
Answer -
(i) (0, 2)
Putting x = 0 and y = 2 in the L.H.Sof the given equation,
x − 2y = 0 −2×2 = − 4 ≠ 4
L.H.S ≠ R.H.S
Therefore, (0, 2) is not a solution of this equation.
(ii) (2, 0)
Putting x = 2 and y = 0 in the L.H.Sof the given equation,
x − 2y = 2 −2 × 0 = 2 ≠ 4
L.H.S ≠ R.H.S
Therefore, (2, 0) is not a solution of this equation.
(iii) (4, 0)
Putting x = 4 and y = 0 in the L.H.Sof the given equation,
x − 2y = 4 −2(0)
= 4 = R.H.S
Therefore, (4, 0) is a solution of this equation.
(iv)
Putting 
and 
in the L.H.S of the given equation, L.H.S ≠ R.H.S
Therefore, is not a solution of this equation.
(v) (1, 1)
Putting x = 1 and y = 1 in the L.H.Sof the given equation,
x − 2y = 1 −2(1) = 1 − 2 = − 1 ≠ 4
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution of this equation.