Mathematics
Find which of the following points lie on the line x - 2y + 5 = 0 :
(i) (1, 3)
(ii) (0, 5)
(iii) (-5, 0)
(iv) (5, 5)
(v) (2, -1.5)
(vi) (-2, -1.5)
Straight Line Eq
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Answer
(i) Substituting x = 1 and y = 3 in the L.H.S. of the equation x - 2y + 5 = 0, we get :
L.H.S. = 1 - 2 × 3 + 5
= 1 - 6 + 5
= -5 + 5
= 0.
Since, L.H.S. = R.H.S.
∴ Point (1, 3) satisfies the equation.
Hence, (1, 3) lies on the line represented by the equation x - 2y + 5 = 0.
(ii) Substituting x = 0 and y = 5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :
L.H.S. = 0 - 2 × 5 + 5
= 0 - 10 + 5
= -5
Since, L.H.S. ≠ R.H.S.
∴ Point (0, 5) does not satisfies the equation.
Hence, (0, 5) does not lies on the line represented by the equation x - 2y + 5 = 0.
(iii) Substituting x = -5 and y = 0 in the L.H.S. of the equation x - 2y + 5 = 0, we get :
L.H.S. = -5 - 2 × 0 + 5
= -5 - 0 + 5
= -5 + 5
= 0.
Since, L.H.S. = R.H.S.
∴ Point (-5, 0) satisfies the equation.
Hence, (-5, 0) lies on the line represented by the equation x - 2y + 5 = 0.
(iv) Substituting x = 5 and y = 5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :
L.H.S. = 5 - 2 × 5 + 5
= 5 - 10 + 5
= 10 - 10
= 0.
Since, L.H.S. = R.H.S.
∴ Point (5, 5) satisfies the equation.
Hence, (5, 5) lies on the line represented by the equation x - 2y + 5 = 0.
(v) Substituting x = 2 and y = -1.5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :
L.H.S. = 2 - 2 × (-1.5) + 5
= 2 + 3 + 5
= 10
Since, L.H.S. ≠ R.H.S.
∴ Point (2, -1.5) does not satisfies the equation.
Hence, (2, -1.5) does not lies on the line represented by the equation x - 2y + 5 = 0.
(vi) Substituting x = -2 and y = -1.5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :
L.H.S. = -2 - 2 × (-1.5) + 5
= -2 + 3 + 5
= 6
Since, L.H.S. ≠ R.H.S.
∴ Point (-2, -1.5) does not satisfies the equation.
Hence, (-2, -1.5) does not lies on the line represented by the equation x - 2y + 5 = 0.
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