State true or false :

(i) the line $\dfrac{x}{2} + \dfrac{y}{3} = 0$ passes through the point (2, 3).

(ii) the line $\dfrac{x}{2} + \dfrac{y}{3} = 0$ passes through the point (4, -6).

(iii) the point (8, 7) lies on the line y - 7 = 0

(iv) the point (-3, 0) lies on the line x + 3 = 0

(v) if the point (2, a) lies on the line 2x - y = 3, then a = 5.

(i) Substituting x = 2 and y = 3 in the L.H.S. of the equation $\dfrac{x}{2} + \dfrac{y}{3} = 0$, we get :

L.H.S. = $\dfrac{2}{2} + \dfrac{3}{3}$

= 1 + 1

= 2

Since, L.H.S. ≠ R.H.S.

∴ The line $\dfrac{x}{2} + \dfrac{y}{3} = 0$ does not passes through the point (2, 3).

Hence, the statement is false.

(ii) Substituting x = 4 and y = -6 in the L.H.S. of the equation $\dfrac{x}{2} + \dfrac{y}{3} = 0$, we get :

L.H.S. = $\dfrac{4}{2} + \dfrac{-6}{3}$

= 2 + (-2)

= 0

Since, L.H.S. = R.H.S.

∴ The line $\dfrac{x}{2} + \dfrac{y}{3} = 0$ passes through the point (4, -6).

Hence, the statement is true.

(iii) Substituting y = 7 in the L.H.S. of the equation y - 7 = 0, we get :

L.H.S. = 7 - 7

= 0.

Since, L.H.S. = R.H.S.

∴ Point (8, 7) lies on the line y - 7 = 0.

Hence, the statement is true.

(iv) Substituting x = -3 in the L.H.S. of the equation x + 3 = 0, we get :

L.H.S. = -3 + 3

= 0.

Since, L.H.S. = R.H.S.

∴ Point (-3, 0) lies on the line x + 3 = 0.

Hence, the statement is true.

(v) Given,

(2, a) lies on the line 2x - y = 3.

∴ Substituting (2, a) in 2x - y = 3, satisfies the equation.

⇒ 2(2) - a = 3

⇒ 4 - a = 3

⇒ a = 4 - 3 = 1.

Hence, the statement is false.