1.)The graph of the linear equation 3x-2y=0 passes through the point

a.) (2/3, -2/3)

b.) (2/3, 3/2)

c.) (1/3, 1/2)

The given equation is 3x - 2y = 0
Put x = 1/3   and y = 1/2 in the LHS of above equation, we get
$$\begin{gathered} \mathrm{LHS} = 3 \times \frac{1}{3} - 2 \times \frac{1}{2} = 1 - 1 = 0 \\ \mathrm{RHS} = 0 \\ \mathrm{so}, \mathrm{LHS} = \mathrm{RHS} \\ \mathrm{so} \mathrm{graph} \mathrm{of} \mathrm{the} \mathrm{equation} \mathrm{passes} \mathrm{through} \left(\frac{1}{3}, \frac{1}{2}\right) \end{gathered}$$

When we substitute the remaining two points , we will see that they will not satisfy the given equation.