Given the equations:

1. Curve: \(xy = 12\)

2. Line: \(2x + y = 11\)

Substitute \(y = 11 - 2x\) from the line equation into the curve equation:

\(x(11 - 2x) = 12\)

\(11x - 2x^2 = 12\)

Rearrange to form a quadratic equation:

\(2x^2 - 11x + 12 = 0\)

Factor the quadratic equation:

\((2x - 3)(x - 4) = 0\)

Thus, \(x = \frac{3}{2}\) or \(x = 4\).

For \(x = \frac{3}{2}\), substitute back to find \(y\):

\(y = 11 - 2 \times \frac{3}{2} = 8\)

For \(x = 4\), substitute back to find \(y\):

\(y = 11 - 2 \times 4 = 3\)

Therefore, the points of intersection are \((1.5, 8)\) and \((4, 3)\).