First, substitute \(k = 2\) into the equations:

Line: \(y = 2x + 6\)

Curve: \(y = x^2 + 3x + 4\)

Set the equations equal to find intersection points:

\(x^2 + 3x + 4 = 2x + 6\)

Simplify to: \(x^2 + x - 2 = 0\)

Factorize: \((x - 1)(x + 2) = 0\)

Solutions: \(x = 1\) and \(x = -2\)

Find corresponding \(y\)-values:

For \(x = 1\), \(y = 2(1) + 6 = 8\)

For \(x = -2\), \(y = 2(-2) + 6 = 2\)

Points \(A(1, 8)\) and \(B(-2, 2)\)

Distance \(AB\):

\(AB = \sqrt{(1 - (-2))^2 + (8 - 2)^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}\)

Midpoint of \(AB\):

\(\left(\frac{1 + (-2)}{2}, \frac{8 + 2}{2}\right) = \left(-\frac{1}{2}, 5\right)\)