To find the coordinates of the midpoint of the line segment $AB$ where the curve $y = 2xy + 5$ and the line $2x + 5y = 1$ intersect, we need to solve for the points of intersection and then calculate the midpoint. Here are the steps:

1. Solve the system of equations:
   $y = 2xy + 5 \quad \text{(Equation 1)}$ $2x + 5y = 1 \quad \text{(Equation 2)}$

2. Substitute Equation 1 into Equation 2 to express $y$ in terms of $x$.

Step 1: Rearrange and Substitute

From Equation 1: $y = 2xy + 5$ This can be rearranged to: $y - 2xy = 5 \quad \Rightarrow \quad y(1 - 2x) = 5 \quad \Rightarrow \quad y = \frac{5}{1 - 2x}$

Substitute this expression for $y$ in Equation 2: $2x + 5\left(\frac{5}{1 - 2x}\right) = 1$

Step 2: Solve for $x$

Simplify and solve the equation: $2x + \frac{25}{1 - 2x} = 1$

Multiply through by $1 - 2x$ to clear the fraction: $2x(1 - 2x) + 25 = 1 - 2x$

Expanding: $2x - 4x^2 + 25 = 1 - 2x$

Rearrange terms: $-4x^2 + 4x + 24 = 0$

Divide through by $-4$: $x^2 - x - 6 = 0$

Step 3: Solve the Quadratic Equation

Factor the quadratic equation: $x^2 - x - 6 = (x - 3)(x + 2) = 0$

The solutions are: $x = 3 \quad \text{or} \quad x = -2$

Step 4: Find Corresponding $y$ Values

For $x = 3$: $y = \frac{5}{1 - 2(3)} = \frac{5}{1 - 6} = \frac{5}{-5} = -1$

For $x = -2$: $y = \frac{5}{1 - 2(-2)} = \frac{5}{1 + 4} = \frac{5}{5} = 1$

Step 5: Find the Midpoint of $AB$

The points of intersection are $A(3, -1)$ and $B(-2, 1)$.

The midpoint $M$ of the line segment $AB$ is calculated as: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Substituting the values: $M = \left(\frac{3 + (-2)}{2}, \frac{-1 + 1}{2}\right) = \left(\frac{1}{2}, 0\right)$

Thus, the coordinates of the midpoint are $\left(\frac{1}{2}, 0\right)$.

Would you like further details or have any questions about this solution? Here are some related questions that might help deepen your understanding:

1. How do you derive the equation of a line from two points?
2. What are other methods to solve quadratic equations?
3. How can you find the point of intersection for two non-linear equations?
4. What are the properties of a quadratic curve?
5. How can a graphical method help in visualizing the solutions?
6. How does changing the coefficients of the equations affect their points of intersection?
7. What happens when a curve and line do not intersect?
8. How do transformations affect the geometry of a line or a curve?

Tip: When solving simultaneous equations involving non-linear terms, substituting one equation into the other to eliminate a variable can often simplify the problem significantly.