A curve has equation y = 2xy+5 and a line has equation 2x +5y= 1.

The curve and the line intersect at the points A and B. Find the coordinates of the midpoint

of the line AB.

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.

Discover how to find the midpoint of the line segment AB where the curve y = 2xy + 5 and the line 2x + 5y = 1 intersect. Includes step-by-step solution with detailed calculations. Suitable for advanced high school students in Grades 10-12.

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