Line y=2x-b cuts the parabola `y=x^(2)-4x` at points A and B. Then the value of b for which `angleAOB` is a right is (where O is origin) _________ .

Explanation

To solve the problem step by step, we will find the value of \( b \) for which the angle \( AOB \) is a right angle, where \( O \) is the origin and \( A \) and \( B \) are the points of intersection of the line and the parabola.

### Step 1: Write the equations of the line and parabola
The given line is:
\[ y = 2x - b \]
The given parabola is:
\[ y = x^2 - 4x \]

### Step 2: Set the equations equal to find points of intersection
To find the points of intersection, we set the equations equal to each other:
\[ 2x - b = x^2 - 4x \]
Rearranging this gives:
\[ x^2 - 6x + b = 0 \]

### Step 3: Identify the coefficients for the condition of a right angle
For the angle \( AOB \) to be a right angle, the product of the slopes of the lines OA and OB must be -1. The slopes of the lines OA and OB can be derived from the points of intersection. The line can be expressed in the form of a quadratic equation, and we need to find the coefficients.

The quadratic equation we have is:
\[ x^2 - 6x + b = 0 \]
The coefficients are:
- Coefficient of \( x^2 \) is \( 1 \)
- Coefficient of \( x \) is \( -6 \)
- Constant term is \( b \)

### Step 4: Use the condition for right angles
For the angle \( AOB \) to be 90 degrees, we can use the condition that the sum of the squares of the coefficients of \( x \) and \( y \) must equal zero:
\[ \text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 0 \]
Here, the coefficient of \( y^2 \) can be derived from the homogenized equation. The homogenized equation is:
\[ 1 - \frac{8}{b} + \frac{1}{b} = 0 \]
This simplifies to:
\[ 1 - \frac{8}{b} + \frac{1}{b} = 0 \]
Combining terms gives:
\[ 1 - \frac{7}{b} = 0 \]

### Step 5: Solve for \( b \)
Solving the equation:
\[ 1 = \frac{7}{b} \implies b = 7 \]

### Final Answer
Thus, the value of \( b \) for which the angle \( AOB \) is a right angle is:
\[ \boxed{7} \]