If `(a,-2)` and `(4,b^(2))` belong to the relation R where `R={(x,y) : x,y in I, y=2x-4}` then find the values of a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given that the points \( (a, -2) \) and \( (4, b^2) \) belong to the relation \( R \) defined by \( R = \{(x, y) : x, y \in \mathbb{I}, y = 2x - 4\} \).

### Step-by-Step Solution:

1. **Understanding the Relation**:
The relation \( R \) is defined such that for any point \( (x, y) \) in \( R \), the \( y \)-coordinate is given by the equation \( y = 2x - 4 \).

2. **Finding the Value of \( a \)**:
Since the point \( (a, -2) \) belongs to \( R \), we can substitute \( y = -2 \) and \( x = a \) into the equation:
\[ -2 = 2a - 4 \]
Rearranging this equation:
\[ 2a = -2 + 4 \]
\[ 2a = 2 \]
Dividing both sides by 2:
\[ a = 1 \]

3. **Finding the Value of \( b \)**:
Next, we consider the point \( (4, b^2) \) which also belongs to \( R \). We substitute \( x = 4 \) and \( y = b^2 \) into the equation:
\[ b^2 = 2(4) - 4 \]
Simplifying this:
\[ b^2 = 8 - 4 \]
\[ b^2 = 4 \]
Taking the square root of both sides gives:
\[ b = \pm 2 \]

4. **Final Values**:
Thus, the values we found are:
\[ a = 1 \quad \text{and} \quad b = \pm 2 \]

### Summary:
- The value of \( a \) is \( 1 \).
- The values of \( b \) are \( 2 \) and \( -2 \).