a and b are real numbers between 0 and 1 `A(a,1),B(1,b)and C(0,0)` are the vertices of a triangle.
If `angleC=90^(@)` then

Options

A. `a+b=0`
B. `a-b=0`
C. `a+b=1`
D. `a-b=1`

Answer

To solve the problem, we need to find the relationship between the real numbers \( a \) and \( b \) given that the triangle formed by the points \( A(a, 1) \), \( B(1, b) \), and \( C(0, 0) \) has a right angle at point \( C \).

### Step-by-Step Solution:

1. **Identify the Points**:
- The points of the triangle are \( A(a, 1) \), \( B(1, b) \), and \( C(0, 0) \).

2. **Find the Slopes**:
- The slope of line segment \( AC \) is calculated as:
\[ \text{slope of } AC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 0}{a - 0} = \frac{1}{a} \]
- The slope of line segment \( BC \) is calculated as:
\[ \text{slope of } BC = \frac{b - 0}{1 - 0} = b \]

3. **Use the Right Angle Condition**:
- Since \( \angle C \) is \( 90^\circ \), the product of the slopes of lines \( AC \) and \( BC \) must be equal to \(-1\):
\[ \text{slope of } AC \cdot \text{slope of } BC = -1 \]
Substituting the slopes we found:
\[ \frac{1}{a} \cdot b = -1 \]

4. **Set Up the Equation**:
- Rearranging the equation gives:
\[ b = -a \]

5. **Consider the Constraints**:
- Since \( a \) and \( b \) are both real numbers between 0 and 1, the equation \( b = -a \) implies that both \( a \) and \( b \) cannot simultaneously satisfy this equation within the given range. Therefore, we conclude that there is no valid solution for \( a \) and \( b \) under the constraints provided.

### Conclusion:
The relationship derived from the right angle condition leads to an inconsistency with the constraints that \( a \) and \( b \) must be between 0 and 1. Thus, no valid values for \( a \) and \( b \) exist under the given conditions.