If a and b are real numbers between 0 and 1 such that the points (a,1),(1,b) and (0,0) form an equilateral triangle, then a,b are

Options

A. `2-sqrt(3),2-sqrt(3)`
B. `sqrt(3)-1,sqrt(3)-1`
C. `sqrt(2)-1,sqrt(2)-1`
D. None of these

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the points \( (a, 1) \), \( (1, b) \), and \( (0, 0) \) form an equilateral triangle.

### Step-by-Step Solution:

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

2. **Calculate the Distances**:
We need to calculate the lengths of the sides of the triangle formed by these points.
- Distance \( AB \):
\[ AB = \sqrt{(1 - a)^2 + (b - 1)^2} \]
- Distance \( BC \):
\[ BC = \sqrt{(1 - 0)^2 + (b - 0)^2} = \sqrt{1 + b^2} \]
- Distance \( CA \):
\[ CA = \sqrt{(a - 0)^2 + (1 - 0)^2} = \sqrt{a^2 + 1} \]

3. **Set Up the Equations**:
For the triangle to be equilateral, all sides must be equal:
\[ AB = BC = CA \]

4. **Equate Distances**:
First, equate \( AB \) and \( BC \):
\[ \sqrt{(1 - a)^2 + (b - 1)^2} = \sqrt{1 + b^2} \]
Squaring both sides:
\[ (1 - a)^2 + (b - 1)^2 = 1 + b^2 \]
Expanding:
\[ (1 - 2a + a^2) + (b^2 - 2b + 1) = 1 + b^2 \]
Simplifying:
\[ 1 - 2a + a^2 - 2b + 1 = 1 \]
\[ a^2 - 2a - 2b + 1 = 0 \quad \text{(Equation 1)} \]

5. **Equate \( BC \) and \( CA \)**:
Next, equate \( BC \) and \( CA \):
\[ \sqrt{1 + b^2} = \sqrt{a^2 + 1} \]
Squaring both sides:
\[ 1 + b^2 = a^2 + 1 \]
Simplifying:
\[ b^2 = a^2 \quad \text{(Equation 2)} \]

6. **Substituting Equation 2 into Equation 1**:
From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = a \quad \text{or} \quad b = -a \]
Since \( a \) and \( b \) are between 0 and 1, we take \( b = a \).
Substitute \( b = a \) into Equation 1:
\[ a^2 - 2a - 2a + 1 = 0 \]
\[ a^2 - 4a + 1 = 0 \]

7. **Solve the Quadratic Equation**:
Using the quadratic formula:
\[ a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ a = \frac{4 \pm \sqrt{16 - 4}}{2} \]
\[ a = \frac{4 \pm \sqrt{12}}{2} \]
\[ a = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ a = 2 \pm \sqrt{3} \]

8. **Determine Valid Values**:
Since \( a \) must be between 0 and 1:
\[ 2 - \sqrt{3} \quad \text{is valid since} \quad 2 - \sqrt{3} \approx 0.268 < 1 \]
The other value \( 2 + \sqrt{3} \) is greater than 1, so we discard it.

9. **Final Result**:
Therefore, \( a = b = 2 - \sqrt{3} \).