Evaluate `lamda`, if three points (0, 0), `(3, sqrt3)` and `(3, lamda)` form an equilateral triangle.

Options

A. `-4`
B. 2
C. `-3`
D. `pmsqrt3`

Answer

To evaluate `lambda` such that the points (0, 0), (3, √3), and (3, λ) form an equilateral triangle, we will follow these steps:

### Step 1: Identify the points
The points given are:
- A(0, 0)
- B(3, √3)
- C(3, λ)

### Step 2: Use the distance formula
The distance formula between two points (x1, y1) and (x2, y2) is given by:
\[ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \]

### Step 3: Calculate the distances
We need to find the distances AB, AC, and BC and set them equal since the triangle is equilateral.

1. **Distance AB**:
\[ AB = \sqrt{(3 - 0)^2 + (\sqrt{3} - 0)^2} = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \]

2. **Distance AC**:
\[ AC = \sqrt{(3 - 0)^2 + (\lambda - 0)^2} = \sqrt{3^2 + \lambda^2} = \sqrt{9 + \lambda^2} \]

3. **Distance BC**:
\[ BC = \sqrt{(3 - 3)^2 + (\lambda - \sqrt{3})^2} = \sqrt{0 + (\lambda - \sqrt{3})^2} = |\lambda - \sqrt{3}| \]

### Step 4: Set distances equal
Since the triangle is equilateral, we have:
\[ AB = AC = BC \]

1. Set AB equal to AC:
\[ 2\sqrt{3} = \sqrt{9 + \lambda^2} \]
Squaring both sides:
\[ (2\sqrt{3})^2 = 9 + \lambda^2 \]
\[ 12 = 9 + \lambda^2 \]
\[ \lambda^2 = 12 - 9 = 3 \]
\[ \lambda = \pm \sqrt{3} \]

### Step 5: Set AB equal to BC
We can also set AB equal to BC to confirm:
\[ 2\sqrt{3} = |\lambda - \sqrt{3}| \]
This gives us two cases:
1. \(\lambda - \sqrt{3} = 2\sqrt{3}\) which leads to \(\lambda = 3\sqrt{3}\)
2. \(\lambda - \sqrt{3} = -2\sqrt{3}\) which leads to \(\lambda = -\sqrt{3}\)

However, since we already found \(\lambda = \pm \sqrt{3}\) from the previous calculation, we conclude:

### Final Answer:
\[ \lambda = \sqrt{3} \text{ or } \lambda = -\sqrt{3} \]