If `A(1,1),B(sqrt(3)+1,2)` and `C(sqrt(3),sqrt(3)+2)` be three vertices of a square then the diagonal through B is

Options

A. `y=(sqrt(3)-2)x+(3-sqrt(3))`
B. `y=0`
C. `y=x`
D. None of these

Answer

To find the diagonal through point B of the square formed by the vertices A(1,1), B(√3 + 1, 2), and C(√3, √3 + 2), we will follow these steps:

### Step 1: Identify the coordinates of the given points
We have the following points:
- A(1, 1)
- B(√3 + 1, 2)
- C(√3, √3 + 2)

### Step 2: Determine the slope of line AC
To find the slope of line AC, we use the slope formula:
\[ m_{AC} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of points A and C:
\[ m_{AC} = \frac{(\sqrt{3} + 2) - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Step 3: Find the slope of the diagonal BD
Since the diagonals of a square are perpendicular, the slope of BD (let's denote it as \( m_{BD} \)) will be the negative reciprocal of \( m_{AC} \):
\[ m_{BD} = -\frac{1}{m_{AC}} = -\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 4: Write the equation of line BD
Using point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting point B(√3 + 1, 2) and the slope \( m_{BD} \):
\[ y - 2 = -\frac{\sqrt{3} - 1}{\sqrt{3} + 1}(x - (\sqrt{3} + 1)) \]

### Step 5: Simplify the equation
Distributing and rearranging gives us the equation of line BD:
\[ y - 2 = -\frac{\sqrt{3} - 1}{\sqrt{3} + 1}x + \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{\sqrt{3} + 1} \]
This simplifies to:
\[ y = -\frac{\sqrt{3} - 1}{\sqrt{3} + 1}x + 2 + \frac{2}{\sqrt{3} + 1} \]

### Step 6: Finalize the equation
This gives us the equation of the diagonal through point B.