The equation of the line passing though the intersection of `x-sqrt(3)y+sqrt(3)-1=0` and `x+y-2=0` and making an angle of `15^@` with the first line is

Options

A. x - y = 0
B. x - y + 1 = 0
C. y = 1
D. `sqrt3 x - y + 1 - sqrt3 = 0`

Answer

To solve the problem, we need to find the equation of a line that passes through the intersection of two given lines and makes an angle of \(15^\circ\) with the first line. Let's break this down step by step.

### Step 1: Find the intersection of the two lines
The two lines given are:
1. \(x - \sqrt{3}y + \sqrt{3} - 1 = 0\) (let's call this Line 1)
2. \(x + y - 2 = 0\) (let's call this Line 2)

To find the intersection, we can solve these equations simultaneously.

From Line 2, we can express \(x\) in terms of \(y\):
\[ x = 2 - y \]

Now, substitute \(x\) in Line 1:
\[ (2 - y) - \sqrt{3}y + \sqrt{3} - 1 = 0 \]
Simplifying this:
\[ 2 - y - \sqrt{3}y + \sqrt{3} - 1 = 0 \]
\[ 1 + \sqrt{3} - (1 + \sqrt{3})y = 0 \]
\[ y(1 + \sqrt{3}) = 1 + \sqrt{3} \]
Thus, we find:
\[ y = 1 \]

Now substituting \(y = 1\) back into Line 2 to find \(x\):
\[ x + 1 - 2 = 0 \implies x = 1 \]

So, the point of intersection is \((1, 1)\).

### Step 2: Find the slope of Line 1
Next, we need to find the slope of Line 1. Rearranging Line 1:
\[ x - \sqrt{3}y + \sqrt{3} - 1 = 0 \implies \sqrt{3}y = x + \sqrt{3} - 1 \]
\[ y = \frac{1}{\sqrt{3}}x + \frac{\sqrt{3} - 1}{\sqrt{3}} \]
The slope \(m_1\) of Line 1 is:
\[ m_1 = \frac{1}{\sqrt{3}} \]

### Step 3: Use the angle formula to find the new slope
We are given that the new line makes an angle of \(15^\circ\) with Line 1. The formula for the tangent of the angle between two lines is:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
where \(m_2\) is the slope of the new line.

Substituting \(\theta = 15^\circ\) and \(m_1 = \frac{1}{\sqrt{3}}\):
\[ \tan 15^\circ = 2 - \sqrt{3} \]
Thus, we have:
\[ 2 - \sqrt{3} = \left| \frac{\frac{1}{\sqrt{3}} - m_2}{1 + \frac{1}{\sqrt{3}} m_2} \right| \]

### Step 4: Solve for \(m_2\)
We will consider both cases for the absolute value.

**Case 1:**
\[ 2 - \sqrt{3} = \frac{\frac{1}{\sqrt{3}} - m_2}{1 + \frac{1}{\sqrt{3}} m_2} \]
Cross-multiplying gives:
\[ (2 - \sqrt{3})(1 + \frac{1}{\sqrt{3}} m_2) = \frac{1}{\sqrt{3}} - m_2 \]

**Case 2:**
\[ 2 - \sqrt{3} = -\left(\frac{\frac{1}{\sqrt{3}} - m_2}{1 + \frac{1}{\sqrt{3}} m_2}\right) \]
This will give us another equation for \(m_2\).

### Step 5: Find the equations of the lines
Once we find the values of \(m_2\), we can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1) = (1, 1)\) is the point of intersection.