Find the equation of line passing through (1,3) and inclined at angle `tan^-1(sqrt2)` with `y=3sqrt2x+1`

Answer

To find the equation of the line passing through the point (1, 3) and inclined at an angle of \( \tan^{-1}(\sqrt{2}) \) with the line given by \( y = 3\sqrt{2}x + 1 \), we will follow these steps:

### Step 1: Identify the slope of the given line
The equation of the given line is \( y = 3\sqrt{2}x + 1 \). The slope \( m_2 \) of this line is \( 3\sqrt{2} \).

**Hint:** The slope of a line in the form \( y = mx + c \) is simply \( m \).

### Step 2: Determine the angle and its tangent
The angle of inclination given is \( \theta = \tan^{-1}(\sqrt{2}) \). Therefore, we have:
\[ \tan \theta = \sqrt{2} \]

**Hint:** Remember that the tangent of the angle gives the ratio of the opposite side to the adjacent side in a right triangle.

### Step 3: Use the formula for the tangent of the angle between two lines
The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( m_2 = 3\sqrt{2} \) and \( \tan \theta = \sqrt{2} \):
\[ \sqrt{2} = \left| \frac{m_1 - 3\sqrt{2}}{1 + 3\sqrt{2} m_1} \right| \]

**Hint:** The absolute value indicates that there are two possible cases for \( m_1 \).

### Step 4: Solve for \( m_1 \) (Case 1)
First, we consider the positive case:
\[ \sqrt{2} = \frac{m_1 - 3\sqrt{2}}{1 + 3\sqrt{2} m_1} \]
Cross-multiplying gives:
\[ \sqrt{2}(1 + 3\sqrt{2} m_1) = m_1 - 3\sqrt{2} \]
Expanding and rearranging:
\[ \sqrt{2} + 6m_1 = m_1 - 3\sqrt{2} \]
\[ 5m_1 = -4\sqrt{2} \]
Thus,
\[ m_1 = -\frac{4\sqrt{2}}{5} \]

**Hint:** When rearranging equations, keep track of the signs and terms carefully.

### Step 5: Solve for \( m_1 \) (Case 2)
Now consider the negative case:
\[ -\sqrt{2} = \frac{m_1 - 3\sqrt{2}}{1 + 3\sqrt{2} m_1} \]
Cross-multiplying gives:
\[ -\sqrt{2}(1 + 3\sqrt{2} m_1) = m_1 - 3\sqrt{2} \]
Expanding and rearranging:
\[ - \sqrt{2} - 3 \cdot 2 m_1 = m_1 - 3\sqrt{2} \]
\[ -6m_1 = -2\sqrt{2} \]
Thus,
\[ m_1 = \frac{2\sqrt{2}}{7} \]

**Hint:** Always check both cases when dealing with absolute values.

### Step 6: Write the equations of the lines
Now we have two slopes for the lines:
1. \( m_1 = -\frac{4\sqrt{2}}{5} \)
2. \( m_1 = \frac{2\sqrt{2}}{7} \)

Using the point-slope form \( y - y_1 = m(x - x_1) \) with the point (1, 3):

**For the first line:**
\[ y - 3 = -\frac{4\sqrt{2}}{5}(x - 1) \]
Expanding:
\[ y - 3 = -\frac{4\sqrt{2}}{5}x + \frac{4\sqrt{2}}{5} \]
\[ y = -\frac{4\sqrt{2}}{5}x + 3 + \frac{4\sqrt{2}}{5} \]
\[ y = -\frac{4\sqrt{2}}{5}x + \frac{15 + 4\sqrt{2}}{5} \]

**For the second line:**
\[ y - 3 = \frac{2\sqrt{2}}{7}(x - 1) \]
Expanding:
\[ y - 3 = \frac{2\sqrt{2}}{7}x - \frac{2\sqrt{2}}{7} \]
\[ y = \frac{2\sqrt{2}}{7}x + 3 - \frac{2\sqrt{2}}{7} \]
\[ y = \frac{2\sqrt{2}}{7}x + \frac{21 - 2\sqrt{2}}{7} \]

### Final Result
The equations of the two lines are:
1. \( y = -\frac{4\sqrt{2}}{5}x + \frac{15 + 4\sqrt{2}}{5} \)
2. \( y = \frac{2\sqrt{2}}{7}x + \frac{21 - 2\sqrt{2}}{7} \)