If a line is passing through `p(2,3)` which intersects the line `x+y=7` at a distance of four units from `P`. Then the slope of line is

Options

A. `(1-sqrt(7))/(1-sqrt(7))`
B. `(sqrt(7)-1)/(sqrt(7)+1)`
C. `(1-sqrt(5))/(1+sqrt(5))`
D. `(sqrt(5)-1))/(sqrt(5)=1)`

Answer

To find the slope of the line passing through the point \( P(2, 3) \) that intersects the line \( x + y = 7 \) at a distance of 4 units from point \( P \), we can follow these steps:

### Step 1: Identify the equation of the line
The given line is \( x + y = 7 \). We can rewrite it in the standard form \( Ax + By + C = 0 \):
\[ x + y - 7 = 0 \]
Here, \( A = 1 \), \( B = 1 \), and \( C = -7 \).

### Step 2: Calculate the slope of the given line
The slope \( m_2 \) of the line \( x + y = 7 \) can be found using the formula \( m = -\frac{A}{B} \):
\[ m_2 = -\frac{1}{1} = -1 \]

### Step 3: Use the distance formula
The distance \( d \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \( A = 1 \), \( B = 1 \), \( C = -7 \), and point \( P(2, 3) \):
\[ d = \frac{|1 \cdot 2 + 1 \cdot 3 - 7|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 7|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 4: Set up the relationship with the distance
We know that the distance from point \( P \) to the line \( x + y = 7 \) is \( \sqrt{2} \), and we need to find the distance to the intersection point which is given as 4 units. Let \( O \) be the foot of the perpendicular from \( P \) to the line \( x + y = 7 \).

Using the Pythagorean theorem in triangle \( POO' \) (where \( O' \) is the intersection point):
\[ PO^2 + OO'^2 = PO'^2 \]
Where \( PO = \sqrt{2} \) and \( PO' = 4 \):
\[ (\sqrt{2})^2 + OO'^2 = 4^2 \]
\[ 2 + OO'^2 = 16 \]
\[ OO'^2 = 14 \quad \Rightarrow \quad OO' = \sqrt{14} \]

### Step 5: Use the tangent formula
Now we can use the tangent formula for the angle \( \theta \) between the two lines:
\[ \tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} \]
Where \( m_1 \) is the slope of the line we want to find. We know \( m_2 = -1 \) and \( \tan \theta = \frac{\sqrt{2}}{\sqrt{14}} \).

### Step 6: Set up the equation
Substituting the values into the tangent formula:
\[ \frac{\sqrt{2}}{\sqrt{14}} = \frac{|m_1 + 1|}{1 - m_1} \]

### Step 7: Solve for \( m_1 \)
Cross-multiplying gives:
\[ \sqrt{2}(1 - m_1) = \sqrt{14}|m_1 + 1| \]
This leads to two cases to solve for \( m_1 \):
1. \( \sqrt{2}(1 - m_1) = \sqrt{14}(m_1 + 1) \)
2. \( \sqrt{2}(1 - m_1) = -\sqrt{14}(m_1 + 1) \)

### Step 8: Solve the equations
1. **First case:**
\[ \sqrt{2} - \sqrt{2}m_1 = \sqrt{14}m_1 + \sqrt{14} \]
Rearranging gives:
\[ m_1(\sqrt{2} + \sqrt{14}) = \sqrt{2} - \sqrt{14} \]
\[ m_1 = \frac{\sqrt{2} - \sqrt{14}}{\sqrt{2} + \sqrt{14}} \]

2. **Second case:**
\[ \sqrt{2} - \sqrt{2}m_1 = -\sqrt{14}m_1 - \sqrt{14} \]
Rearranging gives:
\[ m_1(\sqrt{2} - \sqrt{14}) = \sqrt{2} + \sqrt{14} \]
\[ m_1 = \frac{\sqrt{2} + \sqrt{14}}{\sqrt{2} - \sqrt{14}} \]

### Final Result
Thus, the slopes of the line can be:
\[ m_1 = \frac{\sqrt{2} - \sqrt{14}}{\sqrt{2} + \sqrt{14}} \quad \text{or} \quad m_1 = \frac{\sqrt{2} + \sqrt{14}}{\sqrt{2} - \sqrt{14}} \]