Find all points on x - y + 2 = 0 that lie at a unit distance from the line `12x - 5y + 9 = 0`

Answer

To solve the problem of finding all points on the line \( x - y + 2 = 0 \) that lie at a unit distance from the line \( 12x - 5y + 9 = 0 \), we can follow these steps:

### Step 1: Identify the line equation and the point on the line
The line equation is given as:
\[ x - y + 2 = 0 \]
We can express \( y \) in terms of \( x \):
\[ y = x + 2 \]
Thus, any point on this line can be represented as \( (x, x + 2) \).

### Step 2: Write the distance formula from a point to a line
The distance \( d \) from a point \( (h, k) \) to a line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
For our second line \( 12x - 5y + 9 = 0 \), we have \( A = 12 \), \( B = -5 \), and \( C = 9 \).

### Step 3: Substitute the point into the distance formula
Substituting \( h = x \) and \( k = x + 2 \) into the distance formula, we get:
\[ d = \frac{|12x - 5(x + 2) + 9|}{\sqrt{12^2 + (-5)^2}} \]
Calculating the denominator:
\[ \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]
Thus, the distance becomes:
\[ d = \frac{|12x - 5x - 10 + 9|}{13} = \frac{|7x - 1|}{13} \]

### Step 4: Set the distance equal to 1
Since we want the distance to be 1, we set up the equation:
\[ \frac{|7x - 1|}{13} = 1 \]
Multiplying both sides by 13 gives:
\[ |7x - 1| = 13 \]

### Step 5: Solve the absolute value equation
This absolute value equation gives us two cases to solve:
1. \( 7x - 1 = 13 \)
2. \( 7x - 1 = -13 \)

**Case 1:**
\[ 7x - 1 = 13 \implies 7x = 14 \implies x = 2 \]
Substituting \( x = 2 \) back into \( y = x + 2 \):
\[ y = 2 + 2 = 4 \]
Thus, one point is \( (2, 4) \).

**Case 2:**
\[ 7x - 1 = -13 \implies 7x = -12 \implies x = -\frac{12}{7} \]
Substituting \( x = -\frac{12}{7} \) back into \( y = x + 2 \):
\[ y = -\frac{12}{7} + 2 = -\frac{12}{7} + \frac{14}{7} = \frac{2}{7} \]
Thus, the second point is \( \left(-\frac{12}{7}, \frac{2}{7}\right) \).

### Final Points
The points on the line \( x - y + 2 = 0 \) that lie at a unit distance from the line \( 12x - 5y + 9 = 0 \) are:
\[ (2, 4) \quad \text{and} \quad \left(-\frac{12}{7}, \frac{2}{7}\right) \]