Find points on the line `x + y + 3 = 0` that are at a distance of 5 units from the line `x + 2y + 2 = 0`

Answer

To find points on the line \( x + y + 3 = 0 \) that are at a distance of 5 units from the line \( x + 2y + 2 = 0 \), we can follow these steps:

### Step 1: Understand the equations of the lines
We have two lines:
1. Line 1: \( x + y + 3 = 0 \)
2. Line 2: \( x + 2y + 2 = 0 \)

### Step 2: Express points on Line 1
Any point on Line 1 can be expressed in terms of a parameter \( h \):
Let \( (h, k) \) be a point on Line 1. From the equation of Line 1, we have:
\[ k = -h - 3 \]
Thus, the point can be represented as \( (h, -h - 3) \).

### Step 3: Use the distance formula
The distance \( d \) from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is given by:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
For Line 2 \( x + 2y + 2 = 0 \), we have \( a = 1, b = 2, c = 2 \).

### Step 4: Set up the distance equation
We want the distance from the point \( (h, -h - 3) \) to Line 2 to be 5 units:
\[ 5 = \frac{|1 \cdot h + 2(-h - 3) + 2|}{\sqrt{1^2 + 2^2}} \]
Calculating the denominator:
\[ \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]
So, we can rewrite the distance equation:
\[ 5 = \frac{|h - 2h - 6 + 2|}{\sqrt{5}} \]
This simplifies to:
\[ 5 = \frac{|-h - 4|}{\sqrt{5}} \]

### Step 5: Solve the absolute value equation
Multiplying both sides by \( \sqrt{5} \):
\[ 5\sqrt{5} = |-h - 4| \]
This gives us two cases to consider:
1. \( -h - 4 = 5\sqrt{5} \)
2. \( -h - 4 = -5\sqrt{5} \)

### Step 6: Solve for \( h \)
**Case 1:**
\[ -h - 4 = 5\sqrt{5} \implies -h = 5\sqrt{5} + 4 \implies h = -5\sqrt{5} - 4 \]

**Case 2:**
\[ -h - 4 = -5\sqrt{5} \implies -h = -5\sqrt{5} + 4 \implies h = 5\sqrt{5} - 4 \]

### Step 7: Find corresponding \( k \) values
Using \( k = -h - 3 \):
1. For \( h = -5\sqrt{5} - 4 \):
\[ k = -(-5\sqrt{5} - 4) - 3 = 5\sqrt{5} + 4 - 3 = 5\sqrt{5} + 1 \]

2. For \( h = 5\sqrt{5} - 4 \):
\[ k = -(5\sqrt{5} - 4) - 3 = -5\sqrt{5} + 4 - 3 = -5\sqrt{5} + 1 \]

### Final Points
Thus, the points on the line \( x + y + 3 = 0 \) that are at a distance of 5 units from the line \( x + 2y + 2 = 0 \) are:
1. \( \left(-5\sqrt{5} - 4, 5\sqrt{5} + 1\right) \)
2. \( \left(5\sqrt{5} - 4, -5\sqrt{5} + 1\right) \)