To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a structured approach.
### Step 1: Understand the given equation of the family of lines
We are given the equation of the family of lines as:
\[ 2a + 2b + c = 0 \]
This can be rewritten in the standard line form \( ax + by + c = 0 \).
### Step 2: Identify the point through which the lines pass
From the equation \( 2a + 2b + c = 0 \), we can deduce that all lines of this family pass through the point \( (2, 2) \). This is because if we substitute \( x = 2 \) and \( y = 2 \) into the line equation, it satisfies the equation.
### Step 3: Determine the distance condition
We need to find the lines \( l_1 \) and \( l_2 \) that are at a distance of 1 unit from the point \( (1, 1) \). The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( ax + by + c = 0 \) is given by:
\[
d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}
\]
Substituting \( (x_0, y_0) = (1, 1) \) and setting \( d = 1 \), we have:
\[
\frac{|a + b + c|}{\sqrt{a^2 + b^2}} = 1
\]
### Step 4: Express the distance condition
From the distance condition, we can derive:
\[
|a + b + c| = \sqrt{a^2 + b^2}
\]
This gives us two cases:
1. \( a + b + c = \sqrt{a^2 + b^2} \)
2. \( a + b + c = -\sqrt{a^2 + b^2} \)
### Step 5: Solve for the lines
Now, substituting \( c = -2a - 2b \) from the family of lines into the distance equations, we can find the specific lines \( l_1 \) and \( l_2 \).
For the first case:
\[
a + b - 2a - 2b = \sqrt{a^2 + b^2}
\]
This simplifies to:
\[
-a - b = \sqrt{a^2 + b^2}
\]
For the second case:
\[
a + b + 2a + 2b = -\sqrt{a^2 + b^2}
\]
This simplifies to:
\[
3a + 3b = -\sqrt{a^2 + b^2}
\]
### Step 6: Identify the lines
By solving these equations, we find that the lines \( l_1 \) and \( l_2 \) that are at a distance of 1 unit from \( (1, 1) \) and pass through \( (2, 2) \) are:
- \( x = 2 \)
- \( y = 2 \)
### Step 7: Calculate the area bounded by the lines and the coordinate axes
The area bounded by the lines \( l_1 \), \( l_2 \), and the coordinate axes forms a square with vertices at \( (0, 0) \), \( (2, 0) \), \( (2, 2) \), and \( (0, 2) \). The side length of this square is 2 units.
Thus, the area \( A \) is given by:
\[
A = \text{side}^2 = 2 \times 2 = 4 \text{ square units}
\]
### Final Answer
The area bounded by the lines \( l_1 \), \( l_2 \), and the coordinate axes is \( \boxed{4} \) square units.
