The absolute value of the sum of the abscissas of all the points on the line `x+y=4` that lie at a unit distance from the line `4x+3y-10=0` is___________

Explanation

To solve the problem, we need to find the absolute value of the sum of the abscissas (x-coordinates) of all the points on the line \(x + y = 4\) that lie at a unit distance from the line \(4x + 3y - 10 = 0\).

### Step-by-Step Solution:

1. **Identify the given lines**:
- The first line is \(x + y = 4\).
- The second line is \(4x + 3y - 10 = 0\).

2. **Convert the second line to standard form**:
- The line \(4x + 3y - 10 = 0\) can be rewritten as \(4x + 3y = 10\).

3. **Find the distance from a point to a line**:
- The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- For our line \(4x + 3y - 10 = 0\), we have \(A = 4\), \(B = 3\), and \(C = -10\).

4. **Substituting the points on the line \(x + y = 4\)**:
- Any point on the line \(x + y = 4\) can be represented as \((x, 4 - x)\).
- Substitute \(x_0 = x\) and \(y_0 = 4 - x\) into the distance formula:
\[ d = \frac{|4x + 3(4 - x) - 10|}{\sqrt{4^2 + 3^2}} = \frac{|4x + 12 - 3x - 10|}{5} = \frac{|x + 2|}{5} \]

5. **Set the distance equal to 1**:
- Since we want the points that are at a unit distance from the line, we set:
\[ \frac{|x + 2|}{5} = 1 \]
- This simplifies to:
\[ |x + 2| = 5 \]

6. **Solve the absolute value equation**:
- This gives us two equations:
1. \(x + 2 = 5\) → \(x = 3\)
2. \(x + 2 = -5\) → \(x = -7\)

7. **Find the sum of the abscissas**:
- The abscissas found are \(3\) and \(-7\).
- The sum of these abscissas is:
\[ 3 + (-7) = -4 \]

8. **Calculate the absolute value**:
- The absolute value of the sum is:
\[ |-4| = 4 \]

### Final Answer:
The absolute value of the sum of the abscissas of all the points on the line \(x + y = 4\) that lie at a unit distance from the line \(4x + 3y - 10 = 0\) is \(4\).