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

Answer

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

### Step-by-Step Solution:

1. **Identify the Points on the Line**:
The line equation is \(x + y = 4\). We can express the points on this line in terms of \(x\) (abscissa):
\[ y = 4 - x \]
Therefore, a point on the line can be represented as \((h, 4 - h)\).

2. **Distance from the Given Line**:
The distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(4x + 3y - 10 = 0\), we have \(A = 4\), \(B = 3\), and \(C = -10\). We need to find the distance from the point \((h, 4 - h)\):
\[ d = \frac{|4h + 3(4 - h) - 10|}{\sqrt{4^2 + 3^2}} = \frac{|4h + 12 - 3h - 10|}{\sqrt{16 + 9}} = \frac{|h + 2|}{5} \]

3. **Set the Distance Equal to 1**:
Since we are looking for points that are at a unit distance from the line, we set the distance equal to 1:
\[ \frac{|h + 2|}{5} = 1 \]
Multiplying both sides by 5 gives:
\[ |h + 2| = 5 \]

4. **Solve the Absolute Value Equation**:
This absolute value equation gives us two cases:
- Case 1: \(h + 2 = 5\)
- Case 2: \(h + 2 = -5\)

Solving these:
- From Case 1: \(h = 5 - 2 = 3\)
- From Case 2: \(h = -5 - 2 = -7\)

5. **Calculate the Sum of the Squares of the Abscissas**:
Now we have two values for \(h\): \(3\) and \(-7\). We need to find the sum of their squares:
\[ 3^2 + (-7)^2 = 9 + 49 = 58 \]

### Final Answer:
The sum of the squares of the abscissas is \(58\).