Find all points on the line `x+y=4` that lie at a unit distance from the line `4x+3y=10`.

Answer

To solve the problem of finding all points on the line \( x + y = 4 \) that lie at a unit distance from the line \( 4x + 3y = 10 \), we can follow these steps:

### Step 1: Define the Point on the Line
Let the point on the line \( x + y = 4 \) be represented as \( (h, k) \). Since this point lies on the line, we have:
\[ h + k = 4 \quad \text{(Equation 1)} \]

### Step 2: Use the Distance Formula
The distance \( d \) from a point \( (h, k) \) to the line \( Ax + By + C = 0 \) can be calculated using the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( 4x + 3y - 10 = 0 \), we have \( A = 4 \), \( B = 3 \), and \( C = -10 \). The distance from the point \( (h, k) \) to this line is given by:
\[ d = \frac{|4h + 3k - 10|}{\sqrt{4^2 + 3^2}} = \frac{|4h + 3k - 10|}{5} \]
Since we want this distance to be equal to 1 (unit distance), we set up the equation:
\[ \frac{|4h + 3k - 10|}{5} = 1 \]
Multiplying both sides by 5 gives:
\[ |4h + 3k - 10| = 5 \quad \text{(Equation 2)} \]

### Step 3: Solve the Absolute Value Equation
The absolute value equation \( |4h + 3k - 10| = 5 \) leads to two cases:
1. \( 4h + 3k - 10 = 5 \)
2. \( 4h + 3k - 10 = -5 \)

#### Case 1: \( 4h + 3k - 10 = 5 \)
Rearranging gives:
\[ 4h + 3k = 15 \quad \text{(Equation 3)} \]

#### Case 2: \( 4h + 3k - 10 = -5 \)
Rearranging gives:
\[ 4h + 3k = 5 \quad \text{(Equation 4)} \]

### Step 4: Solve the System of Equations
Now we will solve the pairs of equations formed by Equation 1 with Equations 3 and 4.

#### Solving Equation 1 and Equation 3
From Equation 1:
\[ k = 4 - h \]
Substituting into Equation 3:
\[ 4h + 3(4 - h) = 15 \]
Expanding and simplifying:
\[ 4h + 12 - 3h = 15 \\ h + 12 = 15 \\ h = 3 \]
Substituting back to find \( k \):
\[ k = 4 - 3 = 1 \]
Thus, one point is \( (3, 1) \).

#### Solving Equation 1 and Equation 4
Using Equation 1 again:
\[ k = 4 - h \]
Substituting into Equation 4:
\[ 4h + 3(4 - h) = 5 \]
Expanding and simplifying:
\[ 4h + 12 - 3h = 5 \\ h + 12 = 5 \\ h = -7 \]
Substituting back to find \( k \):
\[ k = 4 - (-7) = 11 \]
Thus, the second point is \( (-7, 11) \).

### Final Answer
The points on the line \( x + y = 4 \) that lie at a unit distance from the line \( 4x + 3y = 10 \) are:
\[ (3, 1) \quad \text{and} \quad (-7, 11) \]