The number of points on the line` x +y=4` which are unit distance apart from the line `2x+2y = 5 `is :

Options

A. 0
B. 1
C. 2
D. `oo`

Answer

To solve the problem, we need to find the number of points on the line \( x + y = 4 \) that are a unit distance away from the line \( 2x + 2y = 5 \).

### Step-by-Step Solution:

1. **Rewrite the second line in slope-intercept form**:
The line \( 2x + 2y = 5 \) can be simplified to:
\[ x + y = \frac{5}{2} \]

2. **Identify the lines**:
We have two lines:
- Line 1: \( x + y = 4 \)
- Line 2: \( x + y = \frac{5}{2} \)

3. **Determine the distance between the two lines**:
The distance \( d \) between two parallel lines of the form \( Ax + By = C_1 \) and \( Ax + By = C_2 \) is given by:
\[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 1 \), \( B = 1 \), \( C_1 = 4 \), and \( C_2 = \frac{5}{2} \).

First, calculate \( |C_1 - C_2| \):
\[ |4 - \frac{5}{2}| = |4 - 2.5| = |1.5| = 1.5 \]

Now, calculate the distance:
\[ d = \frac{1.5}{\sqrt{1^2 + 1^2}} = \frac{1.5}{\sqrt{2}} = \frac{1.5 \sqrt{2}}{2} = \frac{3\sqrt{2}}{4} \]

4. **Check if there are points that are unit distance from the second line**:
We are looking for points that are 1 unit away from the line \( x + y = \frac{5}{2} \).

Since the distance between the two lines is \( \frac{3\sqrt{2}}{4} \), we can see that this distance is greater than 1 unit. Therefore, there are no points on the line \( x + y = 4 \) that are exactly 1 unit away from the line \( x + y = \frac{5}{2} \).

5. **Conclusion**:
Hence, the number of points on the line \( x + y = 4 \) which are unit distance apart from the line \( 2x + 2y = 5 \) is:
\[ \text{Number of points} = 0 \]

### Final Answer:
The number of points is \( 0 \).