The distance between two parallel lines `3x+4y -8 =0 and 3x+4y -3=0,` is given by

Options

A. 4
B. 5
C. 3
D. 1

Answer

To find the distance between the two parallel lines given by the equations \(3x + 4y - 8 = 0\) and \(3x + 4y - 3 = 0\), we can use the formula for the distance \(d\) between two parallel lines of the form \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\):

\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]

### Step-by-Step Solution:

1. **Identify the coefficients:**
From the equations of the lines:
- For the first line \(3x + 4y - 8 = 0\):
- \(A = 3\)
- \(B = 4\)
- \(C_1 = -8\)
- For the second line \(3x + 4y - 3 = 0\):
- \(C_2 = -3\)

2. **Calculate the difference between \(C_2\) and \(C_1\):**
\[ |C_2 - C_1| = |-3 - (-8)| = |-3 + 8| = |5| = 5 \]

3. **Calculate \(A^2 + B^2\):**
\[ A^2 + B^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

4. **Calculate the square root of \(A^2 + B^2\):**
\[ \sqrt{A^2 + B^2} = \sqrt{25} = 5 \]

5. **Substitute values into the distance formula:**
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} = \frac{5}{5} = 1 \]

### Final Answer:
The distance between the two parallel lines is \(1\).

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