The perpendicular distance between the lines `3x+4y=6` and `3x+4y+4=0` is :

Options

A. `1` unit
B. `2` units
C. `3` units
D. None of these

Answer

To find the perpendicular distance between the lines \(3x + 4y = 6\) and \(3x + 4y + 4 = 0\), we can use the formula for the distance between two parallel lines given in the form \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\).

### Step-by-step Solution:

1. **Rewrite the equations in standard form**:
- The first line is given as \(3x + 4y = 6\). We can rewrite this in the standard form:
\[ 3x + 4y - 6 = 0 \]
Here, \(c_1 = -6\).
- The second line is given as \(3x + 4y + 4 = 0\). Here, \(c_2 = 4\).

2. **Identify coefficients**:
- From both equations, we have:
- \(a = 3\)
- \(b = 4\)

3. **Use the distance formula**:
- The formula for the distance \(d\) between the two parallel lines is:
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]
- Substitute the values of \(c_1\) and \(c_2\):
\[ d = \frac{|-6 - 4|}{\sqrt{3^2 + 4^2}} \]

4. **Calculate the numerator**:
- Calculate \(-6 - 4\):
\[ -6 - 4 = -10 \quad \text{so} \quad |-10| = 10 \]

5. **Calculate the denominator**:
- Calculate \(a^2 + b^2\):
\[ 3^2 + 4^2 = 9 + 16 = 25 \]
- Therefore, \(\sqrt{25} = 5\).

6. **Final calculation**:
- Substitute back into the distance formula:
\[ d = \frac{10}{5} = 2 \]

### Conclusion:
The perpendicular distance between the lines \(3x + 4y = 6\) and \(3x + 4y + 4 = 0\) is \(2\) units.

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