The distance between the lines `3x+4y=9` and `6x+8y=15` is

Options

A. `3//2`
B. `3//10`
C. `6`
D. None of these

Answer

To find the distance between the lines given by the equations \(3x + 4y = 9\) and \(6x + 8y = 15\), we can follow these steps:

### Step 1: Check if the lines are parallel
To determine if the lines are parallel, we can rewrite both equations in the standard form \(Ax + By + C = 0\).

1. The first line is:
\[ 3x + 4y - 9 = 0 \]
Here, \(A_1 = 3\), \(B_1 = 4\), and \(C_1 = -9\).

2. The second line is:
\[ 6x + 8y - 15 = 0 \]
Here, \(A_2 = 6\), \(B_2 = 8\), and \(C_2 = -15\).

Now, we can see that the second line can be simplified:
\[ 6x + 8y - 15 = 0 \implies 3x + 4y - \frac{15}{2} = 0 \]
This shows that the second line is a multiple of the first line, confirming that the lines are indeed parallel.

### Step 2: Find the distance between the two parallel lines
The formula for the distance \(d\) between two parallel lines given by \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]

Substituting the values:
- \(C_1 = -9\)
- \(C_2 = -\frac{15}{2}\)
- \(A = 3\)
- \(B = 4\)

Now, calculate \(C_2 - C_1\):
\[ C_2 - C_1 = -\frac{15}{2} - (-9) = -\frac{15}{2} + 9 = -\frac{15}{2} + \frac{18}{2} = \frac{3}{2} \]

Now, calculate \(A^2 + B^2\):
\[ A^2 + B^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

Now substitute these into the distance formula:
\[ d = \frac{|\frac{3}{2}|}{\sqrt{25}} = \frac{\frac{3}{2}}{5} = \frac{3}{10} \]

### Final Answer
The distance between the lines \(3x + 4y = 9\) and \(6x + 8y = 15\) is \(\frac{3}{10}\).

---