Find the distance between 3x + 2y + 7 = 0 and 6x + 4y + 3 = 0.

Answer

To find the distance between the two lines given by the equations \(3x + 2y + 7 = 0\) and \(6x + 4y + 3 = 0\), we can follow these steps:

### Step 1: Check if the lines are parallel
To determine if the lines are parallel, we can compare the coefficients of \(x\) and \(y\) in both equations.

The first line is:
\[ 3x + 2y + 7 = 0 \]
Here, \(A_1 = 3\), \(B_1 = 2\), and \(C_1 = 7\).

The second line is:
\[ 6x + 4y + 3 = 0 \]
Here, \(A_2 = 6\), \(B_2 = 4\), and \(C_2 = 3\).

Now, we check the ratios:
\[ \frac{A_1}{A_2} = \frac{3}{6} = \frac{1}{2} \]
\[ \frac{B_1}{B_2} = \frac{2}{4} = \frac{1}{2} \]

Since \(\frac{A_1}{A_2} = \frac{B_1}{B_2}\), the lines are parallel.

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

### Step 3: Substitute the values into the distance formula
From the first line, we have \(C_1 = 7\) and from the second line, we have \(C_2 = 3\). The coefficients \(A\) and \(B\) can be taken from either line since they are the same for parallel lines:
- \(A = 3\)
- \(B = 2\)

Now, substituting these values into the distance formula:
\[ d = \frac{|7 - (-3)|}{\sqrt{3^2 + 2^2}} = \frac{|7 - 3|}{\sqrt{9 + 4}} = \frac{4}{\sqrt{13}} \]

### Step 4: Simplify the expression
Thus, the distance between the two lines is:
\[ d = \frac{4}{\sqrt{13}} \]

### Final Answer
The distance between the lines \(3x + 2y + 7 = 0\) and \(6x + 4y + 3 = 0\) is \(\frac{4}{\sqrt{13}}\).

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