What is the perpendicular distance between the parallel lines 3x + 4y = 9 and 9x + 12y + 28 = 0 ?

Options

A. `7//3` units
B. `8//3` units
C. `10//3` units
D. `11//3` units

Answer

To find the perpendicular distance between the parallel lines given by the equations \(3x + 4y = 9\) and \(9x + 12y + 28 = 0\), we can follow these steps:

### Step 1: Rewrite the equations in standard form
The first line can be rewritten as:
\[ 3x + 4y - 9 = 0 \]
The second line is already in standard form:
\[ 9x + 12y + 28 = 0 \]

### Step 2: Check if the lines are parallel
To check if the lines are parallel, we need to compare their coefficients. The first line has coefficients \(A_1 = 3\), \(B_1 = 4\), and \(C_1 = -9\). The second line can be simplified by dividing all terms by 3:
\[ 3x + 4y + \frac{28}{3} = 0 \]
This gives us coefficients \(A_2 = 3\), \(B_2 = 4\), and \(C_2 = \frac{28}{3}\). Since the ratios of the coefficients \(A_1/A_2\) and \(B_1/B_2\) are equal, the lines are parallel.

### Step 3: Use the formula for the distance between two parallel lines
The formula for the distance \(d\) between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
Here, \(C_1 = -9\) and \(C_2 = \frac{28}{3}\).

### Step 4: Calculate \(C_2 - C_1\)
First, we need to find \(C_2 - C_1\):
\[ C_2 - C_1 = \frac{28}{3} - (-9) = \frac{28}{3} + 9 = \frac{28}{3} + \frac{27}{3} = \frac{55}{3} \]

### Step 5: Calculate \(A^2 + B^2\)
Next, we calculate \(A^2 + B^2\):
\[ A^2 + B^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

### Step 6: Substitute into the distance formula
Now we can substitute into the distance formula:
\[ d = \frac{\left|\frac{55}{3}\right|}{\sqrt{25}} = \frac{\frac{55}{3}}{5} = \frac{55}{15} = \frac{11}{3} \]

### Final Answer
Thus, the perpendicular distance between the two parallel lines is:
\[ \frac{11}{3} \]

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