The distance between the lines 4x + 3y = 11 and 8x + 6y = 15 is

Options

A. 4
B. `(7)/(10)`
C. `(5)/(7)`
D. 26

Answer

To find the distance between the lines \(4x + 3y = 11\) and \(8x + 6y = 15\), we will follow these steps:

### Step 1: Identify the lines
The given lines are:
1. \(L_1: 4x + 3y = 11\)
2. \(L_2: 8x + 6y = 15\)

### Step 2: Check if the lines are parallel
To determine if the lines are parallel, we can rewrite them in the standard form \(Ax + By + C = 0\):
- For \(L_1\): \(4x + 3y - 11 = 0\)
- For \(L_2\): \(8x + 6y - 15 = 0\)

Now, we can compare the coefficients:
- The coefficients of \(x\) and \(y\) in \(L_2\) are multiples of those in \(L_1\) (specifically, \(2 \times 4 = 8\) and \(2 \times 3 = 6\)). Thus, the lines are parallel.

### Step 3: Calculate the perpendicular distance from the origin to each line
The formula for the perpendicular distance \(P\) from a point to a line \(Ax + By + C = 0\) is given by:
\[ P = \frac{|C|}{\sqrt{A^2 + B^2}} \]

#### For Line \(L_1\):
- \(A = 4\), \(B = 3\), \(C = -11\)
\[ P_1 = \frac{|11|}{\sqrt{4^2 + 3^2}} = \frac{11}{\sqrt{16 + 9}} = \frac{11}{\sqrt{25}} = \frac{11}{5} \]

#### For Line \(L_2\):
- \(A = 8\), \(B = 6\), \(C = -15\)
\[ P_2 = \frac{|15|}{\sqrt{8^2 + 6^2}} = \frac{15}{\sqrt{64 + 36}} = \frac{15}{\sqrt{100}} = \frac{15}{10} = \frac{3}{2} \]

### Step 4: Calculate the distance \(D\) between the two parallel lines
The distance \(D\) between two parallel lines can be calculated using the formula:
\[ D = |P_1 - P_2| \]
Substituting the values we calculated:
\[ D = \left| \frac{11}{5} - \frac{3}{2} \right| \]

### Step 5: Find a common denominator and calculate the difference
The common denominator for \(5\) and \(2\) is \(10\):
\[ D = \left| \frac{11 \times 2}{10} - \frac{3 \times 5}{10} \right| = \left| \frac{22}{10} - \frac{15}{10} \right| = \left| \frac{7}{10} \right| = \frac{7}{10} \]

### Final Answer
The distance between the lines \(4x + 3y = 11\) and \(8x + 6y = 15\) is \(\frac{7}{10}\).

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