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

Options

A. `7/2`
B. `7/3`
C. `7/5`
D. `7/10`

Answer

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

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

### Step 2: Rewrite the second line in the same form as the first
We can simplify the second line \(L_2\) by dividing the entire equation by 2:
\[ L_2: \frac{8x + 6y}{2} = \frac{15}{2} \implies 4x + 3y = \frac{15}{2} \]

### Step 3: Identify coefficients
Now we have:
1. \(L_1: 4x + 3y = 11\)
2. \(L_2: 4x + 3y = \frac{15}{2}\)

From these equations, we can identify:
- \(a = 4\)
- \(b = 3\)
- \(c = -11\) (from \(L_1\))
- \(c_1 = -\frac{15}{2}\) (from \(L_2\))

### Step 4: Use the formula for the distance between parallel lines
The formula for the distance \(d\) between two parallel lines \(ax + by + c = 0\) and \(ax + by + c_1 = 0\) is given by:
\[ d = \frac{|c_1 - c|}{\sqrt{a^2 + b^2}} \]

### Step 5: Substitute the values into the formula
Substituting the values we have:
\[ d = \frac{\left| -\frac{15}{2} - (-11) \right|}{\sqrt{4^2 + 3^2}} \]
Calculating \(c_1 - c\):
\[ -\frac{15}{2} + 11 = -\frac{15}{2} + \frac{22}{2} = \frac{7}{2} \]
Now substituting this back into the distance formula:
\[ d = \frac{\left| \frac{7}{2} \right|}{\sqrt{16 + 9}} = \frac{\frac{7}{2}}{\sqrt{25}} = \frac{\frac{7}{2}}{5} = \frac{7}{10} \]

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

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