The distance between the parallel lnes `y=2x+4 and 6x-3y-5` is (A) `1` (B) `17/sqrt(3)` (C) `7sqrt(5)/15` (D) `3sqrt(5)/15`

Options

A. `17//sqrt(3)`
B. 1
C. `3//sqrt(5)`
D. `17sqrt(15)//15`

Answer

To find the distance between the parallel lines given by the equations \( y = 2x + 4 \) and \( 6x - 3y - 5 = 0 \), we can follow these steps:

### Step 1: Rewrite the second line in slope-intercept form
The second line is given as \( 6x - 3y - 5 = 0 \). We can rearrange this to find \( y \):
\[ 3y = 6x - 5 \implies y = 2x - \frac{5}{3} \]
This shows that the slope of the second line is also \( 2 \), confirming that both lines are parallel.

### Step 2: Convert both equations to standard form
We can rewrite the first line \( y = 2x + 4 \) in standard form:
\[ 2x - y + 4 = 0 \]
For the second line, we already have it in standard form:
\[ 6x - 3y - 5 = 0 \]
To make it easier to compare, we can divide the entire equation by 3:
\[ 2x - y - \frac{5}{3} = 0 \]

### Step 3: Identify coefficients
From the standard forms, we have:
- For the first line: \( A = 2, B = -1, C_1 = 4 \)
- For the second line: \( C_2 = -\frac{5}{3} \)

### Step 4: Use the distance formula for 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:
\[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]
Substituting the values we have:
\[ d = \frac{|4 - (-\frac{5}{3})|}{\sqrt{2^2 + (-1)^2}} \]

### Step 5: Simplify the numerator
Calculating the numerator:
\[ |4 + \frac{5}{3}| = |4 + \frac{5}{3}| = | \frac{12}{3} + \frac{5}{3} | = | \frac{17}{3} | = \frac{17}{3} \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 7: Combine the results
Now substituting back into the distance formula:
\[ d = \frac{\frac{17}{3}}{\sqrt{5}} = \frac{17}{3\sqrt{5}} = \frac{17\sqrt{5}}{15} \]

Thus, the distance between the two parallel lines is:
\[ \boxed{\frac{17\sqrt{5}}{15}} \]