Question

find the distance between the pair of parallel lines with the given equations.
y = \frac{1}{4}x + 2
4y - x = -60
\bigcirc a) 17 units
\bigcirc b) \sqrt{273} or about 16.52 units
\bigcirc c) 4\sqrt{17} or about 16.49 units
\bigcirc d) 4 units

Explanation:

Step1: Rewrite lines in standard form

First, rewrite \( y = \frac{1}{4}x + 2 \) in \( Ax + By + C = 0 \) form:
Subtract \( y \) and 2: \( \frac{1}{4}x - y + 2 = 0 \). Multiply by 4: \( x - 4y + 8 = 0 \).

Second, rewrite \( 4y - x = -60 \) as \( -x + 4y + 60 = 0 \), or multiply by -1: \( x - 4y - 60 = 0 \). Wait, no—original second line: \( 4y - x = -60 \) → \( -x + 4y + 60 = 0 \) → \( x - 4y - 60 = 0 \)? Wait, no: \( 4y - x = -60 \) → \( -x + 4y + 60 = 0 \), or \( x - 4y - 60 = 0 \) (multiplying by -1). Wait, but the first line is \( x - 4y + 8 = 0 \), so \( A = 1 \), \( B = -4 \), \( C_1 = 8 \); second line: \( x - 4y - 60 = 0 \), so \( C_2 = -60 \).

Step2: Apply distance formula for parallel lines

The formula for distance between \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \).

Here, \( A = 1 \), \( B = -4 \), \( C_1 = 8 \), \( C_2 = -60 \).

Compute \( |C_1 - C_2| = |8 - (-60)| = |68| = 68 \).

Compute \( \sqrt{A^2 + B^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \).

Thus, \( d = \frac{68}{\sqrt{17}} \). Rationalize: \( \frac{68\sqrt{17}}{17} = 4\sqrt{17} \). Calculate \( 4\sqrt{17} \approx 4 \times 4.123 \approx 16.49 \).

Answer:

C) \( 4\sqrt{17} \) or about 16.49 units