Question

Let a be the distance between the lines \( -2x + y = 2 \) and \( 2x - y = 2 \), and b be the distance between the lines \( 4x - 3y = 5 \) and \( 6y - 8x = 1 \), then

• A. \( 40b = 11\sqrt{5}a \)
• B. \( 40\sqrt{2}a = 11b \)
• C. \( 11\sqrt{2}b = 40a \)
• D. \( 11\sqrt{2}a = 40b \)

Hint:

To find the distance between two parallel lines, you must first write them in the standard form \( Ax+By+C=0 \) with identical coefficients for \( x \) and \( y \). Then, apply the formula \( d = |C_1 - C_2| / \sqrt{A^2 + B^2} \).

Answer 1

The Correct Option is A

First, calculate distance 'a'. The lines are \( -2x + y = 2 \) and \( 2x - y = 2 \). Let's write them in the form \( Ax+By+C=0 \). 

Line 1: \( 2x - y + 2 = 0 \) 

Line 2: \( 2x - y - 2 = 0 \) 

These lines are parallel. The distance between parallel lines \( Ax+By+C_1=0 \) and \( Ax+By+C_2=0 \) is \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \). \[ a = \frac{|2 - (-2)|}{\sqrt{2^2 + (-1)^2}} = \frac{4}{\sqrt{4+1}} = \frac{4}{\sqrt{5}} \] Next, calculate distance 'b'. The lines are \( 4x - 3y = 5 \) and \( 6y - 8x = 1 \). 

Let's standardize them. 

Line 3: \( 4x - 3y - 5 = 0 \) 

Line 4: \( -8x + 6y - 1 = 0 \). To make the coefficients of x and y match Line 3, divide by -2: \( 4x - 3y + \frac{1}{2} = 0 \). 

These lines are parallel. \[ b = \frac{|-5 - (\frac{1}{2})|}{\sqrt{4^2 + (-3)^2}} = \frac{|-11/2|}{\sqrt{16+9}} = \frac{11/2}{\sqrt{25}} = \frac{11/2}{5} = \frac{11}{10} \] 

Now we need to find the relationship between a and b. We have \( a = \frac{4}{\sqrt{5}} \) and \( b = \frac{11}{10} \). 

From the options, let's check (A): \( 40b = 11\sqrt{5}a \). LHS: \( 40b = 40 \times \frac{11}{10} = 4 \times 11 = 44 \). RHS: \( 11\sqrt{5}a = 11\sqrt{5} \times \frac{4}{\sqrt{5}} = 11 \times 4 = 44 \). LHS = RHS. So, option (A) is correct.