Question

The distance between the lines represented by \[ 16x^2 + 9y^2 + 48x - 24xy - 36y + 35 = 0 \] is ......... units

• A. \( \frac{2}{5} \)
• B. \( \frac{5}{2} \)
• C. \( \frac{3}{5} \)
• D. \( 5 \)

Hint:

To find the distance between two lines, use the formula \( \frac{|2D - 2E|}{\sqrt{A + C}} \) derived from the general equation of a pair of lines.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The quadratic equation in \( x \) and \( y \) represents a pair of straight lines. If they are parallel, we can rewrite the expression as a quadratic in a linear factor.
Step 2: Key Formula or Approach:
Identify the expression as \( (ax + by)^2 + \dots = 0 \). The distance between parallel lines \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \) is \( \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \).
Step 3: Detailed Explanation:
Rearranging the terms:
\( (16x^2 - 24xy + 9y^2) + (48x - 36y) + 35 = 0 \)
\( (4x - 3y)^2 + 12(4x - 3y) + 35 = 0 \)
Let \( 4x - 3y = u \).
\( u^2 + 12u + 35 = 0 \Rightarrow (u + 5)(u + 7) = 0 \).
The lines are:
1) \( 4x - 3y + 5 = 0 \)
2) \( 4x - 3y + 7 = 0 \)
Distance \( d = \frac{|7 - 5|}{\sqrt{4^2 + (-3)^2}} = \frac{2}{\sqrt{16 + 9}} = \frac{2}{5} \).
Step 4: Final Answer:
The distance is \( 2/5 \).