Question

The line \(L_1\) given by \(\frac{x}{p}+\frac{y}{2} = 1\) passes through the point \((5,0)\). The line \(L_2\) given by \(\frac{x}{10}+\frac{y}{q} = 1\) is parallel to \(L_1\). Then the distance between the lines \(L_1\) and \(L_2\) is...

• A. \(\frac{10}{\sqrt{29}}\)
• B. \(\frac{19}{2\sqrt{29}}\)
• C. \(\frac{5}{\sqrt{41}}\)
• D. \(\frac{10}{\sqrt{41}}\)

Hint:

Find p from the given point, find q from parallelism, then use the parallel line distance formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Line \(L_1\) passes through \((5,0)\), which fixes \(p\). Parallel lines have equal slopes, which fixes \(q\). Then 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 2: Key Formula or Approach:
Substitute \((5,0)\) into \(\frac{x}{p} + \frac{y}{2} = 1\).

Step 3: Detailed Explanation:
\(\frac{5}{p} = 1\), so \(p = 5\). Then \(L_1: \frac{x}{5} + \frac{y}{2} = 1\), i.e. \(2x + 5y = 10\).
For \(L_2\): \(\frac{x}{10} + \frac{y}{q} = 1\) has slope \(-\frac{q}{10}\). It must equal the slope \(-\frac25\) of \(L_1\), so \(q = 4\).
\(L_2: 2x + 5y = 20\).
\[ d = \frac{|20 - 10|}{\sqrt{2^2 + 5^2}} = \frac{10}{\sqrt{29}} \]

Final Answer:
The distance is \(\frac{10}{\sqrt{29}}\), option (A). \[ \boxed{\frac{10}{\sqrt{29}}} \]