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: Find the lines:
Writing in the form $2x + 5y = k$: $L_1$ has $k = 10$ (from $p = 5$). $L_2$ must have the same $x$ and $y$ coefficient ratio, which needs $q = 4$ and gives $k = 20$.

Step 2: Distance via a point:
Take the point $(5,0)$ on $L_1$. Its distance from $2x + 5y - 20 = 0$ is $\frac{|10 - 20|}{\sqrt{29}} = \frac{10}{\sqrt{29}}$.

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