Question:
What is the equation of the straight line passing through the point of intersection of the lines \(\frac{x}{2} + \frac{y}{3} = 1\) and \(\frac{x}{3} + \frac{y}{2} = 1\), and parallel to \(4x + 5y – 6 = 0\)?
रेखाओं \(\frac{x}{2} + \frac{y}{3} = 1\) और \(\frac{x}{3} + \frac{y}{2} = 1\) के प्रतिच्छेदन बिंदु से गुजरने वाली और \(4x + 5y – 6 = 0\) के समानांतर सरल रेखा का समीकरण क्या है?
Options:
- \(20x + 25y – 54 = 0\)
- \(25x + 20y – 54 = 0\)
- \(4x + 5y – 54 = 0\)
- \(4x + 5y – 45 = 0\)
- \(20x + 25y – 54 = 0\)
- \(25x + 20y – 54 = 0\)
- \(4x + 5y – 54 = 0\)
- \(4x + 5y – 45 = 0\)
Correct Answer: a)
Explanation:
Given lines:
\(3x + 2y = 6\) and \(2x + 3y = 6\).
Subtracting the equations gives \(x – y = 0 \implies x = y\).
Substitute into \(3x + 2x = 6 \implies 5x = 6 \implies x = 6/5, y = 6/5\).
The line parallel to \(4x + 5y – 6 = 0\) has the form \(4x + 5y + k = 0\).
Passes through \((6/5, 6/5)\):
\[4\left(\frac{6}{5}\right) + 5\left(\frac{6}{5}\right) + k = 0 \implies \frac{24 + 30}{5} + k = 0 \implies k = -\frac{54}{5}\]
Equation: \(4x + 5y – \frac{54}{5} = 0 \implies 20x + 25y – 54 = 0\).
Hence, the correct option is (a).
दी गई रेखाएँ: \(3x + 2y = 6\) तथा \(2x + 3y = 6\)।
प्रतिच्छेदन बिंदु: \(x = 6/5, y = 6/5\)।
\(4x + 5y – 6 = 0\) के समानांतर रेखा का समीकरण: \(4x + 5y + k = 0\)।
बिंदु \((6/5, 6/5)\) रखने पर:
\[4(6/5) + 5(6/5) + k = 0 \implies k = -54/5\]
5 से गुणा करने पर:
\[20x + 25y – 54 = 0\]
अतः सही विकल्प (a) है।