Question:
The equation of the line passing through the point (2, 3) and the point of intersection of lines \(2x – 3y + 7 = 0\) and \(7x + 4y + 2 = 0\) is:
बिंदु (2, 3) और रेखाओं \(2x – 3y + 7 = 0\) तथा \(7x + 4y + 2 = 0\) के प्रतिच्छेदन बिंदु से गुजरने वाली रेखा का समीकरण है:
Options:
- \(21x + 46y – 180 = 0\)
- \(21x – 46y + 96 = 0\)
- \(46x + 21y + 155 = 0\)
- \(46x – 21y – 29 = 0\)
- \(21x + 46y – 180 = 0\)
- \(21x – 46y + 96 = 0\)
- \(46x + 21y + 155 = 0\)
- \(46x – 21y – 29 = 0\)
Correct Answer: b)
Explanation:
The family of lines passing through the intersection is:
\[(2x – 3y + 7) + \lambda(7x + 4y + 2) = 0\]
Since it passes through (2, 3):
\[(2(2) – 3(3) + 7) + \lambda(7(2) + 4(3) + 2) = 0\]
\[(4 – 9 + 7) + \lambda(14 + 12 + 2) = 0 \implies 2 + 28\lambda = 0 \implies \lambda = -\frac{1}{14}\]
Multiplying by 14:
\[14(2x – 3y + 7) – (7x + 4y + 2) = 0\]
\[28x – 42y + 98 – 7x – 4y – 2 = 0 \implies 21x – 46y + 96 = 0\]
Hence, the correct option is (b).
प्रतिच्छेदन बिंदु से गुजरने वाली रेखाओं का कुल:
\[(2x – 3y + 7) + \lambda(7x + 4y + 2) = 0\]
बिंदु (2, 3) रखने पर:
\[2 + 28\lambda = 0 \implies \lambda = -1/14\]
मान रखकर 14 से गुणा करने पर:
\[21x – 46y + 96 = 0\]
अतः सही विकल्प (b) है।