Question:
What is the value of \(\lambda\) if the straight line \((2x + 3y + 4) + \lambda(6x – y + 12) = 0\) is parallel to y-axis?
यदि सरल रेखा \((2x + 3y + 4) + \lambda(6x – y + 12) = 0\) y-अक्ष के समानांतर है, तो \(\lambda\) का मान क्या है?
Options:
- 3
- -6
- 4
- -3
- 3
- -6
- 4
- -3
Correct Answer: a)
Explanation:
Expanding and rearranging the given line into standard form \(Ax + By + C = 0\):
\[(2 + 6\lambda)x + (3 – \lambda)y + (4 + 12\lambda) = 0\]
A straight line is parallel to the y-axis if the coefficient of \(y\) is zero (i.e. of the form \(x = \text{constant}\)):
\[3 – \lambda = 0 \implies \lambda = 3\]
Hence, the correct option is (a).
दी गई रेखा के समीकरण को \(Ax + By + C = 0\) रूप में लिखने पर:
\[(2 + 6\lambda)x + (3 – \lambda)y + (4 + 12\lambda) = 0\]
कोई रेखा y-अक्ष के समानांतर तब होती है जब \(y\) का गुणांक शून्य हो:
\[3 – \lambda = 0 \implies \lambda = 3\]
अतः सही विकल्प (a) है।