Question:

Let \(\theta_1\) be the angle between two lines \(2x + 3y + c_1 = 0\) and \(-x + 5y + c_2 = 0\) and \(\theta_2\) be the angle between two lines \(2x + 3y + c_1 = 0\) and \(-x + 5y + c_3 = 0\), where \(c_1, c_2, c_3\) are any real numbers :
Statement-1 : If \(c_2\) and \(c_3\) are proportional, then \(\theta_1 = \theta_2\).
Statement-2 : \(\theta_1 = \theta_2\) for all \(c_2\) and \(c_3\).

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The angle between two lines depends only on their slopes (coefficients of x and y), not on the constant terms.
Updated On: Oct 31, 2026
  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1.
  • Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation of Statement-1.
  • Statement-1 is false; Statement-2 is true.
  • Statement-1 is true; Statement-2 is false.
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The Correct Option is A

Solution and Explanation

The angle between two lines is given by \(\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|\), where the slopes are fixed by the coefficients of \(x\) and \(y\) only.
Here the lines \(2x+3y+c_1=0\) and \(-x+5y+c_k=0\) have slopes \(-\tfrac{2}{3}\) and \(\tfrac{1}{5}\) whatever \(c_1, c_2, c_3\) are.
\(\tan\theta = \left|\dfrac{-\tfrac23 - \tfrac15}{1 - \tfrac{2}{15}}\right| = \left|\dfrac{-\tfrac{13}{15}}{\tfrac{13}{15}}\right| = 1\), so \(\theta = 45^\circ\) in every case.
Hence \(\theta_1 = \theta_2\) for all \(c_2, c_3\): Statement-2 is true, and Statement-1 (a special case where \(c_2, c_3\) are proportional) is also true. Statement-2 is the general reason for Statement-1, so it is a correct explanation.
Answer: (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1.
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