What is the value of \sin\theta if \theta is the acute angle between the lines whose equations are px+qy=p+q and p(x-y)+q(x+y)=2q?

  1. A. \frac{\sqrt{3}}{2}
  2. B. \frac{3}{4}
  3. C. \frac{1}{2}
  4. D. \frac{1}{\sqrt{2}} ✓

Correct Answer: D. \frac{1}{\sqrt{2}}

Explanation

The slope of the first line is m_1 = -\frac{p}{q}. The second line simplifies to (p+q)x + (q-p)y = 2q, so its slope is m_2 = \frac{p+q}{p-q}. The angle \theta satisfies \tan\theta = |\frac{m_1 - m_2}{1 + m_1m_2}|. Substituting m_1 and m_2, we find m_1 - m_2 = \frac{-(p^2+q^2)}{q(p-q)} and 1 + m_1m_2 = \frac{-(p^2+q^2)}{q(p-q)}. Thus, \tan\theta = |1| = 1. Since \theta is acute, \theta = 45^\circ, and \sin\theta = \frac{1}{\sqrt{2}}.

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