Straight Lines
Linear Inequalities and Regions
Grade 11
Question:
<p>The complete set of real values of 'a' such that the point \(P(a, \sin a)\) lies inside the triangle formed by the lines \(x - 2y + 2 = 0\), \(x + y = 0\) and \(x - y - p = 0\), is:</p>
<p>(a) \(\left(0, \frac{\pi}{6}\right) \cup \left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)</p>
<p>(b) \(\left(\frac{\pi}{2}, \pi\right) \cup \left(\frac{2\pi}{2}, 2\pi\right)\)</p>
<p>(c) \((0, \pi)\)</p>
<p>(d) \(\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Check that a point lies on the correct side of all three lines bounding a triangular region
<p><strong>Solution:</strong> For point \(P(a, \sin a)\) to lie inside the triangle, it must satisfy the inequalities for all three sides of the triangle formed by the given lines.</p><p>Line 1: \(x - 2y + 2 = 0\)</p><p>Line 2: \(x + y = 0\)</p><p>Line 3: \(x - y - p = 0\)</p><p>Substituting \(P(a, \sin a)\) into the inequality conditions and solving for the range of \(a\) where \(\sin a\) is constrained appropriately gives the answer.</p><p>∴ Answer is (a).</p>
Correct Answer: a