Straight Lines
Locus of Points
Grade 11
Question:
<p>If <p>A(\cos a, \sin a)</p>, <p>B(\sin a, -\cos a)</p>, <p>C(1, 2)</p> are the vertices of a <p>\triangle ABC</p>, then the locus of centroid of triangle is</p>
<p>(a) <p>x^2 + y^2 - 2x - 4y + 1 = 0</p></p>
<p>(b) <p>3(x^2 + y^2) - 2x - 4y + 1 = 0</p></p>
<p>(c) <p>x^2 + y^2 - 2x - 4y + 3 = 0</p></p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Use centroid formula and parametric constraint to eliminate the parameter and obtain the locus equation.
<p><strong>Solution:</strong> The centroid of triangle ABC is at <p>G = \left(\frac{\cos a + \sin a + 1}{3}, \frac{\sin a - \cos a + 2}{3}\right)</p>. Let <p>x = \frac{\cos a + \sin a + 1}{3}</p> and <p>y = \frac{\sin a - \cos a + 2}{3}</p>.</p><p>Then <p>3x - 1 = \cos a + \sin a</p> and <p>3y - 2 = \sin a - \cos a</p>.</p><p>Squaring and adding: <p>(3x-1)^2 + (3y-2)^2 = (\cos a + \sin a)^2 + (\sin a - \cos a)^2 = 2</p></p><p>Therefore: <p>3(x^2 + y^2) - 2x - 4y + 1 = 0</p></p>
Correct Answer: B