Question

If \(A(\cos \alpha, \sin \alpha)\), \(B(\sin \alpha, -\cos \alpha)\), and \(C(1, 2)\) are the vertices of \(\triangle ABC\), then find the locus of its centroid.

• A. \(3(x^2 + y^2) - 2x - 4y + 1 = 0\)
• B. \(x^2 + y^2 - 2x - 4y + 1 = 0\)
• C. \(x^2 + y^2 - 2x - 4y + 3 = 0\)
• D. \(2(x^2 + y^2) - 2x - 4y + 5 = 0\)

Hint:

Use centroid formula and trigonometric identities to find locus by eliminating parameter \(\alpha\).

Answer 1

The Correct Option is A

Centroid \(G\) of \(\triangle ABC\) is \[ G = \left(\frac{\cos \alpha + \sin \alpha + 1}{3}, \frac{\sin \alpha - \cos \alpha + 2}{3}\right). \] Let \[ x = \frac{\cos \alpha + \sin \alpha + 1}{3},
y = \frac{\sin \alpha - \cos \alpha + 2}{3}. \] Multiply both sides by 3: \[ 3x - 1 = \cos \alpha + \sin \alpha,
3y - 2 = \sin \alpha - \cos \alpha. \] Add and subtract: \[ (3x - 1) + (3y - 2) = 2 \sin \alpha, \] \[ (3x - 1) - (3y - 2) = 2 \cos \alpha. \] Then, \[ \sin \alpha = \frac{3x + 3y - 3}{2},
\cos \alpha = \frac{3x - 3y + 1}{2}. \] Use the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\): \[ \left(\frac{3x + 3y - 3}{2}\right)^2 + \left(\frac{3x - 3y + 1}{2}\right)^2 = 1. \] Multiply both sides by 4: \[ (3x + 3y - 3)^2 + (3x - 3y + 1)^2 = 4. \] Expand and simplify to get: \[ 3(x^2 + y^2) - 2x - 4y + 1 = 0. \]