Question

The locus of centroid of a triangle whose vertices are (acost,asint),(bsint,−bcost) and (1,0), where t is a parameter, is

A. (3x−1)²+(3y)²=a²−b²
B. (3x−1)²+(3y)²=a²+b²
C. (3x+1)²+(3y)²=a²+b²
D. (3x+1)²+(3y)²=a²−b²

Answer 1

The locus of the centroid of a triangle can be found by finding the average of the x-coordinates and the y-coordinates of the three vertices of the triangle. The correct equation for the locus of the centroid in this case is B. (3x - 1)^2 + (3y)^2 = a^2 + b^2.

Step-by-step explanation:

The locus of the centroid of a triangle can be found by finding the average of the x-coordinates and the average of the y-coordinates of the three vertices of the triangle.

1. Let's find the coordinates of the centroid:
2. The x-coordinate of the centroid, Cx, is the average of the x-coordinates of the vertices: Cx = (acost + bsint + 1) / 3.
3. The y-coordinate of the centroid, Cy, is the average of the y-coordinates of the vertices: Cy = (asint - bcost + 0) / 3.
4. The locus of the centroid is a circle with center (Cx, Cy) and radius which can be calculated using the distance formula: (3x - Cx)^2 + (3y - Cy)^2 = r^2.
5. Simplifying the equation will give us the equation of the locus of the centroid.

The correct answer is (3x - 1)^2 + (3y)^2 = a^2 + b^2.