If `A(cos theta, sin theta), B (sin theta, cos theta), C (1, 2)` are the vertices of `DeltaABC`. Find the locus of its centroid if `theta` varies.

Evaluate |[cos theta, -sin theta], [sin theta, cos theta]|

((sin theta+cos theta)^(2)-1)/(sin theta*cos theta)

A = [(cos theta,-sin theta),(-sin theta,-cos theta)] then find A^(-1) .

If theta is parameter, A=(a cos theta,a sin theta) and B=(b sin theta,-b cos theta)C=(1,0) then the locus of the centroid of Delta ABC

A -= (cos theta , sin theta), B -= (sin theta, -cos theta) are two points. The locus of the centroid of DeltaOAB , where 'O' is the origin is

A=(cos theta, sin theta) and B=(sin theta,-cos theta) are two points. The locus of the centroid of triangleOAB where O is the origin is

Let A(5,12),B(-13cos theta,13sin theta) and -C(13sin theta,13cos theta) are angular points of ABC where theta in R. The locus of orthocentre of DeltaABC is