In the `triangle OAB`, M is the midpoint of AB, C is a point on OM, such that `2 OC = CM`. X is a point on the side OB such that OX = 2XB. The line XC is produced to meet OA in Y. Then `(OY)/(YA)=`

In the DeltaOAB,M is the mid-point of AB,C is a point on OM, such that 2OC=CM. X is a point on the side OB such that OX=2XB. The line XC is produced to meet OA in Y. then, (OY)/(YA) is equal to

In DeltaOAB , L is the midpoint of OA and M is a point on OB such that (OM)/(MB)=2 . P is the mid point of LM and the line AP is produced to meet OB at Q. If bar(OA)=bar(a), bar(OB)=bar(b) then find vectors bar(OP) and bar(AP) interms of bar(a) and bar(b) .

In a triangle OAB, E is the midpoint of OB and D is a point on AB such that AD:DB=2:1 . IF OD and AE intersects at P, then ratio of (OP)/(PD) is equal to

E is the mid point of side OB of triangle OAB . D is a point on AB such that AD : DB = 2 : 1 . If OD and AE intersect at P , then -

C is the midpoint of the line segment bar(AB) and O is any point outside AB, show that vec(OA)+vec(OB)=2vec(OC) .

P, Q, R are the midpoints of the sides AB, BC and CA of the triangle ABC and O is a point within the triangle, then OA + OB + OC =

In a triangle OAC, if B is the mid point of side AC and vec(OA)=veca,vec(OB)=vecb , then what is vec(OC) .

If O is the origin and A and B are points on the line 3x - 4 y + 25 = 0 such that OA = OB = 13 , Then the area of Delta OAB (In sq units) is

Let X be the midpoint of the side AB of triangle ABC. And Y be the mid point of CX. If Z is the point where BY meets the side AC, then AZ:ZC is equal to