Let a vector $\alpha \hat i + \beta \hat j $ be obtained by rotating the vector $\sqrt 3 \hat i + \hat j $ by an angle $45^\circ $ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices $(\alpha , \beta )$, $(0, \beta )$ and (0, 0) is equal to:
(A) $2\sqrt 2 $
(B) $\frac {1}{2}$
(C) 1
(D) $\frac {1}{\sqrt 2 }$ Continue reading Let a vector $\alpha \hat i + \beta \hat j $ be obtained by ….
and 
are the sides of a triangle ABC, then the length of the median through A is:
(2) 
(3) 
(4) 