If `theta` is the angle between the diagonals of a parallelogram `ABCD` whose vertices are `A(0,2), B(2, -1), C(4,0)` and `D(2, 3).` Show that `tan theta = 2.`

Find the angle between the diagonals of parallelogram ABCD whose vertices are A (0, 2), B (2, - 1), C (4, 0) and D (2, 3).

If theta is the angle between the lines joining the points A(0, 0) and B(2, 3), and the points C(2,-2) and D(3,5), show that tan theta=(11)/(23)

The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2,3), B(6,7) and C(8,3) is (a) (0,1) (b) (0,-1) (c) (-1,0) (d) (1,0)

If theta is the angle between the curves xy = 2 and x^(2)+4y = 0 then tan theta =

If vertices of a parallelogram are respectively (0,0), (1,0) ,(2,2) and (1,2) then angle between diagonals is

If vertices of a parallelogram are respectively (0,0), (1,0) ,(2,2) and (1,2) then angle between diagonals is

The vertices of the rectangle ABCD are A(-1, 0), B(2, 0), C(a, b) and D(-1, 4) Then, the length of the diagonal AC is

The vertices of a reactangle ABCD are A (-1, 0) B (2,0), C (a,b) and D(-1, 4). Then the length of the diagonal AC is

If veca=(3,1) and vecb=(1,2) represent the sides of a parallelogram then the angle theta between the diagonals of the paralelogram is given by (A) theta=cos^-1(1/sqrt(5)) (B) theta=cos^-1(2/sqrt(5)) (C) theta=cos^-1 (1/(2sqrt(5))) (D) theta = pi/2

If veca=(3,1) and vecb=(1,2) represent the sides of a parallelogram then the angle theta between the diagonals of the paralelogram is given by (A) theta=cos^-1(1/sqrt(5)) (B) theta=cos^-1(2/sqrt(5)) (C) theta=cos^-1 (1/(2sqrt(5))) (D) theta = pi/2