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

In the parallelogram ABCD,

Slope of the diagonal AC (m₁) = (0-2)/ (4-0) = -1/2

Slope of the diagonal BD  (m₂) =  (3 - (-1))/ (2 - 2 ) = ∞

That means diagonal BD is parallel to the y axis.

Now, m₁ x m₂ = -1

(-1/2) x tanθ = -1

So, tan θ = 2

Hence Proved.