Question -
Answer -
To prove:
The points (2, -1),(0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram
Let us assume thepoints, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.
Now, let us find theslopes
Slope of AB = [(2+1) /(0-2)]
= -3/2
Slope of BC = [(3-2) /(2-0)]
= ½
Slope of CD = [(0-3) /(4-2)]
= -3/2
Slope of DA = [(-1-0)/ (2-4)]
= ½
Thus, AB is parallelto CD and BC is parallel to DA.
Hence proved, thegiven points are the vertices of a parallelogram.
Now, let us find theangle between the diagonals AC and BD.
Let m1 andm2 be the slopes of AC and BD, respectively.
m1 =[(3+1) / (2-2)]
= ∞
m2 =[(0-2) / (4-0)]
= -1/2
Thus, the diagonal ACis parallel to the y-axis.
∠ODB = tan-1 (1/2)
In triangle MND,
∠DMN = π/2 – tan-1 (1/2)
∴ The angle between thediagonals is π/2 – tan-1 (1/2).