Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coord

1.	Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
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 the points, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices. Now, let us find the slopes Slope of AB = [(2 + 1) / (0 - 2)] = -3/2 Slope of BC = [(3 - 2) / (2 - 0)] = 1/2 Slope of CD = [(0 - 3) / (4 - 2)] = -3/2 Slope of DA = [(-1 - 0) / (2 - 4)] = 1/2 Thus, AB is parallel to CD and BC is parallel to DA. Hence proved, the given points are the vertices of a parallelogram. Now, let us find the angle between the diagonals AC and BD. Let m₁ and m₂ be the slopes of AC and BD, respectively. m₁ = [(3 + 1) / (2 - 2)] = ∞ m₂ = [(0 - 2) / (4 - 0)] = -1/2 Thus, the diagonal AC is parallel to the y-axis. ∠ODB = tan^-1 (1/2) In triangle MND, ∠DMN = π/2 – tan^-1 (1/2) ∴ The angle between the diagonals is π/2 – tan^-1 (1/2).

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

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 the points, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.

Now, let us find the slopes

Slope of AB = [(2 + 1) / (0 - 2)]

= -3/2

Slope of BC = [(3 - 2) / (2 - 0)]

= 1/2

Slope of CD = [(0 - 3) / (4 - 2)]

= -3/2

Slope of DA = [(-1 - 0) / (2 - 4)]

= 1/2

Thus, AB is parallel to CD and BC is parallel to DA.

Hence proved, the given points are the vertices of a parallelogram.

Now, let us find the angle between the diagonals AC and BD.

Let m₁ and m₂ be the slopes of AC and BD, respectively.

m₁ = [(3 + 1) / (2 - 2)]

= ∞

m₂ = [(0 - 2) / (4 - 0)]

= -1/2

Thus, the diagonal AC is parallel to the y-axis.

∠ODB = tan^-1 (1/2)

In triangle MND,

∠DMN = π/2 – tan^-1 (1/2)

∴ The angle between the diagonals is π/2 – tan^-1 (1/2).