Question

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 1

Let A(2, −1), B(0, 2), C(2, 3) and D(4, 0) be the vertices.
Slope of AB = \[\frac{2 + 1}{0 - 2} = - \frac{3}{2}\]

Slope of BC = \[\frac{3 - 2}{2 - 0} = \frac{1}{2}\]

Slope of CD = \[\frac{0 - 3}{4 - 2} = - \frac{3}{2}\]

Slope of DA = \[\frac{- 1 - 0}{2 - 4} = \frac{1}{2}\]

Thus, AB is parallel to CD and BC is parallel to DA.
Therefore, the given points are the vertices of a parallelogram.

Now, let us find the angle between the diagonals AC and BD.
Let \[m_1 \text { and } m_2\] be the slopes of AC and BD, respectively.

\[\therefore m_1 = \frac{3 + 1}{2 - 2} = \infty \]

\[ m_2 = \frac{0 - 2}{4 - 0} = - \frac{1}{2}\]

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

\[\therefore \angle ODB = \tan^{- 1} \left( \frac{1}{2} \right)\]

In triangle MND,

\[\angle DMN = \frac{\pi}{2} - \tan^{- 1} \left( \frac{1}{2} \right)\]

Hence, the acute angle between the diagonal is \[\frac{\pi}{2} - \tan^{- 1} \left( \frac{1}{2} \right)\].