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 $m_1\mathrm{and}{m}_2$ be the slopes of AC and BD, respectively.

$$\begin{gathered} \therefore m_1=\frac{3+1}{2-2}=\infty \\ {m}_2=\frac{0-2}{4-0}=-\frac{1}{2} \end{gathered}$$

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

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

In triangle MND,

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

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