Question

Prove that the points (−4, −1), (−2, −4), (4, 0) and (2, 3) are the vertices of a rectangle.

Answer 1

Let A (−4, −1), B (−2, −4), C (4, 0) and D (2, 3) be the given points.
Let us find the lengths of AB, BC, CD and DA

\[AB = \sqrt{\left( - 2 + 4 \right)^2 + \left( - 4 + 1 \right)^2} = \sqrt{13}\]

\[BC = \sqrt{\left( 4 + 2 \right)^2 + \left( 0 + 4 \right)^2} = 2\sqrt{13}\]

\[CD = \sqrt{\left( 2 - 4 \right)^2 + \left( 3 - 0 \right)^2} = \sqrt{13}\]

\[DA = \sqrt{\left( 2 + 4 \right)^2 + \left( 3 + 1 \right)^2} = 2\sqrt{13}\]

\[\therefore\] AB = CD and BC = DA

Now, we have,

\[m_1 = \text { Slope of AB } = \frac{- 4 + 1}{- 2 + 4} = - \frac{3}{2}\]

\[ m_2 = \text { Slope of BC } = \frac{0 + 4}{4 + 2} = \frac{4}{6} = \frac{2}{3}\]

\[ m_3 = \text { Slope of CD  }= \frac{3 - 0}{2 - 4} = - \frac{3}{2}\]

Here,

\[m_1 m_2 = \left( - \frac{3}{2} \right)\left( \frac{2}{3} \right) = - 1 \text { and } m_1 = m_3\]

Therefore, we have,
AB = CD
BC = DA

\[AB \perp BC\]

And, AB is parallel to DC.
Hence, the given points are the vertices of a rectangle.