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