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Question
A quadrilateral has vertices (4, 1), (1, 7), (−6, 0) and (−1, −9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.
Solution
Let A (4, 1), B (1, 7), C (−6, 0) and D (−1, −9) be the vertices of the given quadrilateral.
Let P, Q, R and S be the mid-points of AB, BC, CD and DA, respectively.
So, the coordinates of P, Q, R and S are \[P \left( \frac{5}{2}, 4 \right), Q \left( \frac{- 5}{2}, \frac{7}{2} \right), R \left( \frac{- 7}{2}, \frac{- 9}{2} \right) \text { and }S \left( \frac{3}{2}, - 4 \right)\].
In order to prove that PQRS is a parallelogram, it is sufficient to show that PQ is parallel to RS andPQ is equal to RS.
Now, we have,
Slope of PQ
\[= \frac{\frac{7}{2} - 4}{\frac{- 5}{2} - \frac{5}{2}} = \frac{1}{10}\]
Slope of RS \[= \frac{- 4 + \frac{9}{2}}{\frac{3}{2} + \frac{7}{2}} = \frac{1}{10}\]
Clearly, Slope of PQ = Slope of RS
Therefore, PQ
\[\lVert\] RS \[PQ = \sqrt{\left( - \frac{5}{2} - \frac{5}{2} \right)^2 + \left( \frac{7}{2} - 4 \right)^2} = \frac{\sqrt{101}}{2}\]
\[RS = \sqrt{\left( \frac{3}{2} + \frac{7}{2} \right)^2 + \left( - 4 + \frac{9}{2} \right)^2} = \frac{\sqrt{101}}{2}\]
Therefore, PQ = RS
Thus, PQ \[\lVert\] RS and PQ = RS
Hence, the mid-points of the sides of the given quadrilateral form a parallelogram.
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