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Question
Show that the points A (1, 0), B (5, 3), C (2, 7) and D (−2, 4) are the vertices of a parallelogram.
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Solution
Let A (1, 0); B (5, 3); C (2, 7) and D (-2, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.
We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.
Now to find the mid-point P(x,y) of two points `A(x_1,y_1)`and `B(x_2, y_2)` we use section formula as,
`P(x,y) = ((x_1 + x_2)/2,(y_1 + y_2)/2)`
So the mid-point of the diagonal AC is,
`Q(x,y) = ((1 + 2)/2, (0 + 7)/2)`
`= (3/2, 7/2)`
Similarly mid-point of diagonal BD is,
`R(x,y) = ((5 - 2)/2, (3 + 4)/2)`
`= (3/2, 7/2)`
Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.
Hence ABCD is a parallelogram.
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